MA001 Study Guide

1a. solve linear, rational, quadratic, radical, and absolute value equations in one variable

• Compare the general forms of linear equations, quadratic equations, and absolute value equations in one variable.
• Identify different methods for solving quadratic equations.
• Describe the difference between rational equations and radical equations in one variable.
• Identify strategies for solving rational and radical equations. 

Equations in a single variable can appear in a variety of forms, with different strategies used depending on the nature of the equation. Perhaps the simplest equation is a linear equation, which is an equation in which the variable is only to the first power. The typical form of a linear equation is $a x+b=0$, in which $a$ and $b$ are real numbers and $a \neq 0$. The solution to such an equation is $x=-\frac{b}{a}$.

In contrast, in a quadratic equation the variable is raised to the second power. The general form of a quadratic equation is $a x^{2}+b x+c=0$, where $a, b$, and $c$ are real numbers and $a \neq 0$. Quadratic equations can always be solved using the quadratic formula, $x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$.

Depending on the values of $a, b$, and $c$, it may be efficient to solve the quadratic equation by other methods. For instance, if the equation can be factored into $(x+h)(x+k)=0$, the solutions are found by solving $x+h=0$ and $x+k=0$. Completing the square involves rewriting the equation into the form $(x+h)^{2}=k$, so that the solution becomes $x=-h \pm \sqrt{k}$.

An absolute value equation is of the form $|a x+b|=c$. The solution to such an equation is $a x+b=c$ or $a x+b=-c$.

Rational equations contain at least one rational expression, which is a ratio of two polynomials. In contrast, radical equations contain a variable in the radicand. To solve a rational equation, multiply both sides of the equation by the least common denominator or rewrite the equation to be of the form $\frac{a}{b}=\frac{c}{d}$ and then solve $a d=b c$. To solve a radical equation, isolate the radical and then raise both sides of the equation to the index power of the radical; if the original equation contains more than one radical, this step might need to be repeated. For both rational and radical equations, check the potential solutions in the original equation to ensure they satisfy the original equation.

Review the material in Solving Linear Equations in One Variable, Solve Quadratic Equations by Factoring, and Equations That are Quadratic in Form.

1b. classify solutions to linear, rational, quadratic, radical, and absolute value equations

• When will a linear equation have no solutions? When will it have an infinite number of solutions?
  
• What is the discriminant and how can it be used to determine the nature of the roots of a quadratic equation?
  
• How is it possible to determine if an absolute value equation will have one, two, or no solutions? 
  
• What are extraneous solutions and when might they arise?

When solving equations, situations arise in which specific solutions exist, no solutions exist, or an infinite number of solutions exist. For instance, when solving a linear equation, if the result simplifies to $x=k$, then a specific solution exists. If the result simplifies to a true statement, such as $5=5$ or $x=x$, then there are an infinite number of solutions because every value of $x$ works in the equation. If the result simplifies to a false statement, such as $10=20$, then there are no solutions to the equation.

The discriminant for a quadratic equation is the value $b^{2}-4 a c$, which is the radicand in the quadratic formula. Its value can be used to determine the nature of the roots of the equation: two real roots if the discriminant is greater than 0 and those two roots are rational if the discriminant is a perfect square; one rational root if the discriminant is 0; and no real roots if the discriminant is negative.

The absolute value of a number is always non-negative, with $|x|=x$ when $x \geq 0$ and $|x|=-x$ when $x < 0$. So, the absolute value equation $|a x+b|=c$ has one solution if $c=0$, two solutions if $c > 0$, or no solutions if $c < 0$.

In both rational and radical equations, it is possible to obtain an extraneous solution, that is a solution to the resulting equation when the rational expressions or radicals are eliminated but that does not satisfy the original equation. This often occurs because the extraneous solution causes a denominator in the rational equation to equal zero or the radicand in a radical equation to be negative. So, when solving rational and radical equations, it is essential to check all potential solutions in the original equation.

1b. Review the material in Solving Linear Equations in One Variable, Solve Quadratic Equations by Factoring, and Equations That are Quadratic in Form.

1c. construct linear equations from words

• Identify phrases that are often used to indicate the four operations when translating a verbal sentence to a mathematical equation.

Many real-world situations are represented and modeled by equations. In a linear equation of the form $a x+b=c$, with $a, b$, and $c$ real numbers and $a \neq 0$, the coefficient a typically represents the change in each unit of the variable and $b$ represents a fixed amount in the situation. When translating from a verbal sentence to a mathematical sentence, it is important to identify the variable and its meaning to assist with interpreting the solution to the equation. For instance, consider the following context: An individual opens a savings account with an initial balance of $100 and plans to deposit $75 into the account each month. In how many months will the balance be $1600? The variable of interest is the number of months, which can be represented by the letter $m$. The fixed amount is $100 and represents $b$; the change for each unit of the variable (months) is $75 and represents $a$. So a reasonable equation to model this situation is $75 m+100=1600$.

Although many different words can be used to suggest the appropriate operation, there are some words that regularly appear:

Although many different words can be used to suggest the appropriate operation, there are some words that regularly appear: 

• Addition: sum, exceeds, is more than, increased by
• Subtraction: difference, decreased by, less, less than
• Multiplication: product, multiplied by, twice, three times
• Division: quotient, divided by

The verb is typically translates as the equal sign. 

It is important to watch for situations that indicate the need to group two or more terms in parentheses. Consider the wording and translation of the following two phrases:

• the sum of three times a number and four: $3 n+4$
• three times the sum of a number and four: $3(n+4)$

Review the material in Applications of Linear Equations.

1d. represent linear and absolute value inequalities using standard notation and graphs

• Compare representing an interval on a number line with an inequality, using set-builder notation, or using interval notation.
• Distinguish between intervals that include the endpoints and those that do not.
• What is a compound inequality?

Although a linear equation typically has a single solution, a linear inequality represents an interval on the number line. Consider the two inequalities below:

The first interval can be represented as $x \geq 2$. Notice that the endpoint of the ray is a closed circle to indicate that 2 is part of the solution. In set-builder notation, this inequality would be represented as $\{x: x \geq 2\}$. In interval notation, the inequality would be represented as $[2, \infty)$, with the bracket on the left used to indicate that 2 is part of the solution and the infinity sign, $\infty$, used to indicate that the solution contains all real numbers greater than or equal to 2.

In contrast, the second inequality can be represented as $x < -3$; notice that the open circle at the endpoint indicates that $-3$ is NOT part of the solution. The interval can also be represented as $\{x: x < -3\}$ or as $(-\infty,-3)$, using set-builder or interval notation, respectively. In interval notation, a parenthesis is used rather than a bracket to indicate the endpoint is NOT part of the solution.

A compound inequality contains more than one inequality within the statement, such as $0 < x \leq 10$ as well as $x < 0$ or $x > 10$. In an inequality such as $0 < x \leq 10$, one should always be able to hide the middle portion of the inequality and still have a true statement, namely $0 \leq 10$. This can ensure that one avoids writing statements that make no sense, such as $0 > x > 10$; if the middle inequality is hidden, the resulting inequality $0 > 10$ is meaningless.

Absolute value inequalities lead to compound inequalities. For instance, $|X| < B$ is equivalent to $-B < X < B$. In contrast, $|X| > B$ is equivalent to $X < -B$ or $X > B$.

Review the material in Absolute Value Inequalities.

1e. solve linear and absolute value inequalities

• Identify differences between solving linear inequalities and solving linear equations.

The techniques used in solving linear equations are also used in solving linear inequalities, with one major difference. When solving inequalities, the inequality sign reverses when multiplying or dividing by a negative. For instance, to solve $x+4 < 12$, one subtracts 4 from both sides of the inequality to obtain $x < 8$. In contrast, to solve $-4 x < 12$, one divides both sides of the inequality by $-4$ and then must reverse the inequality sign to obtain $x > -3$.

To solve an absolute value inequality, first remove the absolute value sign using the fact that $|X| < B$ is equivalent to $-B < X < B$ or that $|X| > B$ is equivalent to $X < -B$ or $X > B$. Then solve the compound inequalities using the same techniques as used when solving other inequalities.

Review the material in Absolute Value Inequalities.

1f. perform algebraic operations on complex numbers

• Define the imaginary number $i$ and explain how it is used to simplify square roots of negative numbers.
• Define a complex number and distinguish between the real part and the imaginary part.
• Explain how to perform the operations of addition, subtraction, and multiplication with complex numbers and represent the result in standard form.
• Define a complex conjugate of a complex number and explain how complex conjugates are used to express divisions of complex numbers in standard form.
• Explain how to simplify powers of $i$.

The imaginary number $\boldsymbol{i}$ is used to represent $\sqrt{-1}$ so that $i^{2}=-1$. When simplifying a square root with a negative radicand, the imaginary number is used to represent the negative, so that $\sqrt{-64}=8 i$ and $\sqrt{-12}=2 i \sqrt{3}$. A complex number in standard form is a number of the form $\mathbf{a}+\boldsymbol{b i}$, where $\boldsymbol{a}$ represents the real part and $\boldsymbol{b}$ represents the imaginary part.

Operations with complex numbers can be simplified using techniques similar to those used with other numbers. To add or subtract complex numbers, simply combine the real parts and then combine the imaginary parts. For instance, $(3+5 i)+(7-2 i)=(3+7)+(5-2) i=10+3 i$. To multiply complex numbers, apply the distributive property and then rewrite $i^{2}$ as $-1$. So, $(3+5 i)(7-2 i)=21-6 i+35 i-10 i^{2}=21+29 i-10(-1)=$ $31+29 i.$

Dividing two complex numbers and representing the result in standard form generally requires using the complex conjugate of $a+b i$, which is $a-b i$. That is, to find the complex conjugate of a complex number, find the additive inverse of the imaginary part. Then to find the quotient of two complex numbers, multiply numerator and denominator by the complex conjugate of the denominator and simplify as illustrated below.

$\frac{3+4 i}{2-7 i}=\frac{3+4 i}{2-7 i} \cdot \frac{2+7 i}{2+7 i}=\frac{6+21 i+8 i+28 i^{2}}{4+14 i-14 i-49 i^{2}}=\frac{-22+29 i}{4+49}=-\frac{22}{53}+\frac{29}{53} i$

The imaginary number $i$ has an interesting property related to its powers: $i^{1}=i ; i^{2}=-1 ; i^{3}=-i ; i^{4}=1$. The powers then repeat. So, to simplify a power of $i$, such as $i^{50}$, divide the exponent by $4$ and look at the remainder. So, $50 \div 4$ has a remainder of 2, meaning $i^{50}$ is equivalent to $i^{2}$ which is $-1$.

Review the material in Complex Numbers.

Unit 1 Vocabulary

This vocabulary list includes terms you will need to know to successfully complete the final exam.

• absolute value equation
• complex conjugate
• complex number
• compound inequality
• discriminant
• extraneous solution
• imaginary number, $i$
• imaginary part of a complex number
• linear equation
• quadratic equation
• quadratic formula
• radical equation
• rational equation
• real part of a complex number

2a. summarize the properties of a function, including input and output values, domain, and range using words, function notation, and tables

• Distinguish between a relation and a function. 
• Identify the difference between input and output values. Relate input and output values to independent and dependent variables and domain and range.
• Be able to interpret function notation.
• Determine when a function is one-to-one.

Any set of ordered pairs represents a relation. The first element in the ordered pair represents the input value or independent variable; the second element in the ordered pair represents the output value or dependent variable. The set of all input or independent values describes the domain of the relation; the set of all output or dependent values represents the range of the relation. 

A relation is a function if each input value is associated with a single output value. Compare the following two sets of ordered pairs.

• Set A: $\{(1, \, 2),(1, \, 4),(2, \, 4),(2, \, 8),(3, \, 6),(3, \, 12)\}$
• Set B: $\{(1, \, 2),(2, \, 4),(3, \, 6)\}$

Both Sets $A$ and $B$ are relations. In Set $A$, each input value is associated with two different output values, so Set $A$ is not a function. Its domain is $\{1,2,3\}$ and its range is $\{2, \, 4, \, 6, \, 8,\, 12\}$. In Set $B$, each input value is associated with just one output value so Set B is a function. Its domain is $\{1,2,3\}$ and its range is $\{2, \, 4, \,6\}$.

Functions are typically represented using equations or function notation. In Set B above, each output value is twice the input value. So, if $x$ represents the input value and $y$ represents the output value, the function could be described as $y=2 x$. If $f$ names the function, then the function could also be described as $f(x)=2 x$, read " $f$ of $x$ is twice $x$ ". Notice that the function is described by indicating the input value for the function (in this case $x$) and then the rule applied to that input value (in this case $2$ times $x$).

A function is a one-to-one function if each output value is associated with a single input value. For instance, $\{(1, \, 2),(2, \, 2),(3, \, 2)\}$ is a function because each input value is associated with just one output value; however, it is not one-to-one because the output value 2 is associated with more than one input value. In contrast, $\{(1, \, 2)$, $(2, \, 4),(3, \, 6)\}$ is a function because each input value is associated with just one output value; it is a one-to-one function because each output value is associated with just one input value.

Review the material in Defining and Writing Functions.

2b. evaluate a function given an equation, table, or words

• Be able to evaluate a function represented in function notation.
• Compare evaluating a function using function notation to evaluating a function represented by a table of values.

To evaluate a function, identify the input or independent variable and then apply the function rule to that variable. For instance, given the function $f$ defined as $f(x)=3 x^{2}+5 x^{3}$. The input variable is the variable inside the parentheses, namely $x$, and the rule for the function is $3 x^{2}+5 x^{3}$. So, to evaluate $f(4)$, substitute 4 for $x$ and simplify: $3 \cdot 4^{2}+5 \cdot 4^{3}=3 \cdot 16+5 \cdot 64=368$.

If a function is represented by a table, first identify the input values and the output values in the table. If the input value is actually listed in the table, then match the input value with its associated output value. If the input value is not actually listed in the table, then attempt to find a rule to describe how the input values are associated with the output values and use this rule to evaluate the function for a non-listed input value.

Review the material in Defining and Writing Functions.

2c. identify whether the graph of a relation represents a function

• Explain how to use the vertical line test to determine if the graph of a relation represents a function.
• Explain how to use the horizontal line test to determine if a function is one-to-one. 

In looking at a graph of a relation to determine whether it represents a function, it is important to consider whether each first element of an ordered pair is associated with a single second element of the ordered pair. If so, the relation is a function; if not, the relation is not a function. One quick way to check whether a graph represents a function is to use the vertical line test. If a vertical line intersects a graph in more than one point, then the graph does NOT represent a function because the input value associated with the vertical line is paired with more than one output value. 

Once a graph is determined to represent a function, the horizontal line test can be applied to determine if the function is one-to-one. If the horizontal line intersects the graph in more than one point, then the graph does not represent a one-to-one function because the output value associated with the horizontal line is paired with more than one input value. 

For instance, consider the graph of $y=f(x)$ shown on the coordinate grid below. The vertical line in red intersects the graph in only one point, even if it were to move across the coordinate plane from left to right; so the graph represents a function. However, the horizontal line in green intersects the graph in more than one point, so the function is not one-to-one.

Review the material in Defining and Writing Functions.

2d. use set builder, inequality, and interval notation to express the domain and range of a function defined by an equation, table, graph, or set

• Identify the domain and range of a function expressed in a table or set of ordered pairs.
• Identify the domain and range of a function given an equation. How are restrictions on the domain or range determined?
• Identify the domain and range of a function represented by a graph. 

If a function is expressed in a table or as a set of ordered pairs, then the domain will be the set of input values represented in the table or the set of first elements of the ordered pairs in the set. The range will be the set of output values represented in the table or the set of second elements of the ordered pairs in the set.

When identifying the domain and range of a function from an equation, it is important to consider whether there are any restrictions on the possible input values and the resulting output values. Such restrictions occur if particular input values would make the equation undefined or if particular output values can never be obtained. 

For example, if $f(x)=4 x+5$, the graph is a line; $x$ can represent any real number and the resulting $f(x)$ value can also be any real number. In contrast, consider $g(x)=x^{2}-3$. In this case, although the input $x$ can be any real number, the output value can never be less than $-3$. So the domain would be the set of all real numbers but the range would be $\{y: y \geq-3\}$ or $[-3, \infty)$ when represented in set-builder or interval notation, respectively. If $h(x)=\sqrt{x+4}$, then the domain will be $x \geq-4$ because $x+4 \geq 0$ in order for the square root to be feasible; the domain is $y \geq 0$ because the value of the square root is always non-negative.

Similar processes are used to identify the domain and range of a graph identified on a coordinate plane. Here, one needs to consider whether endpoints on the graph suggest that the function is defined only over particular values or whether both the independent and dependent variables can extend to $\pm \infty$. Compare the two graphs below. In $y=g(x)$, the domain is $\{x:-4$ and the range is $\{y: 0 \leq y \leq 4\}$. The open circle at $(-4, \, 2)$ indicates that $x=-4$ is not part of the domain; although $y=2$ would not be part of the range based on this point, $y=2$ is part of the domain from the point $(0, \, 2)$. However, in $y=h(x)$, the graph continues in both directions to infinity; so, the domain and range are both $(-\infty, \, \infty)$.

Review the material in Finding Domain and Range from Graphs.

2e. construct the graph of a piecewise-defined function

• What is a piecewise-defined function? 
• How do the domain and range help to graph a piecewise-defined function? 

A piecewise-defined function is a function in which the output is defined differently for various values of the domain. The overall domain of the function is the union of the domain of each piece. Similarly, the overall range of the function is the union of the range of each piece. Compare the graphs of the two functions below.

For function $p$ the domain is the set of all real numbers; the range is $[-1, \infty)$. For function $k$, the domain is $(-\infty, \, 0] \cup[3, \, \infty)$; the range is $\{-2, \, 1\}$. Notice that for function $k$, there are some values of $x$ that are not part of the domain and the range consists of just two values.

Review the material in Finding Domain and Range from Graphs.

2f. calculate the average rate of change of a function given a table, graph, or equation

• Explain how to determine the average rate of change of a function over a specified interval.
• Compare the average rate of change of a linear function with the average rate of change of a nonlinear function. 

The average rate of change of a function describes how the output values of a function change over a particular interval of input values. In general, if $y=f(x)$, then

average rate of change of $f=\frac{f\left(x_{2}\right)-f\left(x_{1}\right)}{x_{2}-x_{2}}$

As an example, consider a function $g$ defined as $g(x)=x^{2}-7 x$. To find the average rate of change of $g$ over the interval $[-4,3]$, first find $g(-4)=44$ and $g(3)=-12$. Then the average rate of change for $g$ in this interval is $\frac{-12-44}{3-(-4)}=\frac{-56}{7}=-8$. If the function is represented by a graph or via a table, the average rate of change is found in the same way, by first identifying the output values from the graph or table and then simplifying the fraction related to the average rate of change.

As an example, consider a function $g$ defined as $g(x)=x^{2}-7 x$. To find the average rate of change of $g$ over the interval $[-4,3]$, first find $g(-4)=44$ and $g(3)=-12$. Then the average rate of change for $g$ in this interval is $\frac{-12-44}{3-(-4)}=\frac{-56}{7}=-8$. If the function is represented by a graph or via a table, the average rate of change is found in the same way, by first identifying the output values from the graph or table and then simplifying the fraction related to the average rate of change.

To review, see Calculate the Rate of Change of a Function.

2g. identify the characteristics of a function given its graph, including behavior over an interval and local and absolute extrema

• Identify intervals over which a function is increasing, decreasing, or constant.
• Determine local minimum or local maximum values of a function over a specific interval.
• Compare local minimum/maximum values with absolute minimum/maximum values.

It is often important to determine how a function behaves over a particular interval. If the output values increase as the input values increase, then the function is an increasing function; that is, function $f$ is increasing if $f\left(x_{2}\right) > f\left(x_{1}\right)$ when $x_{2} > x_{1}$. Likewise, if the output values decrease as the input values increase, then the function is a decreasing function; that is, function $f$ is decreasing if $f\left(x_{2}\right)$ when $x_{2} > x_{1}$. If the output values do not change as the input values increase, then the function is a constant function; that is, function $f$ is constant if $f\left(x_{2}\right)=f\left(x_{1}\right)$ when $x_{2} > x_{1}$.

Local minimum or local maximum values occur where the function has a low value or high value, respectively, within a specific interval. If that low value or high value is the lowest value or highest value over the entire domain of the function, then it is an absolute minimum or absolute maximum value of the function, respectively. 

Consider the function shown on the coordinate graph below. The function is decreasing for $(-\infty,-3)$ and $(-1,1)$ and increasing for $(-3,-1)$ and $(1, \infty)$. The function has a local maximum value of 10 at $x=-1$. The function has an absolute minimum value of $-6$ at both $x=-3$ and $x=1$.

Review the material in Calculate the Rate of Change of a Function.

Unit 2 Vocabulary

 This vocabulary list includes terms you will need to know to successfully complete the final exam.

• absolute maximum value of a function
• absolute minimum value of a function
• average rate of change of a function
• constant function
• decreasing function
• dependent variable
• domain
• function
• function notation
• horizontal line test
• increasing function
• independent variable
• input value
• local maximum value of a function
• local minimum value of a function
• one-to-one function
• output value
• piecewise-defined function
• range
• relation
• vertical line test

3a. perform algebraic operations on functions to construct and deconstruct composite functions

• Evaluate the sum, difference, product, and quotient of two functions. 
• Evaluate a composite function. Compare evaluating a composite function with simplifying the sum, difference, product, or quotient of two functions.

Finding the sum, difference, product, or quotient of two functions is equivalent to simplifying those computations with the functions, provided the input value is in the domain of both functions. That is, given two functions $f$ and $g$, the functions $f+g, f-g, f \cdot g$, and $\frac{f}{g}$ for $g \neq 0$ are evaluated by finding the output of each function for the indicated input and applying the required operation.

• $(f+g)(x)=f(x)+g(x)$
• $(f-g)(x)=f(x)-g(x)$
• $(f \cdot g)(x)=f(x) \cdot g(x)$
• $\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}$ for $g(x) \neq 0$

In contrast to the sum, difference, product, or quotient of two functions, a composite function is a function in which the output of one function becomes the input of the other function. The composite of functions $f$ and $g$, denoted as $f \circ g$ and read " $f$ of $g$ ", is defined as $(f \circ g)=f(g(x))$. That is, the output $g(x)$ becomes the input for function $f$.

Consider functions $f$ and $g$ defined as $f(x)=3 x+4$ and $g(x)=x^{2}$.

• $f(3)=13$
• $g(3)=9$
• $(f+g)(3)=f(3)+g(3)=13+9=22$
• $(f-g)(3)=f(3)-g(3)=13-9=4$
• $(f \cdot g)(3)=f(3) \cdot g(3)=13 \cdot 9=117$
• $\left(\frac{f}{g}\right)(3)=\frac{f(3)}{g(3)}=\frac{13}{9}$
• $(f \circ g)(3)=f(g(3))=f(9)=31$
• $(g \circ f)(3)=g(f(3))=g(13)=169$

Review the material in Creating and Evaluating Composite Functions.

3b. describe the properties of a composite function

• Determine whether the composition of two functions is a commutative operation.
• Explain how to find the domain of a composite function.
• How is function composition used to determine if two functions are inverses of each other?

In general, function composition is not commutative. For instance, given functions $f$ and $g$ defined as $f(x)=3 x+4$ and $g(x)=x^{2}$. Then $(f \circ g)(3)=f(g(3))=f(9)=31$ and $(g \circ f)(3)=g(f(3))=g(13)=169$. So, $(f \circ g)(x) \neq(g \circ f)(x)$.

Finding the domain of a composite function is a two-step process. To find the domain of $(f \circ g)(x)=f(g(x))$, it is important to know the domain of $f$ and $g$ and to determine those values in the range of $g$ that are in the domain of $f$. That is, the output of $g$ becomes the input or domain of $f$.

Suppose functions $h$ and $k$ are defined as $h(x)=\sqrt{x^{2}-4}$ and $k(x)=x^{2}$. For function $h$, the domain is $(-\infty,-2] \cup[2, \infty)$ and the range is $[0, \infty)$. The domain of function $k$ is all real numbers and the range is all nonnegative real numbers, that is, the domain is $(-\infty, \infty)$ and the range is $[0, \infty)$. Then, to find the domain of $h \circ k$, consider those values in the domain of function $k$ whose resulting output values are in the domain of function $h$; that is, of all real numbers in the domain of function $k$, only select those values whose output is in the domain of function $h$. Hence, the domain of $h \circ k$ is $(-\infty,-\sqrt{2}] \cup[\sqrt{2}, \infty)$. In contrast, the domain for $k \circ h$ will be those values in the domain of function $h$ whose output values are in the domain of function $k$. The only real numbers from the domain of function $k$ that are feasible in function $h$ are those in the domain of function $h$. So the domain of $k \circ h$ is $(-\infty,-2] \cup[2, \infty)$.

For two functions $p$ and $q$, if $p(q(x))=q(p(x))=x$ for all values of $x$ in the input of both $p$ and $q$, then the two functions $p$ and $q$ are inverses of each other. That is, inverse functions undo each other.

Review the material in Creating and Evaluating Composite Functions and Inverse Functions.

3c. identify basic graphical transformations of functions, including shifts, compressions, stretches, and reflections

• Describe the result of a vertical shift on the graph of a function. How is the equation $y=f(x)$ modified to indicate a vertical shift?
• Describe the result of a horizontal shift on the graph of a function. How is the equation $y=f(x)$ modified to indicate a horizontal shift?
• Describe the result of a stretch or compression on the graph of a function. How is the equation $y=f(x)$ modified to indicate a stretch or compression?
• Compare the result of a horizontal reflection with the result of a vertical reflection. How is the equation $y=f(x)$ modified to indicate either a vertical or horizontal reflection?

Functions can be transformed in a number of ways, including shifts left or right, shifts up or down, stretches or compressions, or reflections across either the x- or y-axis. Each type of transformation can be represented by an appropriate change in the definition of the function. 

One common transformation is a translation, that is, either a vertical shift (up or down) or a horizontal shift (left or right). A vertical shift is denoted by adding a constant to the function, modifying $y=f(x)$ to $y=f(x)+k$ with a positive value of $k$ indicating a shift up and a negative value of $k$ indicating a shift down. A horizontal shift is denoted by adjusting the input value of the function, modifying $y=f(x)$ to $y=f(x-h)$ with a positive value of $h$ indicating a shift right and a negative value of $h$ indicating a shift left. Thus, the graph of $y=$ $(x+5)^{2}+6$ represents a transformation of the graph of $y=x^{2}$ when shifted to the left 5 units and up 6 units. Notice, that to determine the horizontal shift, it is important to think of the equation as $y=[x-(-5)]^{2}+6$.

A second common transformation is a stretch or compression of a graph. A vertical stretch or compression of the graph of $y=f(x)$ is denoted by $y=a f(x)$, with the transformation being a stretch if $a > 1$ and a compression if $0 < a < 1$. Likewise, a horizontal stretch or compression is denoted by $y=f(b x)$ with the transformation being a stretch by a factor of $\frac{1}{b}$ if $0 < b < 1$ and a compression by a factor of $\frac{1}{b}$ if $b > 1$. Thus, the graph of $y=3 x^{2}$ represents a vertical stretch of the graph of $y=x^{2}$ by a factor of 3 ; the graph of $y=(5 x)^{2}$ represents a horizontal compression of the graph of $y=x^{2}$ by a factor of $\frac{1}{5}$.

A third common transformation is a reflection either vertically across the $x$-axis or horizontally across the $y$ axis. A vertical reflection across the $x$-axis is denoted by $y=-f(x)$; a horizontal reflection across the $y$-axis is denoted by $y=f(-x)$. Thus, the graph of $y=-x^{2}$ is a reflection of the graph of $y=x^{2}$ across the $x$-axis; the graph of $y=(-x)^{3}$ is a reflection of the graph of $y=x^{3}$ across the $y$-axis.

When multiple transformations are incorporated into the graph or equation for the function, it is important to interpret them in a particular order just as computations are completed in a particular order. For vertical shifts and stretches/compressions in the form $y=a f(x)+k$, complete the vertical stretch and then complete the vertical shift. For horizontal shifts and stretches/compressions in the form $y=a f(b(x-h))$, complete the horizontal stretch/compression and then the horizontal shift; if necessary, rewrite the function so that it is in this form.

Review the material in Graphing Functions Using Vertical and Horizontal Shifts.

3d. determine whether a function is even, odd, or neither

• Identify the characteristics of an even function and an odd function.
• What test can be used to determine if a function is even, odd, or neither?

Functions sometimes have symmetry. A function symmetric about the $y$-axis is an even function; a function symmetric about the origin is an odd function. If $y=f(x)=f(-x)$, then the function is an even function. In contrast, if $y=f(x)=-f(-x)$, then the function is an odd function. A function can be even, odd, or neither.

Consider functions $f, g$, and $h$ as shown below. Function $f$ defined as $f(x)=x^{4}$ is an even function; for every $x$, $f(x)=f(-x)$ and the graph is symmetric about the $y$-axis. Function $g$ defined as $g(x)=x^{5}$ is an odd function; for every $x, f(x)=-f(-x)$ and the graph is symmetric about the origin. Function $h$ defined as $h(x)=(x+1)^{2}-3$ is neither even nor odd; $h(x) \neq h(-x)$ and $h(x) \neq-h(-x)$ nor is the graph symmetric to either the $y$-axis or the origin.

Review the material in Graphing Functions Using Vertical and Horizontal Shifts.

3e. verify the inverse of a function

• Determine the conditions under which a function has an inverse which is a function.
• Explain how to verify if two functions are inverses of each other algebraically. 

An inverse function is a function whose input is the output of the original function and whose output is the input of the original function. The inverse of function $f$ is typically denoted $f^{-1}$, but this notation does not mean the reciprocal of $f$. If the ordered pair $(7, \, 10)$ is an element of the original function, then the ordered pair $(10,7)$ would be an element of the inverse function.

A function only has an inverse if the function is one-to-one so that each input value has just one output value and each output value has just one input value. If the function is not one-to-one, then there is at least one output value paired with two different input values. So, switching the input values and output values means the new input value would be paired with two different output values, so the new set of ordered pairs is not a function.

Two functions are inverses of each other if $\left(f \circ f^{-1}\right)(x)=x$ for all $x$ in the domain of $f^{-1}$ and $\left(f^{-1} \circ f\right)(x)=x$ for all $x$ in the domain of $f$. For example, let $f(x)=4 x+3$ and $g(x)=4 x-3$. Because $(f \circ g)(5)=f(g(5))=$ $f(17)=71$, the two functions are not inverses of each other. In contrast, if $h(x)=\frac{x-3}{4}$, then $(f \circ h)(5)=$ $f(h(5))=f\left(\frac{1}{2}\right)=5$ and $(h \circ f)(5)=h(f(5))=h(23)=\frac{23-3}{4}=5$. Thus, functions $f$ and $h$ are inverses of each other.

Review material in Inverse Functions.

3f. define the inverse of a function and its domain and range using algebraic operations

• Explain how to determine the domain and range of an inverse function.
• Identify how the inverse of a function can be determined algebraically.

In general, the domain of a function represents the range of its inverse; likewise, the range of the original function is the domain of the inverse. If the original function is not one-to-one, then a restriction must be placed on the function to identify an appropriate inverse function; the restriction can be any interval over which the restricted function is one-to-one.

If a one-to-one function is represented by a table, then find the inverse of the function by reversing the input and output values. If a one-to-one function is represented by an equation $y=f(x)$, find the inverse by reversing the input and output values and solving for a new value of $y$. For example, let a function $f$ be defined as $y=f(x)=7 x+8$. To find the inverse, reverse $x$ and $y$ and then solve for $y: x=7 y+8$ so $y=\frac{x-8}{7}$. So, the inverse function $f^{-1}$ is defined as $f^{-1}(x)=\frac{x-8}{7}$.

Review material in Inverse Functions.

3g. given a one-to-one function construct the graph of its inverse

Specify the equation of the identity line.

Describe how the identity line can be used to graph the inverse of a function.

The identity line has equation $y=x$. If the point $(a, b)$ is on the graph of a function, then the point $(b, a)$ is on the graph of its inverse. These two points are reflections of each other across the identity line. So, the identity line is helpful in graphing a function and its inverse. If a function is one-to-one so that it has an inverse, the graph of its inverse is the reflection of the graph of the function across the identity line $y=x$.

Consider the graph of a function and its inverse on the coordinate grid below. Observe that the ordered pairs $(3,0)$ and $(12,3)$ are both on the graph of function $f$; thus the ordered pairs $(0,3)$ and $(3,12)$ must be on the graph of the inverse function $f^{-1}$.

Review material in Inverse Functions.

Unit 3 Vocabulary

This vocabulary list includes terms you will need to know to successfully complete the final exam.

• composite function
• compression of a graph
• even function
• horizontal shift
• horizontal reflection
• identity line
• inverse function
• odd function
• reflection
• stretch of a graph
• translation
• vertical reflection
• vertical shift

4a. graph points and lines on the Cartesian coordinate plane

• Compare graphing ordered pairs on a coordinate plane with graphing a line on the coordinate plane.

The Cartesian coordinate plane is constructed using two axes that intersect at a $90^{\circ}$ angle. The horizontal axis is typically labeled as the $\boldsymbol{x}-axis$; the vertical axis is typically labeled as the $\boldsymbol{y}-axis$. The point of intersection of the two axes is the origin. Graphing then uses ordered pairs in the form $(x, y)=(a, b)$, with the origin having coordinates $(0,0)$. The first coordinate, or $x$-coordinate, determines the distance left or right from the origin depending on whether $a < 0$ or $a > 0$, respectively. Similarly, the second coordinate, or $y$-coordinate, determines the distance up or down, depending on whether $b > 0$ or $b < 0$, respectively. Thus, the point $(3,5)$ is the point found by moving 3 units to the right of the origin and 5 units up.

To graph a line, find a set of ordered pairs that lie on the equation of the line. Recall from geometry that any two points determine a line. It is often wise to find at least three points that appear to satisfy the equation of the line when graphing to catch any possible errors in identifying ordered pairs on the line.

The Cartesian coordinate plane below shows the two axes, the origin, the point $(3,5)$, and three points that satisfy the equation of the line $y=-\frac{3}{2} x+1$.

Review material in Representations of Linear Functions.

4b. represent a linear function using words, tables, graphs, and function notation

• Define a linear function and express a linear function using function notation.
• Identify characteristics of a linear function when the function is described in words or via a table.

A linear function is a function whose graph is a line. The rule for the linear function only contains the first power of the input variable. In general, a linear function $f$ is described by $\boldsymbol{y}=\boldsymbol{f}(\boldsymbol{x})=\boldsymbol{m} \boldsymbol{x} \boldsymbol{+} \boldsymbol{b}$.

One characteristic of a linear function is that it has a constant rate of change of the output variable in relation to the input variable. In function form, this rate of change is denoted by $m$ and represents the slope of the line. The starting value of the linear function is the constant, denoted by $b$ in the function form; this value corresponds to the ordered pair $(0, \, b)$ which is the ordered pair associated with the $\boldsymbol{y}-intercept$ of the graph, or the $y$-value where the graph intersects the $y$-axis. So, a linear function can be identified in a real-world context if there is some starting value and if there is a constant rate of change in the situation. For instance, both situations below can be represented by a linear function.

• An individual opens a savings account with $\$ 100$ and deposits $\$ 50$ each week. If $w$ represents the number of weeks, what function $d(w)$ represents the amount in the account after $w$ weeks if no withdrawals are made? $d(w)=50 w+100$
• A bag of rice contains 50 pounds and a restaurant uses 8 pounds of rice per day. If $x$ represents the number of days, what function $r(x)$ represents the amount of rice left after $x$ days? $r(x)=50-8 x$

If a linear function is represented in a table, the constant rate of change should be visible in the values shown in the table. If the input value changes by a constant amount, then the output value must also change by a constant amount. But be careful! This check can only be applied if the input values change by a constant amount. In the table below, the input values, $x$, increase by 2 from left to right; the output values $f(x)$ increase by 10 from left to right; the output values $g(x)$ do not change by a constant amount from left to right. Hence, $f(x)$ represents a linear function but $g(x)$ does not.

Review material in Representations of Linear Functions.

4c. describe the behavior of a linear function over an interval

• Identify a linear function as increasing, decreasing, or constant.
• Determine the value of the rate of change which leads to an increasing, decreasing, or constant linear function.

As mentioned in objective $4 \mathrm{~b}$, a linear function is one in which there is a constant rate of change. Unless a linear function is restricted to particular values, such as a context in which only non-negative values make sense, both the domain and range of a linear function are represented by the interval $(-\infty, \infty)$. Hence, a linear function is always increasing, always decreasing, or constant.

Recall that a linear function is represented as $f(x)=m x+b$, where $m$ represents the slope or the rate of change. Thus, the linear function is increasing when $m > 0$, decreasing when $m < 0$, and constant when $m=0$. A constant linear function is represented by a horizontal line. On the coordinate grid below, the blue line is increasing, the red line is decreasing, and the black line is constant.

Review material in Representations of Linear Functions.

4d. interpret the slope of a line using words

• Compute the slope of a line.
• Interpret the slope of a line as a change in y for a particular change in x. 
• Contrast a line with zero slope to a line whose slope is undefined.
• Compare the slopes of two lines that are parallel with the slopes of two lines that are perpendicular.

The slope of a line represents the rate of change for the line. For the linear function $f(x)=m x+b$, the slope is represented by $m$ and is found by the following formula:

$m=\frac{\text { change in output }}{\text { change in input }}=\frac{f\left(x_{2}\right)-f\left(x_{1}\right)}{x_{2}-x_{1}}$

When the slope is positive, the line increases from left to right; when the slope is negative, the line decreases from left to right; when the slope is zero, the line is constant, or horizontal. A slope of $\frac{2}{3}$ can be interpreted in several ways, all with equivalent meanings:

• an increase of $\frac{2}{3}$ in the output for every increase of 1 in the input
• a rise of 2 units for every run of 3 units
• up 2 units and to the right 3 units
• down 2 units and to the left 3 units

Horizontal lines have a constant rate of change; that is, there is no change in the output values even when there is a change in the input values. So, the slope of a horizontal line is 0. In contrast, for a vertical line, there is a change in output values even when there is no change in input values. A vertical line is NOT a function, and the slope of a vertical line is undefined.

Two lines are parallel if they are in the same plane and never intersect; parallel lines have the same rate of change, meaning they have the same slope. In contrast, two lines are perpendicular if they intersect at a $90^{\circ}$ angle; the slopes of perpendicular lines are negative reciprocals of each other, meaning that the product of the

slopes of perpendicular lines is $-1$ assuming neither line is vertical. For example, if a line has a slope of $\frac{7}{3}$, a line parallel to it also has a slope of $\frac{7}{3}$; a line perpendicular to it will have a slope of $-\frac{3}{7}$.

Review material in Representations of Linear Functions.

4e. construct the equation of a linear function given two points, a table, or words

• Compare finding the equation of a linear function when given the slope and intercept, two points, or a table of values.
• Explain how characteristics in a context, such as rate of change and starting value, can be used to write a linear function to describe the context.

The slope-intercept form of an equation for a line is $\boldsymbol{y}=\boldsymbol{m x}+\boldsymbol{b}$, where $m$ is the slope and $b$ is the $y$-intercept. If both the slope and the $y$-intercept are known, then these values can be substituted into the slope-intercept form to represent the equation of the line. For instance, if a line has slope 6 and $y$-intercept 2, then the equation of the line is $y=6 x+2$.

If the slope or the $y$-intercept is not given, then the equation of the line can be found by using several steps. Suppose that two points on the line are given. First, find the slope of the line between those two points. Second, substitute the values of the coordinates from one of the ordered pairs into the slope-intercept form to find $b$. Third, use the values of $m$ and $b$ found in the first and second steps to write an equation for the line. For example, suppose $(3,1)$ and $(12,16)$ are two points on a line.

• First, find the slope between these two points: $m=\frac{16-1}{12-3}=\frac{15}{9}=\frac{5}{3}$
• Second: use one of the points as values for $x$ and $y$ together with $m$ to find $b$. Using the ordered pair $(3,1)$

$1=\frac{5}{3} \cdot 3+b$ so $b=-4$

• Third, use $m$ and $b$ to write an equation: $y=\frac{5}{3} x-4$

If the values for a linear function are provided in a table, choose any two sets of input/output values and follow the three steps just listed. If one input value is 0, then the corresponding output value is the $y$-intercept; so, this input/output pair should be chosen when feasible.

In a contextual situation, try to identify the starting value, which represents the $y$-intercept or value for $b$, as well as the constant rate of change, which represents the slope or value for $m$. Consider the following context:

• To join a gym, there is an initial membership fee of $50 and then a monthly fee of $10. Write an equation that gives the total amount paid, $y$, after $x$ months of membership.

The starting value is $b=50$ and the constant change is $m=10$. So, an equation to describe the context is $y=10 x+50$

At times, contextual situations that are linear in nature are best modeled by equations of the form $A x+B y=C$. For instance, consider the following situation:

• At a market, grapes cost $2.49 per pound and apples are \$1.99 per pound. How many pounds $g$ of grapes and $p$ of apples can be purchased for $11.45?

The total cost of $g$ pounds of grapes is $2.49 g$; the cost for $p$ pounds of apples is $1.99 p$. Then, the situation is modeled by the equation $2.49 g+1.99 p=11.45$.

Review material in Representations of Linear Functions.

4f. sketch the graph of a linear function using points, the slope and intercept, equations, and transformations

• Compare graphing a linear function using points, the slope and intercept, or an equation.
• Explain how transformations can be applied to the toolkit linear function $y=x$ to $\mathrm{graph} y=m x+b$.

Several techniques can be used to graph a linear function. Although any technique can be used, one technique may be more efficient than another in a particular situation. If specific points are given, simply connect the line through those points.

If points on the line are not given, then points need to be found. If the slope and $y$-intercept are known, then the linear equation can be graphed by plotting the ordered pair associated with the $y$-intercept and then using the slope as rise over run to find additional points. Given a line with $y$-intercept 2 and slope $\frac{3}{4}$. First graph the point $(0, \, 2)$. Then interpret the slope as a rise of 3 units for a run of 4 units. That is, move 4 units to the right and 3 units up to find the point $(4,5)$; again, move 4 units to the right and 3 units up to obtain the point $\left(8, \, 8\right)$. Now draw the line connecting the points as shown below.

If an equation is given rather than the $y$-intercept and slope, first find points that lie on the line. Although a line can be constructed using just two points, it is useful to find at least three points that lie on the line to catch any errors in evaluating the linear function. It is often efficient to substitute $x=0$ into the equation to find $y$; this value of $y$ is the $y$-intercept. Likewise, it can be useful to substitute $y=0$ into the equation to find $x$; this value of $x$ is the $\boldsymbol{x}-intercept$, namely where the graph crosses the $x$-axis. If substituting either $x=0$ or $y=0$ results in a fraction for the other value, then one might not want to use those values because fractional values are more difficult to plot. Instead, choose a value for one of the variables so that the other variable has an integer value.

For instance, if $3 x+4 y=24$, substituting $x=0$ generates the ordered pair $(0,6)$; substituting $y=0$ gives the ordered pair $(8,0)$; if $x=4$, then $y=3$ to give the ordered pair $(4, \, 3)$. Now draw the line connecting these three points. However, if $5 x+3 y=11$, then substituting $x=0$ gives a fractional value for $y$ and substituting $y=0$ gives a fractional value for $x$. So, try to find values that give integer coordinates. Some such points are (1, 2), $(4, \, -3),(7, \, -8)$; connect the line through these points.

Recall that $f(x)=m x$ is the graph of $f(x)=x$ when stretched vertically by a factor of $m$ if $m > 0$ or compressed vertically by a factor of $m$ if $0 < m < 1$; additionally, if $m < 0$, then the line is reflected vertically across the $x$ axis. Likewise, the graph of $f(x)=x+b$ is the graph of $f(x)=x$ when shifted up if $b > 0$ or down if $b < 0$. When graphing using transformations, any transformations must be completed in a particular order; apply transformations related to $m$ first and then apply transformations related to $b$, similar to multiplying first in the order of operations before adding.

Consider graphing $f(x)=-3 x-4$ using transformations. First, graph $f(x)=x$ (the blue line); two points on this line are $(0, \, 0)$ and $(1, \, 1)$. Second, stretch the line vertically by a factor of 3 (the red line) to obtain the line through the corresponding points $(0, \, 0)$ and $(1, \, 3)$. Third, reflect the line vertically across the $x$-axis (the green line) to obtain the line through the corresponding points $(0, \, 0)$ and $(1, \, -3)$. Finally, shift the line down by 4 units to obtain the graph of $f(x)=-3 x-4$, that is, the black line through the corresponding points $(0,-4)$ and $(1, \, -7)$.

Review material in Representations of Linear Functions.

4g. construct the equation for a linear function given a graph

• Explain how to find an equation for a linear function from a graph.
• Compare the equations for a vertical line with that for a horizontal line.

To find an equation for a line given the graph of the line, use the same techniques as described in objective 4 e. That is, one needs either the slope and $y$-intercept or two points. If the $y$-intercept is an integer and easily identified from the graph, then the value of $b$ in $y=m x+b$ is known. Then, find the slope, perhaps by identifying another ordered pair with integer coordinates and then using the slope formula to find $m$. Once both $m$ and $b$ are known, the equation can be written in the form $y=m x+b$.

If the $y$-intercept is not easily identified, try to find two ordered pairs whose coordinates are integers. Find the slope of the line between those two points. Use the coordinates of one ordered pair in the equation $y=m x+b$ to find $b$ and write an equation for the line.

There are two special cases. A horizontal line represents a linear function. All points on a horizontal line have the same $y$-coordinate. Thus, an equation for a horizontal line is $y=b$. Although a vertical line does not represent a linear function, it is possible to write an equation for a vertical line. All points on a vertical line have the same $x$-coordinate, so an equation for a vertical line is $x=h$.

Review material in Representations of Linear Functions.

4h. make predictions using linear models constructed from words and data

• Explain how to construct a linear model to describe data representing a real-world situation.
• Identify a regression line that fits a set of data.
• Describe how a linear model can be used to predict a future outcome.

Many real-world situations can be modeled by a linear equation. Such situations have a starting value and a constant rate of change of the output variable in relation to the input variable. At times, the data may not lie perfectly along a straight line even though the data appear to be linear in nature. In such cases, draw a line that seems to fit the data with about as many points above the line as below the line; such a line is called a regression line. Then, write an equation for this regression line using the techniques in objectives 4e and 4g. Evaluate the equation for an input value not expressed in the data set to predict the corresponding output variable. In real-world situations, it is important to consider a reasonable domain and range for the set of data.

For instance, consider the set of data in the table below that represent the number of students in a school over several years. The data are then graphed on the coordinate plane and a black line is drawn through the data. For the line that was drawn, the vertical intercept is roughly 425 and the slope of the line is roughly $\frac{75}{6} \approx 12.5$ using the ordered pairs $(0, \, 425)$ and $(6,500)$. So, a possible equation to model the data is $s=12.5 n+425.$ In year 4, the equation predicts a population of 475, which is reasonable given that this population is between the population in year 3 (440 students) and the population in year 5 (485 students). The equation predicts a population of about 675 in year 20. There is no guarantee that the line drawn by hand is the best line to model the data. Finding the best line, called the least-squares regression line is discussed in objective $4 j$.

Review the material in Building Linear Models from Words and Finding the Line of Best Fit.

4i. interpret whether scatter diagrams represent a linear relation

• Explain how the correlation coefficient provides insight into whether a set of data represents a linear relation.

When a set of data is plotted on a coordinate grid, the data may suggest a linear trend, a trend best described by another type of function, or no trend at all. One tool to describe a trend in the data is the correlation coefficient, typically denoted by $\boldsymbol{r}$. Most technology tools that attempt to find an equation to model the data will generate the correlation coefficient.

The correlation coefficient has a value in the interval $-1 \leq r \leq 1$. A positive value of $r$ indicates a relationship with a positive slope; a negative value of $r$ indicates a relationship with a negative slope; a value of $r=0$ indicates no relationship in the data. The closer $r$ is to 1 or $-1$, the stronger the linear relationship; the closer $r$ is to 0, the more scattered the data. Typically, the correlation coefficient is not exactly 1 or $-1$ as such values would indicate a perfect linear relationship with no variability. In most real-world situations that generate data, there is not a perfect relationship.

Review material in Finding the Line of Best Fit.

4j. use a graphing utility to construct a linear regression line given a data set

• Apply a graphing utility to find a least-squares regression line for a set of data.

Most graphing utilities, such as graphing calculators or statistical software, have built-in capabilities for finding a regression line for a set of data. The utilities find the least-squares regression line or the line of best fit, using procedures beyond the scope of this course. Generally, the input values are placed in one statistical list and the output values are placed in a second statistical list. Select the linear regression (linreg) option to find the regression line. Obtain the value of $r$ to determine the strength of the linear relationship.

Caution! Most graphing utilities are able to determine regression equations that model data for a variety of functions, including linear, quadratic, cubic, or logarithmic. Make sure to select linear regression (linreg) and not logarithmic regression (lnreg).

Refer back to the data in objective $4 \mathrm{~h}$. One graphing utility gives the regression equation as $s \approx 19 n+384$ with a value of $0.994$ for the correlation coefficient, indicating a strong positive relationship. Using this equation, we would predict the population for year 4 to be 460 students and for year 20 to be 764 students. These values might be a bit more realistic than those obtained from the regression line that was simply "eye-balled" in objective 4h.

Review material in Finding the Line of Best Fit.

4k. analyze how changes in data affect a regression line

• Explain how a change in the data set might affect the equation for a regression line for the data.

As might be expected, changes in a data set can result in varied changes in a regression line. If the data have a strong linear relationship, then changes in the data may not result in major changes in the regression line; this is particularly true if changes in the data occur in the middle of the data set. In contrast, if the data have a more modest linear relationship, then changes in the data might result in major changes in the equation for the regression line, particularly if the changes in the data occur toward one of the ends of the data set. 

To explore how data changes might affect the regression line, use your graphing utility to recompute the regression line and the related correlation coefficient.

Review material in Finding the Line of Best Fit.

4l. distinguish between interpolation and extrapolation

• Contrast interpolation with a data set and extrapolation with a data set.

One reason to obtain a regression line for a set of data is to enable predictions with its use. That is, the equation is used to obtain output values for input values that are not part of the actual data set. When the prediction is made within the domain and range of the data set, the process is called interpolation. When the prediction is made outside the domain and range of the data set, the process is called extrapolation. For the data set describing the number of students at a school over several years, predicting the student population at year 4 involves interpolation because the given data set has input values from 1 to 15. Predicting the student population for year 20 involves extrapolation because this input value is beyond the data set. As might be expected, extrapolation has a greater tendency to be incorrect than does interpolation. Extrapolation can often lead to unreasonable values because prediction is attempted beyond values where the model makes sense.

Review material in Finding the Line of Best Fit.

Unit 4 Vocabulary

This vocabulary list includes terms you will need to know to successfully complete the final exam.

• correlation coefficient, $r$
• extrapolation
• function notation for a linear function
• horizontal axis, $x$-axis
• interpolation
• least-squares regression line
• linear function
• origin
• parallel lines
• perpendicular lines
• regression line
• slope
• slope-intercept form of a linear equation, $y=m x+b$ $x$-intercept
• vertical axis, $y$-axis
• $y$-intercept

5a. identify the properties of power, quadratic, and polynomial functions given a graph or an equation including end behavior, domain, and range, and local behaviors by using algebraic operations, interval notation, and words

• Compare the forms of quadratic functions, power functions, and polynomial functions.
• For a quadratic function, identify the shape of the graph, the vertex, and axis of symmetry. Compare the $y$-intercept with the vertex. How does the leading coefficient determine whether the graph opens up or down?
• Compare the domain and range for quadratic functions, power functions, and polynomial functions. 
• For a power function, explain how the exponent on the variable determines the end behavior.
• For a polynomial function, identify the degree and leading coefficient of the function. Identify the end behavior of a polynomial function and the number of turning points. Explain how to connect the turning points to local minimum or local maximum points.

A quadratic function is a function of the form $f(x)=a x^{2}+b x+c$, with $a, b$, and $c$ real numbers and $a \neq 0$; this form of the function is known as the general form of a quadratic function. The graph of a quadratic function is a parabola that opens up if $a > 0$ and opens down if $a < 0$. A quadratic equation can also be represented in the form $f(x)=a(x-h)^{2}+k$, known as the standard form or vertex form of a quadratic function. The ordered pair $(h, k)$ is the vertex of the parabola, that is, the minimum point or the maximum point of the parabola, depending on whether the parabola opens up or down, respectively; the value of $h$ can be found as $h=-\frac{b}{2 a}$. The quadratic function or parabola is symmetric about the vertical line drawn through the vertex; this line is called the axis of symmetry.

The $y$-intercept is the value of $y$ where the parabola crosses the $y$-axis, that is, it is the value of $y$ when $x=0$. Similarly, the $x$-intercept is the value of $x$ where the parabola crosses the $x$-axis, that is, the value of $x$ when $y=0$; the values of the $x$-intercept are also known as the zeros of the quadratic function. The zeros of the quadratic function can be found using the quadratic formula: $x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} ;$ there are two zeros if the parabola crosses the $x$-axis in two points, one zero if the vertex is on the $x$-axis, or no zeros if the parabola does not intersect the $x$-axis.

The graph below shows these important points marked on the quadratic function.

The table below compares the domain and range as well as the behavior of the quadratic function depending on whether the parabola opens up or down. 

A power function is a function of the form $f(x)=k x^{p}$ where $k$ and $p$ are real numbers; $k$ is the coefficient of the power function. Notice that in a power function the variable is raised to a power. Both the basic linear function $f(x)=x$ and the basic quadratic function $f(x)=x^{2}$ are examples of power functions.

Power functions in which the exponent is a non-negative integer have specific properties of interest. If the exponent $p$ is even, then the power function is an even function, meaning it is symmetric about the $y$-axis. The function opens up if $k > 0$ and down if $k < 0$. So, the domain is $(-\infty, \infty)$ and the range is $[0, \infty)$. The end behavior of the function is the behavior of the function as $x \rightarrow \infty$ or as $x \rightarrow-\infty$. For an even power function, if $k > 0$ the function approaches $\infty$ as $x \rightarrow \infty$ or as $x \rightarrow-\infty$; in contrast, if $k < 0$, the function approaches $-\infty$ as $x \rightarrow \infty$ or as $x \rightarrow-\infty$.

A similar analysis can be applied to power functions in which the non-negative integer is odd. In this case, the function is an odd function, meaning it is symmetric about the origin. If $k > 0$, then the function approaches $\infty$ as $x \rightarrow \infty$ and approaches $-\infty$ as $x \rightarrow-\infty$; in contrast, if $k < 0$, the function approaches $\infty$ as $x \rightarrow-\infty$ and approaches $-\infty$ as $x \rightarrow \infty$. In this case, both the domain and range are $(-\infty, \, \infty)$.

The graphs below represent an even power function and an odd power function, both for $k > 0$.

A polynomial function is a function of the form $f(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{2} x^{2}+a_{1} x+a_{0}$ where $n$ is a non-negative integer, each $a_{i}$ is a real number, and $a_{n} \neq 0$. Every $a_{i} x^{i}$ is called a term of the polynomial. The greatest value of the exponent is the degree of the polynomial and the term with the greatest value of the exponent is the leading term. The end behavior of the polynomial function is determined by the end behavior of the leading term. Notice that the leading term is an even or odd power function, so the end behavior of the polynomial function is the same as the end behavior of an even or odd power function as described above. The domain of a polynomial function is $(-\infty, \infty)$; the range can vary depending on the degree of the polynomial and whether it has any absolute maximum or absolute minimum points.

Polynomial functions are continuous and smooth, meaning there are no breaks in the graph and no sharp corners, respectively. Depending on the degree of the polynomial and the values of the non-leading terms, the polynomial may have various turning points, which are the local minimum and local maximum points, that is, the points at which the polynomial changes from decreasing to increasing or increasing to decreasing, respectively. For a polynomial with degree $n$, there are at most $n-1$ of these turning points. As with other functions, the $y$-intercept is where the polynomial crosses the $y$-axis; the $x$-intercepts or zeros are where the polynomial crosses the $x$-axis. A polynomial with degree $n$ has at most $n$ distinct zeros or $x$-intercepts.

For instance, the polynomial function $f(x)=4 x^{7}-3 x^{5}+4 x^{2}-8$ is a polynomial of degree 7. It has at most 7 zeros and 6 turning points. However, these values are the maximum number of zeros and turning points. As shown by the graph of $y=f(x)$ below, the polynomial function may not have this many zeros or turning points. In this case, the function has 2 turning points, one a local minimum and one a local maximum, and only 1 zero or $x$-intercept. One of the turning points occurs at the point $(0, \, -8)$ associated with the $y$-intercept.

Review material in Understanding How the Graphs of Parabolas are Related to Their Quadratic Functions, Power Functions, and Polynomial Functions.

5b. relate elements of the standard form of a quadratic function with corresponding transformations to its graph

• Describe how the standard or vertex form of a quadratic function provides insights into the use of transformations to graph a quadratic function.

Recall that the standard or vertex form of a quadratic function has the form $f(x)=a(x-h)^{2}+k$. The coefficient $a$ describes the vertical stretch or compression of the graph of $f(x)=x^{2}$, depending on whether $a > 1$ or $0 < a < 1$, respectively. If $a < 0$, the graph is reflected vertically across the $x$-axis, meaning the parabola opens down. The vertex has coordinates $(h, \, k)$ indicating a shift of the function $f(x)=x^{2}$ by $h$ units horizontally and $k$ units vertically.

Consider the function $f(x)=-3(x+1)^{2}+5$. The graph of this function is the graph of $f(x)=x^{2}$ when stretched vertically by a factor of $3$ ; the graph is then reflected vertically across the $x$-axis. Then, because the vertex is at $(-1, \, 5)$, the graph will be shifted 1 unit to the left and 5 units up. For instance, the point $(1, \,  1)$ on $f(x)=x^{2}$ (the red graph) corresponds to $(1,3)$ after the vertical stretch, then corresponds to $(1,-3)$ after reflection across the $x$-axis, and then corresponds to $(0, \, 2)$ after a shift of 1 unit to the left and 5 units up. You can verify that $(0, \, 2)$ is on the final (black) graph of $f(x)=-3(x+1)^{2}+5$.

Review material in Understanding How the Graphs of Parabolas are Related to Their Quadratic Functions.

5c. construct the equation of a power, quadratic, or polynomial function given its graph

• Explain how the characteristics of the graph of a quadratic function can be used to construct an equation for the function.
• Use the properties of the graphs of power functions to construct equations for those power functions.
• Describe how to use the zeros of a polynomial function to construct an equation for a polynomial function when given its graph.

Given the graph of a parabola, which is a quadratic function, the following techniques can be used to construct the corresponding equation. 

• If the vertex can be easily identified, then relate its coordinates to the values $(h, \, k)$. Now write the equation $y=a(x-h)^{2}+k$. Select another point on the parabola, preferably one with integer coordinates, and substitute those values for $x$ and $y$, together with the values for $h$ and $k$, to find the value for $a$.
• If the parabola has zeros at $(p, \, 0)$ and $(q, \, 0)$, then write the equation $y=a(x-p)(x-q)$. Select another point on the parabola, preferably one with integer coordinates, and substitute those values for $x$ and $y$ to find the value for $a$.
• If there is a single zero at $(p, 0)$, then start with the equation $y=a(x-p)^{2}$. Select another point on the parabola, preferably one with integer coordinates, and substitute those values for $x$ and $y$ to find the value for $a$.

Consider the two quadratic graphs shown below. To find the equation for $y=f(x)$, observe that the vertex is located at $(4,-6)$ and that $(0, \, 2)$ is also on the graph. So, substitute $(0,2)$ in the equation $y=a(x-4)^{2}-6$ to find $2=a \cdot 16-6$, so $a=\frac{1}{2}$. Then $y=f(x)=\frac{1}{2}(x-4)^{2}-6$. To find the equation for $y=g(x)$, observe that the zeros are $-5$ and $-1$, so start with the equation $y=a(x+5)(x+1)$. Another point on the graph is $(0,-5)$. Substituting these values into the equation leads to $-5=a(5)(1)$, so $a=-1$. Then $y=g(x)=-(x+5)(x+1)$, which could be written in general form as $y=g(x)=-x^{2}-6 x-5$.

Recall that the graph of a power function is a function of the form $f(x)=k x^{p}$. The end behavior of the power function, together with several ordered pairs on the function, can be used to determine if the function is an even power function or an odd power function, that is, a power function in which the exponent is a nonnegative integer.

Consider the two functions graphed below. The behavior of $s(x)$ suggests it is an odd power function. The points $(1, \, 2),(2, \, 16),(-1, \, -2)$, and $(-2, \, 16)$ appear to be on the graph. So, it appears the equation is $s(x)=2 x^{3}$. In contrast, the behavior of $t(x)$ suggests it is an even power function that has been reflected over the $x$-axis, meaning the coefficient is negative. The points $(2, \, -8)$ and $(-2, \, -8)$ are clearly on the graph and it appears that the points $\left(1,-\frac{1}{2}\right)$ and $\left(-1,-\frac{1}{2}\right)$ are also on the graph. This suggests the equation is $t(x)=-\frac{1}{2} x^{4}$.

If the exponent is not a non-negative integer, then one needs to consider the shape of basic graphs such as $y=x^{1 / 2}, y=x^{1 / 3}$, or $y=x^{-1}$ as a starting point for identifying an equation. Using specific ordered pairs on the graph, together with the basic shapes and transformation concepts, can help determine a feasible equation.

To find the equation of a polynomial function from its graph, start by locating the zeros of the graph. Recall that a polynomial function of degree $n$ has at most $n-1$ turning points and $n$ zeros. These properties can be used to determine the potential degree of the polynomial and the multiplicity of a zero; if $p$ is a zero, then the multiplicity of the zero is the number of times that the factor $(x-p)$ appears in the equation. If a zero has an even multiplicity, then the graph bounces at the zero and does not cross the $x$-axis at that point; if a zero has an odd multiplicity, then the graph crosses the $x$-axis at that point. Use the zeros to write the polynomial in factored form with a leading coefficient of $a$. Find another point on the graph, preferably with integer coordinates, and substitute into the factored form of the equation to find the value of $a$.

Consider the polynomial function shown on the graph. There are two turning points, so the graph is at most degree $3$. The function has zeros at $-1$ and at $5$; the zero at $-1$ bounces off the $x$-axis, so it has an even multiplicity while the zero at 5 crosses the $x$-axis and has an odd multiplicity. Because the degree is at most $3$, the multiplicity at $-1$ must be 2. So, start with the equation $y=h(x)=a(x+1)^{2}(x-5)$. The point (3, -16) appears to be on the graph, so use these coordinates to find $a=\frac{1}{2}$. Then the equation for the polynomial function is $y=h(x)=\frac{1}{2}(x+1)^{2}(x-5)$.

Review material in Understanding How the Graphs of Parabolas are Related to Their Quadratic Functions, Power Functions, Polynomial Functions, and Graphing Polynomial Functions.

5d. apply the intermediate value theorem to determine the approximate location of a zero of a polynomial

• Explain the Intermediate Value Theorem. How can the Intermediate Value Theorem be used to identify intervals in which a zero to a polynomial function must exist?

Recall that a polynomial function is a continuous function, that is, it is a function with no breaks and is defined for all real numbers. The Intermediate Value Theorem is useful for such continuous functions. If $a < b$ and $f(a) \neq f(b)$, then $f(x)$ takes on every value between $f(a)$ and $f(b)$ at least once. For example, suppose a polynomial exists and contains ordered pairs $(5, \, 8)$ and $(9, \, 17)$. Then, for every output value of the function between $8$ and $17$, there is some input between 5 and 9 that results in that output value.

The Intermediate Value Theorem can be used to approximate the location of zeros of a polynomial function. If in some interval $[a, b]$, the values of $f(a)$ and $f(b)$ have opposite signs, meaning that one of these values is above the $x$-axis and the other is below the $x$-axis, then the function must cross the $x$-axis for some value of $c$ in the interval $a < c < b$. This means that $c$ is a zero of the polynomial function because $f(c)=0$. As an example, suppose for some polynomial function that $f(2)=-8$ and $f(5)=10$. Then there is at least one value $c$ in the interval $2 < c < 5$ for which $f(c)=0$; this value of $c$ is a zero of the polynomial.

Review the material in Graphing Polynomial Functions.

5e. evaluate polynomial functions using the remainder and factor theorems

• Contrast the results of the Remainder Theorem and the Factor Theorem when applied to a polynomial function.
• Explain how the Factor Theorem can be used to identify zeros of a polynomial function.

The Remainder Theorem states that if a polynomial function $p(x)$ is divided by the factor $x-k$, the result is the same as evaluating the polynomial for the input value $k$. If $p(k)=0$, then the factor $x-k$ divides the polynomial evenly, meaning that $x-k$ is a factor of the polynomial and $k$ is a zero of the polynomial. This is the special case of the Remainder Theorem known as the Factor Theorem. That is, the Factor Theorem states that if $f(k)=0$ then $(x-k)$ is a factor of the polynomial and $k$ is a zero.

Given the polynomial function $f(x)=4 x^{5}-8 x^{3}+4 x$ and the table of values for the function. Because $f(-1)=0$ and $f(1)=0$, both $x+1$ and $x-1$ are factors of $f(x)$ and $-1$ and 1 are zeros of the function. Because $f(2)=72,2$ is not a zero of the polynomial function; if the polynomial is divided by $x-2$, the remainder is $72$.

Review the material in Identify the x-Intercepts of Polynomial Functions whose Equations are Factorable.

5f. apply the rational zeros theorem, the fundamental theorem of algebra, and the linear factorization theorem to identify zeros of polynomials

• State the Rational Zeros Theorem. How is this theorem used to find zeros of a polynomial?
• Explain how the Linear Factorization Theorem is used to identify zeros of a polynomial.
• What is the Fundamental Theorem of Algebra? How does it help with finding zeros of a polynomial function?
• How are zeros and complex conjugates connected?

The Rational Zeros Theorem, sometimes called the Rational Root Theorem, can be applied to polynomials that have integer coefficients. If the polynomial has rational roots in the form $\frac{p}{q}$, then $p$ is a factor of the constant term and $q$ is a factor of the leading coefficient of the polynomial. This theorem is often used in conjunction with the Factor Theorem from objective 5 e. Evaluate the function for each of the possible rational roots; if the value of the function for any $\frac{p}{q}$ is $0$, then $\frac{p}{q}$ is a root or zero of the polynomial.

For example, consider the polynomial function $f(x)=6 x^{3}+31 x^{2}+3 x-10$. Factors of the leading coefficient are $1,2,3,6$; these are possible values for $q$. Factors of the constant are $1,2,5,10$; these are possible values for $p$. So, possible values of $\frac{p}{q}$ are $\pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}, \pm 2, \pm \frac{2}{3}, \pm 5, \pm \frac{5}{2}, \pm \frac{5}{3}, \pm \frac{5}{6}, \pm 10, \pm \frac{10}{3}$. Given that the polynomial has degree $3$, it has at most 3 zeros. Evaluate the function for each of these values until three roots are found. Evaluating the function for these possible values leads to $f\left(-\frac{2}{3}\right)=f\left(\frac{1}{2}\right)=f(-5)=0$, so the zeros of the polynomial are $-\frac{2}{3}, \frac{1}{2}$, and $-5$.

The Fundamental Theorem of Algebra states that every polynomial has at least one complex zero; so, every polynomial of degree $n > 0$ has $n$ zeros as long as the multiplicity of a zero is considered. Be careful! Not every polynomial has a complex zero of the form $a+b i$ where $b \neq 0$. The real numbers are a subset of the complex numbers, so it might be that the complex zero is of the form $a+b i$ with $b=0$; this is the case when all the zeros of a polynomial are real numbers.

The Linear Factorization Theorem states that a polynomial of degree $n$ has $n$ factors of the form $x-c$ where $c$ is a complex number. If a polynomial has real coefficients and if $a+b i$ with $b \neq 0$ is a zero of the polynomial, the Complex Conjugate Theorem states that $a-b i$ will also be a zero of the polynomial. This theorem has come into use in previous sections when solving quadratic equations; complex roots always occur in conjugate pairs.

These two theorems can be used together to find the equation for a polynomial function if the roots are known. For instance, suppose a fifth degree polynomial has roots $1, \, -4, \, 2+5 i$, and $6$. Because roots must occur in complex conjugate pairs, $2-5 i$ must also be a zero. So, the equation is $f(x)=a(x-1)(x+4)(x-6)(x-[2+5 i])(x-[2-5 i])$. The value of the coefficient $a$ is found by substituting in the coordinates of a specific point on the polynomial function that is not a zero.

Review the material in Identify the x-Intercepts of Polynomial Functions whose Equations are Factorable.

5g. confirm numbers of positive and negative roots of polynomials using Descartes' rule of signs

• Explain how Descartes' Rule of Signs is used to determine the number of positive or negative roots of a polynomial with real coefficients.

Descartes' Rule of Signs provides a test that can be used with other theorems to help in finding the zeros of a polynomial function. Suppose a polynomial $f(x)$ has real coefficients and the terms are written in decreasing order of the exponents. The number of positive zeros equals the number of changes in sign (positive to negative or vice versa) for $f(x)$ or is less than this amount by an even number. The number of negative zeros equals the number of changes in sign (positive to negative or vice versa) for $f(-x)$ or is less than this amount by an even number. Using this information together with the fact that a polynomial of degree $n$ can have at most $n$ zeros helps to know how many positive or negative roots need to be found.

For instance, consider the polynomial $f(x)=3 x^{8}-7 x^{4}+x^{2}-3 x-1$. All the coefficients are real numbers and the exponents are in decreasing order. So, at most there are 8 zeros. There are three sign changes in $f(x)$, so there are 3 or 1 positive roots. Now look at $f(-x)=3 x^{8}-7 x^{4}+x^{2}+3 x-1$; there are three sign changes in $f(-x)$ so there will be 3 or 1 negative roots. The graph of the function below shows 3 negative roots and 1 positive root.

Review the material in Identify the x-Intercepts of Polynomial Functions whose Equations are Factorable.

Unit 5 Vocabulary

This vocabulary list includes terms you will need to know to successfully complete the final exam.

• axis of symmetry of a quadratic function
  
• Complex Conjugate Theorem
• continuous function
• degree of a polynomial
• Descartes' Rule of Signs
• end behavior of a function
• Factor Theorem
• Fundamental Theorem of Algebra
• general form of a quadratic function
• Intermediate Value Theorem
• leading term
• linear Factorization Theorem
• multiplicity of a zero
• polynomial function
• power function
• quadratic function
• Rational Zeros Theorem
• Remainder Theorem
• smooth function
• standard form of a quadratic function
• term of a polynomial
• vertex form of a quadratic function
• vertex of a parabola
• zero of a function

6a. illustrate the properties of rational functions, including long-run behavior and local behaviors using arrow notation

• Describe the form of a rational function. 
• Compare the basic reciprocal function and the basic squared reciprocal function.
• Given the graph of a rational function, describe the end behavior of the function. Compare the end behavior to behavior of the function near values not in the domain of the function. 

A rational function is a function that is the quotient of two polynomial functions. Two basic rational functions are the reciprocal function, $f(x)=\frac{1}{x}$, and the squared reciprocal function, $g(x)=\frac{1}{x^{2}}$. For both of these basic functions, the domain is all real numbers such that $x \neq 0$. For function $f$, the range is all real numbers such that $y \neq 0$; for function $g$, the range is real numbers such that $y > 0$.

The graphs of these two basic rational functions exhibit several important properties. Neither graph crosses the $x$-axis or the $y$-axis. The end behavior of the function describes the behavior of the function as $x$ gets very large in value, denoted $x \rightarrow \infty$, or as $x$ gets very small in value, denoted $x \rightarrow-\infty$. For function $f$, as $x \rightarrow \infty, f(x)$ $\rightarrow 0$; as $x \rightarrow-\infty$, it is also the case that $f(x) \rightarrow 0$. The same is true for function $g$; as $x \rightarrow \pm \infty, g(x) \rightarrow 0$.

Although $x \neq 0$ for either function, the two functions behave differently near $x=0$. For function $f$, as $x$ approaches 0 from the right, denoted $x \rightarrow 0^{+}$, the function gets very large, denoted $f(x) \rightarrow \infty$; as $x$ approaches 0 from the left, denoted $x \rightarrow 0^{-}$, the functions gets very small in value, denoted $f(x) \rightarrow-\infty$. In contrast, for function $g$, the function gets large in value when $x$ approaches 0 from either the left or right, that is, as $x \rightarrow 0^{+}$, $g(x) \rightarrow \infty$ and as $x \rightarrow 0^{-}, g(x) \rightarrow \infty$.

Review material in End Behavior and Local Behavior of Rational Functions.

6b. identify vertical, horizontal, and slant asymptotes of rational functions given a graph or an equation

• Compare vertical and horizontal asymptotes and the conditions under which they arise.
• Determine when slant asymptotes occur.

A vertical asymptote to a graph is a vertical line that the function does not cross and for which the behavior of the function tends toward $\pm \infty$ as $x$ approaches the vertical line from either the right or the left. Vertical asymptotes will occur where the denominator of the rational function is undefined. But be careful! Not every value for which the denominator is zero will lead to a vertical asymptote. If the value of a function is undefined at $x=a$, then it is essential for the value of the function to approach $\pm \infty$ in order for $x=a$ to be a vertical asymptote. For both $f(x)=\frac{1}{x}$ and $g(x)=\frac{1}{x^{2}}$ graphed in objective 6 a, the line $x=0$ is a vertical asymptote.

A horizontal asymptote to a graph is a horizontal line that the function approaches as $x \rightarrow \pm \infty$. For both $f(x)=\frac{1}{x}$ and $g(x)=\frac{1}{x^{2}}$ graphed in objective 6 a, the horizontal line $y=0$ is a horizontal asymptote. Because horizontal asymptotes are determined by the behavior of the function when $x \rightarrow \pm \infty$, equations for horizontal asymptotes depend on the end behavior of the polynomial functions that comprise the rational function. Recall from Unit 5 that the end behavior of a polynomial function is determined by the behavior of the leading term. So, this leads to three situations for the end behavior of a rational function.

• If the degree of the numerator is less than the degree of the denominator, then there is a horizontal asymptote at $y=0$ because the denominator will increase in value faster than the numerator as $x$ gets very large in magnitude.
• If the degree of the numerator equals the degree of the denominator, then there is a horizontal asymptote at $y=$ ratio of the leading coefficients.
• If the degree of the numerator is one more than the degree of the denominator, then the end behavior approaches the line $y=k x$, where $k$ is the ratio of the leading coefficients. This asymptote is not horizontal but is a slant asymptote.

For example, consider the function $f(x)=\frac{6 x+7}{2 x-10}$. A vertical asymptote occurs at $x=5$; for this value of $x$, the denominator is undefined and $f(x) \rightarrow \infty$ as $x \rightarrow 5^{+}$; similarly, $f(x) \rightarrow-\infty$ as $x \rightarrow 5^{-}$. The numerator and denominator are both degree $1$ polynomials, so there is a horizontal asymptote at $y=\frac{6}{2}=3$.

In contrast, consider the function $g(x)=\frac{x-3}{(x-3)(x+2)}=\frac{1}{x+2}$. Although the function is undefined for $x=3$ and $x=-2$, there is only a vertical asymptote at $x=-2$ because only for this value does the function approach $\pm \infty$ as $x$ approaches $2$  from either the left or the right. For the simplified function, the degree of the numerator is $0$  and the degree of the denominator is $1$; so there is a horizontal asymptote at $y=0$.

Review material in End Behavior and Local Behavior of Rational Functions.

6c. describe the domain of a rational function using standard notation given an equation or a graph

• Explain how to find the domain of a rational function from its equation.
• How can the domain of a rational function be found from its graph?

A rational function is the quotient of two polynomial functions. In general, the domain of a polynomial function is all real numbers. However, in a fraction, the denominator cannot be zero. So, the domain of a rational function will be all real numbers that do NOT cause the denominator of the rational function to equal $0$. Then, to find the domain, set the polynomial function in the denominator to zero and solve. The domain is all real numbers except the zeros of that polynomial function.

For instance, consider the rational function $f(x)=\frac{(2 x+5)(4 x-3)}{(x+1)(x-7)(3 x+1)}$. To find the domain, first solve $0=(x+1)(x-7)(3 x+1)$, which results in $x=-1,7$, or $-\frac{1}{3}$. Then the domain is $\left\{x: x \neq-1,-\frac{1}{3}, 7\right\}$.

If given a graph, to find the domain look for any vertical asymptotes or any input values where there is a hole, or removable discontinuity. The input values corresponding to the vertical asymptote or the hole are not in the domain. Consider the rational function graphed below. It appears there is a vertical asymptote at $x=6$. There is also a hole at $x=-3$. So the domain is $\{x: x \neq-3 \, or \, 6\}$.

For instance, consider the rational function $f(x)=\frac{(2 x+5)(4 x-3)}{(x+1)(x-7)(3 x+1)}$. To find the domain, first solve $0=(x+1)(x-7)(3 x+1)$, which results in $x=-1,7$, or $-\frac{1}{3}$. Then the domain is $\left\{x: x \neq-1,-\frac{1}{3}, 7\right\}$.

If given a graph, to find the domain look for any vertical asymptotes or any input values where there is a hole, or removable discontinuity. The input values corresponding to the vertical asymptote or the hole are not in the domain. Consider the rational function graphed below. It appears there is a vertical asymptote at $x=6$. There is also a hole at $x=-3$. So the domain is $\{x: x \neq-3 \, or \, 6\}$.

The $\boldsymbol{y}$-intercept of a rational function occurs where the function crosses the $y$-axis. The $y$-intercept has coordinates $(0, k)$. To find the $y$-intercept, evaluate the function for $x=0$.

In contrast, the $\boldsymbol{x}$-intercept of a rational function occurs where the function crosses the $x$-axis. The $x$ intercept has coordinates $(h, 0)$, so the $x$-intercept is found by setting the function equal to $0$ and solving for the corresponding input value. Because the fraction representing the rational function is 0 when the numerator is $0$, the $x$-intercept is found by setting the polynomial in the numerator equal to $0$ and solving for the variable. For instance, consider $g(x)=\frac{(2 x+1)(x-4)}{(x+5)}$. To find the $y$-intercept, set $x=0$ and evaluate the function; the $y$ intercept occurs for $y=-\frac{4}{5}$ and has coordinates $\left(0,-\frac{4}{5}\right)$. To find the $x$-intercepts, find those values of $x$ for which the numerator is $0$, namely $x=-\frac{1}{2}$ and $x=4$, corresponding to coordinates $\left(-\frac{1}{2}, 0\right)$ and $(4,0)$.

A removable discontinuity for a rational function occurs for an input value that is not in the domain of the function but for which the function does not approach $\pm \infty$ as $x$ approaches this input value from the left or the right. Such a discontinuity occurs when there is a common factor in the numerator and denominator of the polynomial function. Although the input value is not in the domain, the value of the rational function can be determined once the common factor is removed. A removable discontinuity is represented by a hole in the graph, such as the hole shown in the graph in objective $6 c$. The function represented by that graph is $f(x)=$ $\frac{(x-15)(x+3)}{(x-6)(x+3)}=\frac{x-15}{x-6}$. Notice that $x=-3$ is not in the domain of the function; however, when $x=-3$, the function can be evaluated as $f(x)=\frac{-3-15}{-3-6}=2$. So, there is a hole in the graph at $(-3,2)$; the discontinuity that would have occurred at $x=-3$ and that could have resulted in a vertical asymptote has been removed.

Review material in End Behavior and Local Behavior of Rational Functions.

6d. identify and graph removable discontinuities and intercepts of a rational function given an equation

• Explain how to find the x-intercept and y-intercept of a rational function.
• What is a removable discontinuity? How is a removable discontinuity shown on a graph of a rational function?

The $\boldsymbol{y}$-intercept of a rational function occurs where the function crosses the $y$-axis. The $y$-intercept has coordinates $(0, k)$. To find the $y$-intercept, evaluate the function for $x=0$.

In contrast, the $\boldsymbol{x}$-intercept of a rational function occurs where the function crosses the $x$-axis. The $x$ intercept has coordinates $(h, 0)$, so the $x$-intercept is found by setting the function equal to $0$ and solving for the corresponding input value. Because the fraction representing the rational function is $0$ when the numerator is $0$, the $x$-intercept is found by setting the polynomial in the numerator equal to 0 and solving for the variable. For instance, consider $g(x)=\frac{(2 x+1)(x-4)}{(x+5)}$. To find the $y$-intercept, set $x=0$ and evaluate the function; the $y$ intercept occurs for $y=-\frac{4}{5}$ and has coordinates $\left(0,-\frac{4}{5}\right)$. To find the $x$-intercepts, find those values of $x$ for which the numerator is 0, namely $x=-\frac{1}{2}$ and $x=4$, corresponding to coordinates $\left(-\frac{1}{2}, 0\right)$ and $(4,0)$.

A removable discontinuity for a rational function occurs for an input value that is not in the domain of the function but for which the function does not approach $\pm \infty$ as $x$ approaches this input value from the left or the right. Such a discontinuity occurs when there is a common factor in the numerator and denominator of the polynomial function. Although the input value is not in the domain, the value of the rational function can be determined once the common factor is removed. A removable discontinuity is represented by a hole in the graph, such as the hole shown in the graph in objective $6 c$. The function represented by that graph is $f(x)=$ $\frac{(x-15)(x+3)}{(x-6)(x+3)}=\frac{x-15}{x-6}$. Notice that $x=-3$ is not in the domain of the function; however, when $x=-3$, the function can be evaluated as $f(x)=\frac{-3-15}{-3-6}=2$. So, there is a hole in the graph at $(-3,2)$; the discontinuity that would have occurred at $x=-3$ and that could have resulted in a vertical asymptote has been removed.

Review material in End Behavior and Local Behavior of Rational Functions.

6e. construct a graph or an equation of a rational function using its properties

• Construct a graph of a rational function using intercepts, asymptotes, and end behavior.
• Given the graph of a rational function, use the intercepts, asymptotes, and end behavior to construct a possible equation for the function.

The properties of a rational function can be used to sketch its graph. Start by determining the $x$ - and $y$ intercepts using the techniques described in objective $6 d$; for any $x$-intercepts, consider the multiplicity of the zero to determine whether the function bounces off the $x$-axis at the intercept or crosses the axis. Determine values that make the denominator zero and use techniques in objectives $6 b$ and $6 d$ to determine whether these values lead to a vertical asymptote or a removable discontinuity. Compare the degree of the numerator to the degree of the denominator to determine whether there is a horizontal or slant asymptote to know the behavior of the function as $x \rightarrow \pm \infty$.

For instance, consider the function $f(x)=\frac{3\left(x^{2}-16\right)}{(x+4)(x-6)}$. To construct the graph, first factor the numerator and remove the common factor; then analyze the resulting $f(x)=\frac{3(x-4)}{x-6}$. The $y$-intercept is at $(0,2)$. The numerator is zero for $3(x-4)=0$ or $x=4$, so the $x$-intercept is at $(4,0)$; the multiplicity of this zero is $1$ because the exponent on the factor $(x-4)$ is $1$, meaning the graph crosses the axis at the intercept. The denominator of the original function is zero for $x=-4$ and $x=6$. Because the factor $(x+4)$ was common to the numerator, there is a removable discontinuity at $\left(-4, \frac{12}{5}\right)$; as $x \rightarrow 6^{+}, f(x) \rightarrow \infty$ and as $x \rightarrow 6^{-}, f(x) \rightarrow-\infty$. The numerator and denominator have the same degree, so there is a horizontal asymptote at the ratio of the leading coefficients, namely $y=3$; thus, as $x \rightarrow \pm \infty, f(x) \rightarrow 3$. The graph of the function is shown below.

If given a graph, work backward to generate a potential equation. For instance, $x$-intercepts lead to factors for the numerator. Vertical asymptotes lead to factors in the denominator; the behavior of the function near the asymptote gives insight into whether the exponent on this factor is even or odd. If the $x$-axis, namely $y=0$, is a horizontal asymptote, the degree of the numerator is less than the degree of the denominator; if there is a horizontal asymptote at $y=b$, the numerator and denominator have the same degree and $b$ is the ratio of the leading coefficients; if there is a slant asymptote at $y=m x$, the degree of the numerator is one greater than the degree of the denominator. For any potential equation, check that the appropriate $y$-intercept is obtained when the function is evaluated for $x=0$ and adjust constants as needed. The coefficient for a vertical stretch or compression of the graph can then be finalized by identifying the coordinates for a specific point on the graph and substituting into the equation.

For instance, consider the graph of $g(x)$ below.

There is a $x$-intercept at $(-1,0)$, so $(x+1)$ is a factor in the numerator. There is a vertical asymptote at $x=0$ and the function approaches $\infty$ on one side of the asymptote and $-\infty$ on the other side, so $(x-0)$ is a factor in the denominator. There is also a vertical asymptote at $x=4$ and the function approaches $\infty$ as $x$ approaches $4$  from either side of the asymptote; so, $(x-4)$ is not only a factor in the denominator but must be a factor to an even power. The line $y=0$ is a horizontal asymptote, so the degree of the numerator is less than the degree of the denominator. As a start, write $g(x)=\frac{x+1}{x(x-4)^{2}}$. Checking the function value when $x=6$ suggests that there is no vertical stretch or compression.

Review material in End Behavior and Local Behavior of Rational Functions.

Unit 6 Vocabulary

This vocabulary list includes terms you will need to know to successfully complete the final exam.

• end behavior
• horizontal asymptote
• rational function
• reciprocal function
• removable discontinuity
• slant asymptote
• squared reciprocal function
• $x$-intercept of a rational function
• vertical asymptote
• $y$-intercept of a rational function

7a. evaluate exponential and logarithmic functions

• Compare the rate of change of a linear function to an exponential function.
• Define an exponential function. What are possible values for the base of an exponential function?
• Explain how to evaluate an exponential function with base $e$.
• Define a logarithmic function. 
• Compare common logarithms to natural logarithms.
• Compare evaluating exponential functions to evaluating logarithmic functions. 

An exponential function is a function of the form $f(x)=a b^{x}$, where $a$ and $b$ are positive real numbers and $b \neq 1$; the coefficient a represents the starting value and the base $b$ represents growth factor. Notice that in an exponential function, the variable is the exponent; this contrasts with a power function in which the variable is the base. Exponential functions change by a constant percentage for equal increments of the input value; in contrast, linear functions change by a constant amount for equal increments of the input value. This is equivalent to saying that there is a multiplicative rate of change for an exponential function in contrast to an additive rate of change for a linear function. When the base $b > 1$, the exponential function represents growth or increase over time; if $0 < b < 1$, the exponential function represents decay or decrease over time.

To evaluate an exponential function, use the normal rules of algebra to simplify the result. For instance, given the function $f$ with $f(x)=3 \cdot 4^{x}$. Then $f(2)=3 \cdot 4^{2}=3 \cdot 16=48$ and $f(-2)=3 \cdot 4^{-2}=3 \cdot \frac{1}{16}=\frac{3}{16}$. Similarly if function $g$ is defined as $g(x)=\left(\frac{1}{4}\right)^{x}$, then $g(2)=\left(\frac{1}{4}\right)^{2}=\frac{1}{16}$ and $g(-2)=\left(\frac{1}{4}\right)^{-2}=16$.

The base $e$ is a special case as a base in exponential functions with many applications in the real world. The base e represents the value of $\left(1+\frac{1}{n}\right)^{n}$ as $n \rightarrow \infty; e \approx 2.71828$. Most scientific and graphing calculators have a special key for $e$ that can be used to evaluate exponential functions with base $e$.

A logarithmic function is a function of the form $y=\log _{b} x$, where $x > 0, b > 0$, and $b \neq 1$; this logarithmic function is read "the logarithm with base $b$ of $x^{\prime \prime}$ and is equivalent to $b^{y}=x$. Typically, a logarithmic function is evaluated by rewriting it in exponential form and then solving based on what is known about exponential relationships. For instance, to evaluate $y=\log _{3} 81$, rewrite the logarithm as $3^{y}=81$; so $y=4$ meaning that $\log _{3} 81=4$. Likewise, to evaluate $y=\log _{5} \frac{1}{125}$ rewrite in exponential form as $5^{y}=\frac{1}{125}$; because $y=-3$, $\log _{5} \frac{1}{125}=-3$.

There are two special logarithms that appear in many applications. Logarithms with base 10 are called common logarithms; rather than writing $y=\log _{10} x$, common logarithms are generally written as $y=\log _{\text {in }} x$. Natural logarithms are logarithms with base $e$, and these logarithms appear regularly in calculus and many scientific applications. Natural logarithms are typically written as $y=\ln x$ rather than as $y=\log _{e} x$. Most scientific and graphing calculators have special keys to evaluate common logarithms (log) or natural logarithms $(\ln)$.

Review this material in Properties of Exponential Functions and Characterisitics of Graphs of Logarithmic Functions.

7b. define the equation for exponential and logarithmic functions given a graph or data points

• Describe how the equation of an exponential function can be written by knowing the initial value and growth factor.
• Given a set of data points for an exponential or logarithmic function, describe how to obtain an equation for the function through those points. 
• Given the graph of an exponential or logarithmic function, explain how to obtain an equation for the graph.

Determining the equation of an exponential function depends on the information that is given. In some contexts, the initial value is known, corresponding to $a$ in the definition of an exponential function, and the growth factor is also known, corresponding to the base $b$ in the exponential function. Consider the following situation.

• The population of a rural area is 1200 people. If the population is expected to grow by $2 \%$ a year, what equation describes the population after $t$ years?

In this context, the initial value is 1200, so $a=1200$. A growth of $2 \%$ per year means that the population is the base population $+2 \%$ of the base population, or $1.02$ times the base population; this means $b=1.02$. Then an equation for the population after $t /$ years is $f(t)=1200 \cdot 1.02^{t}$. After 3 years, the equation predicts a population of $1200 \cdot 1.02^{3} \approx 1273$ people.

If two data points are given, then these can be substituted into the form of an exponential function to find values for $a$ and $b$. If one of those points corresponds to the $y$-intercept, the work is simplified because the $y$ intercept corresponds to $a$. For instance, consider an exponential function containing the ordered pairs $(0,7)$ and $(2,63)$. The first ordered pair corresponds to the $y$-intercept, so $a=7$. Then substitute the second ordered pair in the equation $f(x)=7 b^{x}$ to get $63=7 b^{2}$; this leads to $b^{2}=9$, or $b=3$ because the base must be positive. Thus, the desired equation is $f(x)=7 \cdot 3^{x}$.

If neither point corresponds to the $y$-intercept, this process of substitution must be repeated more than once. Suppose an exponential function contains the ordered pairs $(1,2)$ and $\left(5, \frac{1}{8}\right)$. Substituting these coordinates into the form of an exponential function yields $2=a b^{1}$ and $\frac{1}{8}=a b^{5}$. The first equation leads to $\frac{2}{a}=b$, which can be substituted into the second equation to obtain $\frac{1}{8}=a \cdot\left(\frac{2}{a}\right)^{5}$ or $a=4$; this means that $b=\frac{1}{2}$. Then the final equation is $g(x)=4 \cdot\left(\frac{1}{2}\right)^{x}$.

If a graph of an exponential function is given rather than a context or a set of ordered pairs, the techniques described above are still used. Identify two points on the graph, preferably with one of the points being the $y$ intercept. Then substitute to find $a$ and $b$ for the equation $f(x)=a b^{x}$.

If the points are on a logarithmic function, similar techniques are used. A logarithmic function does not have a $y$-intercept, unless there has been some type of transformation. However, the function does have a $x$-intercept at $(1,0)$; if the $x$-intercept is at some other point $(h, 0)$, this is a clue that the function has been shifted horizontally $h$ units from $(1,0)$. Because $y=\log _{b} x$ is equivalent to $b^{y}=x$, this relationship is used, together with knowledge of powers of common positive integers, to find a potential equation.

Consider a logarithmic function which contains the ordered pairs $(36,2)$ and $(216,3)$. If $y=\log _{b} x$, then these ordered pairs can be substituted into the equivalent exponential form to yield $36=b^{2}$ and $216=b^{3}$. These two equations lead to $b=6$, meaning the logarithmic equation is $y=\log _{6} x$.

Now consider the logarithmic function whose graph is shown below. Notice that the $x$-intercept occurs at $(-1,0)$, meaning the $x$-intercept of the basic logarithmic function $(1,0)$ has been translated 2 units to the left. This provides a starting point of $y=\log _{b}(x+2)$. The point $(8,1)$ also appears to be on the graph. So, substitute this point into the exponential equivalent of the logarithmic equation to obtain $b^{1}=8+2$, so $b=10$. The logarithmic function is the common logarithm function $f(x)=\log (x+2)$.

Review this material in Properties of Exponential Functions and Characterisitics of Graphs of Logarithmic Functions.

7c. identify properties of exponential and logarithmic equations, including asymptotes, long run, and local behavior

• Identify the properties of exponential functions, including intercepts, asymptotes, and end behavior.
• Identify the properties of logarithmic functions, including intercepts, asymptotes, and end behavior.
• Explain how the properties of the logarithmic function are related to those of the corresponding exponential function.

The two basic exponential functions are graphed below: $f(x)=b^{x}$ for $b > 1$ and $g(x)=b^{x}$ for $0 < b < 1.$ These two functions have similar properties:

• Both functions are one-to-one functions.
• For both functions, the $y$-intercept is at $(0, \, \mid 1)$.
• Neither function has a $x$-intercept.
• Both functions contain the point $(1, \, b)$.
• For both functions, there is a horizontal asymptote at $y=0$. For function $f$ with $b > 1, f(x) \rightarrow 0$ as $x \rightarrow$ $-\infty$; for function $g$ with $0 < b < 1, g(x) \rightarrow 0$ as $x \rightarrow \infty$.
• For both functions, the domain is $(-\infty, \infty)$; the range is $(0, \, \infty)$.
• For function $f$ with $b > 1, f(x) \rightarrow \infty$ as $x \rightarrow \infty$; for function $g$ with $0 < b < 1, g(x) \rightarrow 0$ as $x \rightarrow \infty$.
• For $b > 1$, the function is always increasing; when $0 < b < 1$, the function is always decreasing.

The logarithmic function $y=f(x)=\log _{b} x$ is equivalent to $b^{y}=x$, so the two functions are inverses of each other. This means the domain of the logarithmic function is the range of the exponential function; likewise, the range of the logarithmic function is the domain of the exponential function. Because the two functions are inverses of each, the graph of the logarithmic function is the reflection of the graph of the exponential function across the line $y=x$ as reviewed in objective $3 \mathrm{~g}$.

Two basic logarithmic functions $p$ and $q$ are graphed below. The logarithmic function $y=p(x)=\log _{b} x$ for $b > 1$ is the inverse of the exponential function $y=f(x)$ graphed above. Likewise, the logarithmic function $y=$ $q(x)=\log _{b} x$ for $0 < b < 1$ is the inverse of the exponential function $y=g(x)$ graphed above. Both graphs have similar properties which correspond to the properties of the related exponential function.

• Both functions are one-to-one functions.
• Neither function has a $y$-intercept.
• Both functions have a $x$-intercept at $(1,0)$.
• Both functions contain the point $(b, 1)$.
• For both functions, there is a vertical asymptote at $x=0$. For function $p$ with $b > 1, p(x) \rightarrow-\infty$ as $x \rightarrow$ $0^{+}$; for function $q$ with $0 < b < 1, q(x) \rightarrow \infty$ as $x \rightarrow 0^{+}$.
• For both functions, the domain is $(0,  \,  \infty)$; the range is $(-\infty,  \,  \infty)$.
• For function $p,  \,  p(x) \rightarrow \infty$ as $x \rightarrow \infty$; for function $q,  \,  q(x) \rightarrow-\infty$ as $x \rightarrow \infty$.
• When $b > 1$, the function is always increasing; when $0 < b < 1$, the function is always decreasing.

Review this material in Characteristics of Graphs of Exponential Functions and Characterisitics of Graphs of Logarithmic Functions.

7d. graph exponential and logarithmic equations using transformations

• Describe how transformations such as vertical stretches or compressions, horizontal or vertical shifts, or vertical reflections can be used to graph exponential functions. 
• Compare the use of transformations when graphing exponential functions to their use when graphing logarithmic functions.

Graphing exponential and logarithmic functions using transformations involves the same techniques used in earlier chapters, particularly in objective 3c. Those same transformations apply to the concepts associated with the exponential function, such as the $y$-intercept, horizontal asymptote, and range. Consider the function $f(x)=a b^{x+c}+d$. When compared to the parent function $f(x)=b^{x}$ for $b > 1$, the transformed function has been

• Shifted horizontally by $c$ units (to the left if $c > 0$, to the right if $c < 0$).
• Stretched vertically by $a$ if $|a| > 1$ or compressed vertically if $0 < |a| < 1$.
• Shifted vertically by $d$ units (up if $d > 0$, down if $d < 0$).
• Reflected vertically across the $x$-axis if $a < 0$.

For instance, consider the function $y=f(x)=2 \cdot 4^{x+1}-3$. To graph, first graph the parent function $g(x)=4^{x}$ (the red graph below); two points on this graph are $(0,1)$ and $(1,4)$. Apply the transformations thinking about the order of operations. Shift the graph to the left 1 unit because $c=1$ (the blue graph); the points $(0,1)$ and $(1,  \,  4)$ now correspond to $(-1,1)$ and $(0,4)$. Apply a vertical stretch by a factor of 2 because $a=2$ (the green graph); points $(-1,1)$ and $(0,4)$ correspond to $(-1, \, 2)$ and $(0.8)$. Finally, shift the graph down 3 units because $d=-3$ (the black graph); the points $(-1,2)$ and $(0,8)$ correspond to points $(-1, \, -1)$ and $(0,5)$ on the final graph. The domain is still $(-\infty, \infty)$ but the range is now $(-3, \,  \infty)$; there is a horizontal asymptote at $y=-3$.

Similar analysis is applied to graph a translated logarithmic function. Consider the function $f(x)=a l o g_{b}(x+$ $c)+d$. When compared to the parent function $f(x)=\log _{b} x$ for $b > 1$, the transformed function has been

• Shifted vertically by $d$ units (up if $d > 0$, down if $d < 0$).
• Shifted horizontally by $c$ units (to the left if $c > 0$, to the right if $c < 0$).
• Stretched vertically by $a$ if $|a| > 1$ or compressed vertically if $0 < |a| < 1$.
• Reflected horizontally about the $x$-axis if $a < 0$.

The $x$-intercept, domain, and equation for a vertical asymptote will be transformed in similar ways.

Consider the function $y=f(x)=-\log (x-4)+3$. First graph the parent function $g(x)=\log x($ the red graph below); the points $(1, \, 0)$ and $(10, \, 1)$ are on this graph. Apply the transformations thinking about the order of operations. Reflect the graph vertically over the $x$-axis (the blue graph); the points $(1,0)$ and $(10,1)$ correspond to $(1, \, 0)$ and $(10, \, -1)$. Because $c=-4$, the graph now shifts horizontally 4 units to the right (the green graph); the points $(1, \, 0)$ and $(10, \, -1)$ correspond to $(5, \, 0)$ and $(14, \, -1)$. Finally, because $d=3$, shift the graph vertically up 3 units (the black graph); the points $(5, \, 0)$ and $(14, \, -1)$ correspond to $(5, \, 3)$ and $(14, \, 2)$ on the final graph. The vertical asymptote for function $g$ is translated 4 units to the right for function $f$ and is at $x=4$. The domain is now $(4,  \, \infty)$ and the range is still $(-\infty, \infty)$.

Review this material in Characteristics of Graphs of Exponential Functions and Characterisitics of Graphs of Logarithmic Functions.

7e. identify the domain and range of exponential and logarithmic functions

• Identify the domain and range of the basic exponential and logarithmic functions.
• Explain how the domain and range of the basic exponential and logarithmic functions change when transformations are applied to the basic functions.

As described in objective 7c, the basic exponential function $f(x)=b^{x}$ for $b > 0, b \neq 1$ has domain $(-\infty, \infty)$ and range $(0, \infty)$. This means that the basic logarithmic function $g(x)=\log _{b} x$ for $x > 0, b > 0$, and $b \neq 1$ has domain $(0, \infty)$ and range $(-\infty, \infty)$. A transformation of the basic exponential function will not affect the domain, but the range will adjust based on any vertical shifts. A transformation of the basic logarithmic function will not affect the range, but will affect the domain because the input of a logarithmic function must be positive.

For instance, consider the exponential function $y=5 \cdot 7^{x+4}-6$. The domain is still $(-\infty, \infty)$; however, the range will be $(-6, \infty)$. Observe that $5 \cdot 7^{x+4} > 0$ for all values of $x$, meaning that $y > -6$.

Given the logarithmic function $y=6 \log _{5}(x-8)+4$. The input for the logarithm must be positive, that is, $x-8 > 0$; so, the domain is $(8, \infty)$. The range continues to be $(-\infty, \infty)$.

Review this material in Characteristics of Graphs of Exponential Functions and Characterisitics of Graphs of Logarithmic Functions.

7f. summarize the inverse relationship between exponential and logarithmic functions

• Describe the inverse relationship between exponential and logarithmic functions.
• What properties are explained by the fact that exponential and logarithmic functions are inverses of each other?

As indicated in objectives 7 a and 7c, exponential functions and logarithmic functions are inverses of each other. This relationship is used to evaluate and rewrite logarithmic functions in their equivalent exponential form. That is, $y=\log _{b} x$ for $x > 0, b > 0$, and $b \neq 1$ is equivalent to $b^{y}=x$ for $b > 0$ and $b \neq 1$. Given that both functions are inverses of each other, they are both one-to-one functions. The domain and range of the exponential function are the range and domain, respectively, of the logarithmic function.

Review the material in Properties of Logarithms.

Unit 7 Vocabulary

This vocabulary list includes terms you will need to know to successfully complete the final exam.

• base of an exponential function
• common logarithm
• $e$
• exponential function
• logarithmic function
• natural logarithm

8a. apply properties of logarithms to rewrite and simplify logarithmic expressions, including the product, quotient, and power rules and the change-of-base formula

• Explain how to use logarithmic properties to rewrite the logarithm of a sum, the logarithm of a product, and the logarithm of a power.
• Describe how the power rule can simplify $\log _{b} 1, \log _{b} b$, and $\log _{b} b^{n}$.
• Rewrite a logarithm with a base other than $10$ or $e$ using the change of base formula. How is this formula helpful in using a calculator to evaluate a logarithm?
• Given an expression with multiple terms containing logarithms, use the logarithmic rules to rewrite the expression as a single logarithm.

Before the widespread availability of scientific and graphing calculators, complicated calculations were often completed using tables of logarithms by converting multiplications to additions, divisions to subtractions, and powers to basic multiplications by using the following properties of logarithms for bases $b > 0, b \neq 1$.

• Product rule for logarithms: $\log _{b}(M \cdot N)=\log _{b} M+\log _{b} N$
• Quotient rule for logarithms: $\log _{b} \frac{M}{N}=\log _{b} M-\log _{b} N$
• Power rule for logarithms: $\log _{b} M^{N}=N \cdot \log _{b} M$

Consider the expression $\log \frac{x y^{4}}{z}$ for $x, y$, and $z$ positive real numbers. Using these three logarithmic rules, this expression can be rewritten as $\log \frac{x y^{4}}{z}=\log x+4 \cdot \log y-\log z$. These rules can also be applied in reverse to rewrite an expression such as $\log 5-\log 7+3 \log x$ as $\log \frac{5 x^{3}}{7}$.

These rules can be used, together with the related exponential equation, to evaluate three special cases.

• $\log _{b} 1=0$
• $\log _{b} b=1$
• $\log _{b} b^{n}=n$

Scientific and graphing calculators have special keys for common logarithms (base 10) and natural logarithms (base e). Sometimes, however, there is a need to evaluate a logarithm with a different base. The change-ofbase formula can be used to rewrite a logarithm from one base to a quotient of logarithms with another base.

• Change-of-base formula for logarithms: For $M > 0, n > 0, b > 0, n \neq 1, b \neq 1, \log _{b} M=\frac{\log _{n} M}{\log _{n} b}$.

So, if the base $b$ is not 10 or $e$, then choose $n$ to be 10 or $e$ so that the logarithm can be evaluated using a calculator. For instance, $\log _{4} 21=\frac{\log 21}{\log 4} \approx \frac{1.322}{0.602} \approx 2.196$. This result makes sense because $y=\log _{4} 21$ is equivalent to $4^{y}=21$, so $2 < y < 3$. Also, evaluating $4^{2.196} \approx 20.995$.

Review material in Using the Definition of a Logarithm to Solve Logarithmic Equations.

8b. solve exponential equations using like bases and logarithms

• Explain how exponential equations can be solved if the exponential expressions on both sides of the equation have the same base. How can this technique be used even if the exponential expressions do not have the same base?
• How can logarithms be used to solve exponential equations? What are extraneous solutions?

Some types of exponential equations can be solved using known algebraic techniques. One such instance occurs when two algebraic expressions have the same base; in that case, the exponents are set equal to each other.

• For $b > 0$ and $b \neq 1, b^{M}=b^{N}$ if and only if $M=N$.

Sometimes the bases are already written in a form that indicates they are equal. For instance, if $3^{2 x-5}=$ $3^{x+7}$, then $2 x-5=x+7$, so $x=12$. However, in other cases, the bases are not the same but can be rewritten so they are the same. Consider the equation $3^{x-4}=81^{x+1}$. Although the bases are not currently the same, 81 is a power of 3, so rewrite $81^{x+1}$ as $\left(3^{4}\right)^{x+1}=3^{4 x+4}$. Now, $3^{x-4}=81^{x+1}$ can be rewritten as $3^{x-4}=3^{4 x+4}$ so $x-4=4 x+4$ or $x=-\frac{8}{3}$.

At times, however, there is no way to rewrite expressions so the exponential expressions have the same base. In that case, an appropriate technique is to take the logarithm of both sides of the equation, applying the rules from objective $8 \mathrm{a}$ as needed. Because calculators have dedicated keys for common logarithms and natural logarithms, usually base 10 or base $e$ should be used as the base of the logarithm to be applied to each side of the equation. To solve $6^{x+2}=7^{3 x}$, take the common logarithm of both sides and then apply the rules of logarithms from objective 8a.

$\log \left(6^{x+2}\right)=\log \left(7^{3 x}\right)$

$(x+2) \log 6=(3 x) \log 7$

$2 \log 6=x(3 \log 7-\log 6)$

$x=\frac{2 \log 6}{3 \log 7-\log 6} \approx \frac{1.5563}{2.5353-0.7782} \approx 0.8857$

Whenever the logarithm of both sides is taken, it is essential to check the solution to ensure that none of the potential solutions are extraneous, that is, a solution obtained during the algebraic process that does not solve the original equation. In this case, $6^{0.8857+2}=7^{3 \cdot 0.8857}$, or $175.9993 \approx 175.9989$, close enough given potential rounding errors.

Review material in Solving Exponential Equations Using Logarithms.

8c. solve logarithmic equations using the definition of a logarithm and the one-to-one property of logarithms

• Explain how to use the definition of a logarithm, together with the rules of logarithms, to solve a logarithmic equation.
• Explain how the one-to-one property of logarithms can be used to solve logarithmic equations. Compare the use of this property to the use of the same base property when solving exponential equations.

Two techniques are often helpful in solving logarithmic equations. The first uses the fact that the logarithmic equation $y=\log _{b} x$ is equivalent to $b^{y}=x$. So, start by using the rules of logarithms to rewrite multiple logarithmic terms as a single logarithm and then rewrite the result in exponential form and solve.

Consider the equation $\log (4 x+4)+\log (x-14)=3$. Because the left side of the equation is the sum of two logarithms, it can be rewritten as the logarithm of a product: $\log [(4 x+4)(x-14)]$. Then rewrite the logarithmic equation into exponential form:

$\log [(4 x+4)(x-14)]=3$ is equivalent to $(4 x+4)(x-14)=10^{3}$

So, $4 x^{2}-52 x-56=1000$, or $x^{2}-13 x-264=0$. Factoring or using the quadratic formula leads to $x=24$ or $x=-11$, but $x=-11$ is an extraneous solution because it is not possible to take the logarithm of a negative number. Checking the solution $x=24$ results in $\log 100+\log 10=2+1=3$, so it checks.

The technique used to solve an exponential equation if two exponential expressions have the same base has an analogy that can be used to solve a logarithmic equation if two logarithmic expressions have the same base.

• For $b > 0, b \neq 1$ and $M, N > 0, \log _{b} M=\log _{b} N$ if and only if $M=N$.

Consider the equation $\log (6 x+8)=\log (2 x+28)$. Both logarithmic expressions have base 10, so $6 x+8=2 x+28$, or $x=5$. Checking this value in the equation leads to $\log 38=\log 38$

Review material in Solving Exponential Equations Using Logarithms.

8d. solve applications of exponential and logarithmic equations, including half-life and Newton's law of cooling

• Write the formula for compound interest and describe the meanings of the component parts of the formula. 
• Compare the compound interest formula with the formula for continuous growth or decay.
• Identify the exponential equation used for half-life and identify the meaning of all parts of the equation.
• Identify an equation for doubling time.
• Be able to write and use the formula for Newton's law of cooling.

Exponential and logarithmic functions appear in many real-world situations. Solving real-world applications uses the same techniques as described in objectives 8b and 8c. But some applications appear so commonly that they deserve special attention.

One such application is compound interest, determined by the equation $A(t)=P\left(1+\frac{r}{n}\right)^{n t}$, with $A(t)$ representing the amount in the account after $t$ years, $P$ representing the initial principal or amount, $r$ representing the decimal for the interest rate, and $n$ representing the compounding periods per year (i.e., 2 for semiannually, 4 for quarterly, 12 for monthly). Suppose $\$ 100$ is deposited into an account at an interest rate of $3 \%$ compounded twice a year. After 5 years, the amount in the account would be $100\left(1+\frac{0.03}{2}\right)^{2 \cdot 5} \approx \$ 107.73$.

Many situations modify the compound interest formula to be continuous growth or decay by using $e$. In this case, the equation becomes $A(t)=A_{0} e^{k t}$, where $A(t)$ represents the amount after time $t, A_{0}$ is the initial value, and $k$ is the growth factor per unit time. If $k > 0$, the situation models continuous growth; if $k < 0$, the situation models continuous decay. For example, assume a population of 1000 bacteria decreases by $3 \%$ every 5 hours. To find the number of bacteria that exist after 20 hours, first recognize that 20 hours is equivalent to 4 of the 5 hour time periods over which the $3 \%$ applies. Then the number of bacteria left is $1000 e^{-0.03 \cdot 4} \approx 887$ bacteria.

An important scientific application using exponential equations is that of half-life, or the time period in which half of a radioactive substance decays. An equation for half-life is $A(t)=A_{0}\left(\frac{1}{2}\right)^{t / T}$, where $A_{0}$ represents the initial amount, $t$ represents the time period of interest, $T$ represents the half-life, and $A(t)$ represents the amount remaining after time $t$. Notice that the exponent $\frac{t}{T}$ represents how many half-lives occur in the $t$ time period. For instance, the half-life of gallium-67 is about 80 hours. Then after 240 hours, an initial quantity of 10 micrograms would decay so that only $10\left(\frac{1}{2}\right)^{240 / 80} \approx 1.25$ micrograms.

In contrast to half-life, some applications are interested in doubling time, that is, the amount of time it takes for a quantity to double. The equation for doubling time $t$ is $t=\frac{\ln 2}{k}$, where $k$ is the growth factor per unit time. So, if a quantity has a growth factor of $6 \%$ per year, then it will double in $\frac{\ln 2}{0.06} \approx 11.55$ years.

Newton's law of cooling provides insight into how quickly a hot object will cool to approximate the surrounding temperature:

• $T(t)=A e^{k t}+T_{s}$, where $T(t)$ is the temperature after time $t, A$ is the difference between the initial temperature of an object and the surrounding temperature, $k$ is the cooling rate, and $T_{s}$ is the temperature of the surrounding air.

Such an equation is solved using the same techniques for solving exponential equations as previously reviewed.

Suppose a boiling pot of soup at a temperature of $100^{\circ}$ is placed on a table in a room whose temperature is $72^{\circ}$. If the temperature of the soup is $90^{\circ}$ after 5 minutes, what equation can be used to find the temperature for any time $t$? In this case, $A=100-72=28$ and $T_{s}=72$. Because $T(5)=90$, the equation becomes $90=$ $28 e^{5 k}+72$, so $e^{5 k}=\frac{90-72}{28}$ leading to $k \approx-0.088$. So, the equation becomes $T(t)=28 e^{-0.088 t}+72$.

Review material in Models of Exponential Growth and Decay, Solving Applied Problems Using Exponential and Logarithmic Equations, and Use Data to Build a Logarithmic Model.

8e. differentiate between exponential growth and decay

• Contrast exponential growth with exponential decay.

The equation $y=A_{0} e^{k t}$, where $A_{0}$ is the initial value, $k$ is the growth factor, and $t$ is the time, represents an exponential model that characterizes growth if $k > 0$ or decay if $k < 0$. An exponential growth function is always increasing; in contrast, an exponential decay function is always decreasing. For both exponential growth and decay, the domain is $(-\infty, \infty)$ and the range is $(0, \infty)$. Both functions have a $y$-intercept at $\left(0, A_{0}\right)$ and a horizontal asymptote at $y=0$. Neither function has a $x$-intercept. The graphs of an exponential growth and exponential decay are similar to the graphs in objective 7c.

Examples of exponential growth are compound interest and population growth that is a percentage over time. Examples of exponential decay are half-life and Newton's law of cooling.

$f(x)=\frac{c}{1+a e^{-b x}}$

In this equation, $c$ is the carrying capacity, $\frac{c}{1+a}$ is the initial value, and $b$ is determined by the growth factor; because $\frac{c}{1+a}$ is the value of $f(0)$, knowing the value of the intercept enables the value of the coefficient a to be determined.

Consider the following problem.

• A middle school has a student population of 1500 students. Four students hear a rumor; 9 hours later, 32 students have heard the rumor. What is an equation for the number of students who will have heard the rumor after $t$ hours?

It makes sense to model this problem with a logistic growth model because the maximum number of students who can have heard the rumor is 1500; this means $c=1500$. The two points on the curve $(0,4)$ and $(9,32)$ can be used to find an equation. The point $(0,4)$ leads to $4=\frac{1500}{1+a}$ or $a=374$. Then, use the point $(9,32)$ to obtain $32=\frac{1500}{1+374 \cdot e^{-9 b}}$ which leads to $e^{-9 b}=\frac{1468}{11968}$ so $b=\frac{\ln (1468 / 11968)}{-9} \approx 0.233$. Thus, the equation is $f(x)=$ $\frac{1500}{1+374 e^{-0.233 x}}$.

Review material in Use Data to Build a Logarithmic Model.

8f. identify the key components of a logistic growth model

• Describe an equation for a logistic model. How does a logistic model differ from an exponential model?
• What is the carrying capacity of a logistic model?

An exponential growth model suggests that growth continues forever as indicated by the fact that an exponential function increases without bound as the input value increases without bound. But in the real world, growth can generally not continue in this manner forever but reaches a limiting value and then begins to slow. So, many real-world situations involving exponential growth would be better modeled by a logistic growth function, which is exponential initially but decreases as the input reaches an upper bound or limiting value known as the carrying capacity. The equation for a logistic growth model is

$f(x)=\frac{c}{1+a e^{-b x}}$

In this equation, $c$ is the carrying capacity, $\frac{c}{1+a}$ is the initial value, and $b$ is determined by the growth factor; because $\frac{c}{1+a}$ is the value of $f(0)$, knowing the value of the intercept enables the value of the coefficient $a$ to be determined.

Consider the following problem.

• A middle school has a student population of 1500 students. Four students hear a rumor; 9 hours later, 32 students have heard the rumor. What is an equation for the number of students who will have heard the rumor after $t$ hours?

It makes sense to model this problem with a logistic growth model because the maximum number of students who can have heard the rumor is 1500 ; this means $c=1500$. The two points on the curve $(0,4)$ and $(9,32)$ can be used to find an equation. The point $(0,4)$ leads to $4=\frac{1500}{1+a}$ or $a=374$. Then, use the point (9, 32) to obtain $32=\frac{1500}{1+374 \cdot e^{-9 b}}$ which leads to $e^{-9 b}=\frac{1468}{11968}$ so $b=\frac{\ln (1468 / 11968)}{-9} \approx 0.233$. Thus, the equation is $f(x)=$ $\frac{1500}{1+374 e^{-0.233 x}}$.

Review material in Use Data to Build a Logarithmic Model.

8g. build exponential and logarithmic regression models from data, and assess their fit

• Given a set of data that appears to be exponential, find an appropriate exponential regression equation. Use the equation to predict the output for other input values.
• Given a set of data that appears to be logarithmic, find an appropriate regression equation. Use the equation to predict the output for other input values.

In objective $4 \mathrm{~h}$, there were procedures for finding a linear regression to fit a set of data, often using a graphing utility. The correlation coefficient $r$ provides insights into how well the model fits the data; the closer $r$ is to 1 or 1, the better the model fits the data. Similar procedures are used to fit regression models where the data appear to grow or decay exponentially or to fit a logarithmic function. Most graphing calculators or comparable software utilities have options for an exponential regression (ExpReg) or natural logarithmic regression ($\mathrm{LnReg}$); once again, the value of $r$ provides insight into how well the equation fits the data, with better models having values of $r$ closer to 1 or $-1$. After a regression equation is found, it can be used to predict new output values by substituting input values into the equation.

Consider the data in the table below and the graphs of those data values. The data for function $f$ appear to be exponential while that for function $g$ appear to be logarithmic.

When the data are entered into a graphing calculator, the following regression equations are obtained.

• $f(x)=0.54 \cdot 2.95^{x}$ with $r=0.9999$
• $\mathrm{g}(x)=7.94-1.25 \cdot \ln x$ with $r=-0.996$

These equations can be used to predict the output for other input values. For instance, $f(10) \approx 26,953$ and $g(10) \approx 5.062.$

Review material in Models of Exponential Growth and Decay.

Unit 8 Vocabulary

This vocabulary list includes terms you will need to know to successfully complete the final exam.

• carrying capacity
• change-of-base formula for logarithms
• compound interest formula
• continuous growth or decay
• doubling time
• extraneous solution
• half-life
• logistic growth model
• Newton's law of cooling
• power rule for logarithms
• product rule for logarithms
• quotient rule for logarithms

9a. solve systems of linear equations in two and three variables using graphs, substitution, and addition

• Compare solving systems of linear equations in two variables via graphs, substitution, or addition.
• How might graphs, substitution, and addition be used to solve systems of linear equations in three variables?

There are often contexts that require the use of more than one equation to be solved at the same time. A system of linear equations consists of two or more linear equations with two or more variables that must be solved at the same time. The solution is the ordered pair or ordered triple whose values for the variables satisfy each equation at the same time. For a system to have the potential for a unique solution, there should be as many linear equations as there are variables.

Consider the following system of three linear equations in three variables.

$\begin{aligned} &2 x+5 y-7 z=-39 \\ &4 x+y+z=3 \\ &9 x-2 y-2 z=11 \end{aligned}$

The ordered triple $(1, \, -4, \, 3)$ is a solution to the system. When the coordinates are substituted into each of the three equations, each equation is a true statement.

There are three techniques that are often used to solve a linear system. Consider the system below of two linear equations in two variables.

$\begin{aligned} &5 x-4 y=-3 \\ &6 x+y=8 \end{aligned}$

The first technique is to graph both lines and then find the coordinates of the point of intersection. The coordinate grid below shows the graph of both lines. The two lines intersect at the point $(1, \, 2)$, which is the solution to the system.

Sometimes it is not easy to graph the lines or the coordinates of the intersection are not integers. In this situation, it helps to have algebraic techniques. One algebraic technique is the substitution method. Solve for one of the variables and then substitute for that variable in the other equation and solve. For the system above, the second equation can be easily solved for $y$ to obtain $y=8-6 x$. Now substitute this value for $y$ into the first equation.

$5 x-4(8-6 x)=-3 \text { leads to } 5 x-32+24 x=-3$

Then $29 x=29$, so $x=1$. Substitute this value for $x$ into $y=8-6 x$ to get $y=2$. So the solution is $(1, \, 2)$ as found on the graph.

If solving for one of the variables leads to fractional coefficients, it may be more efficient to use a second algebraic technique, specifically the addition method. The goal here is to modify one or more of the equations so that the coefficients of one of the variables are opposites of each other. Then, when the equations are added, this variable is eliminated and a solution to the other variable can be found. For the system earlier, it might help to adjust the second equation so that the coefficient for the variable $y$ is $-4$, the opposite of the coefficient of the variable $y$ for the first equation. Multiplying the second equation by 4 yields $24 x+4 y=32$. When this equation is added to the first equation, the result becomes $29 x=29$, so $x=1$. Substituting this value into either equation yields $y=2$, for the solution $(1,2)$ found by both of the other two methods.

For the addition method, it is possible to adjust the equations of the previous system so that the coefficients of the variable $x$ are opposites; given the coefficients of 5 and 6 from the two equations, the least common multiple is 30. So, multiply the first equation by 6 to get $30 x-24 y=-18$ and multiply the second equation by $-5$ to get $-30 x-5 y=-40$. Adding these two equations yields $-29 y=-58$, so $y=2$. Substituting into either equation then gives $x=1$. Although it does not matter which variable is eliminated when adding, sometimes eliminating one variable is more efficient than eliminating the other.

The same three techniques can be used to solve a system of three linear equations in three variables. However, these techniques are more complicated when three variables are involved. The solution to a system of three linear equations in three variables is an ordered triple $(x, y, z)$. So, trying to graph the solution means trying to graph in 3-space, which is hard to accomplish on paper. The substitution method is not very efficient. So, generally the addition method is used to solve a system of three linear equations in three variables. However, this method will need to be used more than once. Take two of the equations and use the addition method to eliminate one of the variables. Then take a different two equations and eliminate the same variable. The result is a system of two equations in two variables, and so the addition method is applied again.

Review material in Analyzing the Solution to a System in Two Variables and Solve Systems with Three Variables.

9b. categorize solutions of systems of linear equations

• Compare the solution in a consistent system and an inconsistent system. 
• Contrast the solution in an independent system and a dependent system. 

A consistent system has at least one solution; an inconsistent system has no solution. There are two types of consistent systems; an independent system has a single solution whereas a dependent system has an infinite number of solutions. For a system of two linear equations in two variables, these different types of systems can be described graphically. An independent system has a single solution so it consists of two lines with different slopes that intersect in a single point. A dependent system consists of two lines with the same slope and the same y-intercept; that is, the two lines are coincident so that every point on one line satisfies the second line. This means that the solution to a dependent system of two linear equations is actually a line. An inconsistent system consists of two lines with the same slope but different y-intercepts; the two lines are parallel and never intersect. When solving algebraically, if the process leads to a statement that is always true, then the system is a dependent system; if the process leads to a statement that is false, then the system is an inconsistent system. 

Similar analysis describes the nature of the solutions to three linear equations in three variables. An independent system consists of three planes that intersect in a single point with coordinates $(x, \, y, \, z)$. A dependent system of three linear equations consists of three coincident planes, three planes that intersect in a line, or two coincident planes that intersect the third plane in a line. A dependent system can be recognized in an algebraic process when a true statement is obtained. An inconsistent system has no solution and consists of three parallel planes, two parallel lines that intersect with a third plane, or three planes in which each plane intersects the other two planes but the intersections are not at the same point. An inconsistent system can be recognized in an algebraic process when a false statement is obtained.

Review material in Analyzing the Solution to a System in Two Variables and Solve Systems with Three Variables.

9c. solve systems of nonlinear equations using substitution and elimination

• Be able to use substitution or elimination to solve a nonlinear system of equations in two variables.

At times, a system of two equations in two variables may consist of an equation that is not a linear equation, that is, one of the variables has a power greater than 1 or the two variables are multiplied. Such a system is a system of nonlinear equations. However, the techniques of substitution and addition (elimination) used with linear systems can still be applied. As with linear systems, there are several possibilities for the solution.

One of the simplest nonlinear systems consists of a quadratic equation and a linear equation, that is, an equation whose graph is a parabola and an equation whose graph is a line. In this case, three possibilities exist: the line might intersect the parabola in two points; the line might be tangent to the parabola, intersecting it in just one point; or the line might not intersect the parabola at all so that there is no solution.

Consider the system below and its graph.

$\begin{aligned} &y=-2 x+10 \\ &y=(x-2)^{2}+3^{\square} \end{aligned}$

Substituting the value for $y$ from the line into the value of $y$ for the parabola leads to the following.

$-2 x+10=(x-2)^{2}+3$ or $x^{2}-2 x-3=0$ leading to $x=3$ or $x=-1$

So the solutions are the ordered pairs $(3,4)$ and $(-1,12)$. Both ordered pairs are visible on the graph and satisfy both equations.

A second common nonlinear system consists of a circle and a line. Again, there are three possibilities: two solutions if the line intersects the circle in two points; one solution if the line is tangent to the circle; and no solutions if the line does not intersect the circle.

When a line is one of the equations in the nonlinear system, substitution is typically the most efficient method to use. However, if the nonlinear system consists of two parabolas or two equations with both variables to a power greater than 1 (e.g., the equation of a circle), then it is likely more efficient to use addition to eliminate one of the variables, just as is done with a linear system.

Review material in Algebraic Methods for Solving Systems of Non-Linear Equations.

9d. graph single and systems of nonlinear inequalities

• Compare graphing an inequality to graphing an equation.
• Explain how to graph the solution to a system of inequalities.

The graph of an equation consists of those ordered pairs that make the equation true. In contrast, the graph of an inequality consists of a region. In fact, the graph of an equation $y=f(x)$ divides the coordinate grid into three regions: those ordered pairs that make the equation true, that is, ordered pairs for which $y=f(x)$; those ordered pairs that when substituted into $f(x)$ lead to $y < f(x)$; and those ordered pairs that when substituted into $f(x)$ lead to $y > f(x)$. If $y \leq f(x)$ or $y \geq f(x)$ then ordered pairs on the equation are also part of the solution; in this case, the equation is graphed with a solid curve. If the inequality is a strict inequality (that is, $<$ or $>$), then the equation is graphed with a dashed curve.

To graph an inequality, first graph the equation associated with the inequality, using either a solid or dashed curve as appropriate. Then choose a point not on the equation and substitute into the inequality. If the inequality is true, then the region containing that ordered pair is the solution; if the inequality is not true, then the region on the other side of the equation is the solution.

For instance, the graph shown is the graph of $y < x^{2}$. The ordered pair $(0,1)$ leads to $1 < 0$ which is false; so, the region containing $(0,1)$ is not part of the solution. In contrast, the ordered pair $(2,0)$ leads to $0 < 4$ which is true; so, the solution is the region containing $(2,0)$, denoted by shading the region (the purple region). The parabola itself is dotted to indicate that ordered pairs on the parabola are not part of the solution.

A system of inequalities is a system containing inequalities rather than equations. To solve such a system, graph the solution to each inequality separately, and then determine where the solutions overlap. The solution to the system is represented by the overlap.

Consider the system of inequalities graphed below.

$x^{2}+y^{2} < 4^{\square}$and $y < x-2$

The solution to the system is the dark green region that is both inside the circle and below the line.

Review material in Algebraic Methods for Solving Systems of Non-Linear Equations.

9e. construct an augmented matrix from a system of linear equations

• Identify an augmented matrix for a system of linear equations.
• Given an augmented matrix, write a system of linear equations.

An augmented matrix for a system is a matrix that contains the coefficients of the variables and the constants, with a vertical line used to separate the coefficients from the constants. For a system of two equations in two variables, the matrix is a $2 \times 3$ matrix, meaning there are 2 rows and 3 columns. For a system of three equations in three variables, the matrix is a $3 \times 4$ matrix, meaning there are 3 rows and 4 columns.

Consider systems $A$ and $B$ below.

A. $\begin{array}{ll} 5 x+7 y=-11 & \text { B. } 4 x-y+2 z=5 \\ 3 x+2 y=8 & 7 x+4 y-2 z=3 \\ & 7 x \end{array}$

The augmented matrices for these two systems are shown below.

A. $\left[\begin{array}{cc|c}5 & 7 & -11 \\ 1 & -2 & 8\end{array}\right]$                      B. $\left[\begin{array}{ccc|c}4 & -1 & 2 & 5 \\ 3 & 1 & 1 & 2 \\ 7 & 4 & -2 & 3\end{array}\right]$

It is also possible to write a system when given the augmented matrix. For instance, given the matrix below.

$\left[\begin{array}{rr|c} 9 & 4 & -6 \\ 5 & -1 & -13 \end{array}\right]$

The associated system consists of the two equations $9 x+4 y=-6$ and $5 x-y=-13$.

Once an augmented matrix is obtained, it can be used to solve a system by working only with the coefficients. Using the same procedures as used in the addition method, work on the rows of the matrix to get the matrix in a form which has $1 \mathrm{~s}$ on the diagonal and 0s below the diagonal. Then the values of the variables can be identified. For instance, for the augmented matrix above, apply the following steps.

• Multiply the first row by $\frac{1}{9}$ to obtain the matrix $\quad\left[\begin{array}{cc|c}1 & \frac{4}{9} & \frac{-6}{9} \\ 5 & -1 & -13\end{array}\right]$
• Now replace the second row with the result of multiplying the first row by $-5$ and adding it to the second row to obtain the matrix $\left[\begin{array}{ccc} 1 & \frac{4}{9} & -6 \\ 0 & \frac{-29}{9} & -87 \\ 0 & \frac{-87}{9} \end{array}\right]$
• Solve for the variables. $-\frac{29}{9} y=-\frac{87}{9}$, so $y=3$. Substituting this value of $y$ into the equation of the first row leads to $x+\frac{4}{9} \cdot 3=-\frac{6}{9}$, so $x=-2$. The solution to the system is the ordered pair $(-2, \, 3)$.

Review the material in Algebraic Methods for Solving Systems of Non-Linear Equations.

Unit 9 Vocabulary

This vocabulary list includes terms you will need to know to successfully complete the final exam.

• addition method for solving systems
• augmented matrix
• consistent system
• dependent system
• inconsistent system
• independent system
• solution to a system
• substitution method for solving systems
• system of inequalities
• system of linear equations
• system of nonlinear equations

10a. write the equation of ellipses, hyperbolas, and parabolas in standard form

• Compare the equations of ellipses, hyperbolas, and parabolas in standard form. 
• Compare how the equations of ellipses, hyperbolas, and parabolas are modified depending on their orientation.
• Identify the critical components of ellipses, hyperbolas, and parabolas, such as axes of symmetry, vertices, center, directrices, or asymptotes, as appropriate. How do these critical components help to write an equation for the ellipse, hyperbola, or parabola?

An ellipse is the set of all points $(x, y)$ in the plane such that the sum of their distances from two fixed points, called foci, is a constant. The standard forms of an ellipse depend on whether the center is at the origin or off the origin and on whether the ellipse is oriented horizontally or vertically, corresponding to whether the ellipse is longer in the direction of the $x$-axis or in the direction of the $y$-axis, respectively. The critical components of the ellipse help identify the appropriate standard form of the equation as indicated in the table below.

Although the table seems to have a lot of information, there are many parallels between the columns. The vertices (endpoints) are along the major axis or longer axis; the co-vertices (also endpoints) are along the minor axis or shorter axis. Both the major and minor axes are axes of symmetry for the ellipse; the center of the ellipse is at the intersection of the major axis and minor axis. The foci are always along the major axis and are the two points that play a part in the definition of an ellipse. The larger denominator is paired with the variable that indicates the major axis. Thus, if the larger denominator is connected with the variable $x$, then the major axis is along the $x$-axis or parallel to the $x$-axis so that the ellipse is longer horizontally. If the larger denominator is connected with the variable $y$, then the major axis is along the $y$-axis or parallel to the $y$-axis, and the ellipse is longer vertically. If $a=b$, then the ellipse becomes a circle.

Given an ellipse with center at $(4,-3)$, vertices at $(-9,-3)$ and $(17,-3)$, and foci at $(-1,-3)$ and $(9,-3)$. How can the equation be determined? Notice that the $y$-coordinates of the two vertices are the same. This means that the ellipse is longer horizontally and the major axis is parallel to the $x$-axis. The length of the major axis is $17-(-9)=26$, so $a=13$. The distance from the center to the foci is 5, so $c=5$; using $a=13$ and $c=5$ leads to $b=12$. Because the center is at $(4,-3)$, this means $h=4$ and $k=-3$. Using all this information leads to an equation for the ellipse of $\frac{(x-4)^{2}}{169}+\frac{(y+3)^{2}}{144}=1$.

Similar analyses can be applied to a hyperbola. A hyperbola consists of those ordered pairs $(x, y)$ in the plane such that the difference of their distances from two fixed points, the foci, is a positive constant.

The table below describes the critical components of a hyperbola. A careful look at this table shows many analogies to the comparable table for the ellipse.

As with an ellipse, a hyperbola has two axes of symmetry which are the transverse axis and the conjugate axis. The foci are along the transverse axis and the vertices are at the endpoints of the transverse axis; the vertices of the transverse axis are the vertices of the hyperbola. The conjugate axis is perpendicular to the transverse axis, and the co-vertices are the endpoints of the conjugate axis; the co-vertices are not points on the hyperbola. The transverse and conjugate axes intersect at the center of the hyperbola. A hyperbola also has two asymptotes that intersect at the center of the hyperbola. 

Given a hyperbola with center at $(0,0)$, vertices at $(0,-8)$ and $(0,8)$, and conjugate axis of length 6. To find an equation for this hyperbola, first note that the $x$-coordinates of the vertices are the same, so the hyperbola is oriented with the transverse axis along the $y$-axis. This means the hyperbola opens up and down. The coordinates of the vertices yield $a=8$. The length of the conjugate axis is $2 b=6$, so $b=3$. Putting all these pieces together gives an equation of $\frac{y^{2}}{64}-\frac{x^{2}}{9}=1$.

The final curve to be studied here is the parabola. A parabola is the set of all ordered pairs $(x, y)$ in the plane that are at the same distance from a fixed line, called the directrix, and a fixed point not on the line, called the focus. As noted in objectives $5 a$ and $5 b$, a parabola has an axis of symmetry that goes through the vertex; recall that the vertex is the minimum or maximum point of the parabola. Although the focus is not a point on the parabola, there is a segment containing the focus that is sometimes of importance. The latus rectum is a segment parallel to the directrix that contains the focus; the endpoints of the latus rectum are on the parabola.

The table below contains an analysis of the critical components of a parabola similar to those for an ellipse and a hyperbola. 

Given a parabola with vertex $(2, \, -5)$ and directrix $y=-8$. Because the equation of the directrix is a horizontal line, the parabola opens either up or down. The equation of the directrix leads to $-8=-5-p$, so $p=3$; because $p > 0$, the parabola opens up. The equation is then $(x-2)^{2}=12(y+5)$.

Review this material in Writing Equations of Ellipses, Writing Equations of Hyperbolas, and Parabolas Centered at the Origin.

10b. graph ellipses, hyperbolas, and parabolas centered on and off the origin

• Compare the graphs of ellipses, hyperbolas, and parabolas centered at the origin.
• How do the graphs of ellipses, hyperbolas, and parabolas transform when the center is not at the origin?

To graph an equation whose graph is an ellipse, hyperbola, or parabola, it is important to start by analyzing the form of the equation. If both $x$ and $y$ are to the second power and are added, the graph is an ellipse; if both $x$ and $y$ are to the second power and are subtracted, the graph is a hyperbola; if only one of $x$ or $y$ is to the second power, the graph is a parabola. If the graph is an ellipse, determine whether the major axis is horizontal or vertical; then identify coordinates of any vertices or co-vertices, and the coordinates of the center. If the graph is a hyperbola, determine whether the transverse axis is horizontal or vertical to know whether the hyperbola opens left/right or up/down, respectively; use the coordinates of the vertices and the center and the equation of the asymptotes to complete the graph. If the graph is a parabola, determine the axis of symmetry to decide if the parabola opens up/down or left/right; then find the coordinates of the vertex and possibly the endpoints of the latus rectum to complete the graph.

Given the equation $\frac{x^{2}}{25}+\frac{y^{2}}{36}=1$ to be graphed. Both $x$ and $y$ are to the second power and are added, so the equation represents an ellipse; the larger number is connected to the $y$ variable, so the ellipse is longer in the vertical direction. No number is subtracted from either $x$ or $y$, so the ellipse is centered at the origin $(0, \,0)$. Because $36=6^{2}$ and $25=5^{2}, a=6$ and $b=5$. Then, the vertices have coordinates $(0, \,-6)$ and $(0, \, 6)$ and the co-vertices have coordinates $(-5, \,0)$ and $(5, \,0)$. The graph is shown below.

To graph the equation $(x-1)^{2}-\frac{(y+3)^{2}}{16}=1$, notice that both $x$ and $y$ are to the second power and the variables are subtracted. So, the graph is that of a hyperbola with center at $(1,-3)$; the order of the variables indicates that the transverse axis is parallel to the $x$-axis. The denominator connected to the $x$-variable is 1, so $a=1$; likewise, $b^{2}=16$, so $b=4$. So, the coordinates of the vertices are $(0,-3)$ and $(2, \, -3)$. The graph is shown below. The asymptotes in red have equations $y=\pm 4(x-1)-3$; the two asymptotes intersect at $(1$, -3) which is the center of the hyperbola.

Given the equation $y^{2}=8 x$ to be graphed. Notice that only one of the variables is to the second power, so the equation is of the form for a parabola. Because $y$ is the squared variable, the parabola opens to the left or the right. The vertex is at $(0,  \,  0)$ because no transformations are applied to either variable; the axis of symmetry has equation $y=0$. The coefficient of 8 leads to $8=4 p$, or $p=2$; because $p > 0$, the parabola opens to the right. The latus rectum has endpoints $(2, \, -4)$ and $(2, \, 4)$. The graph of the parabola is shown below. The directrix in red has equation $x=-2$.

Review this material in Writing Equations of Ellipses, Writing Equations of Hyperbolas, and Parabolas Centered at the Origin.

Unit 10 Vocabulary

This vocabulary list includes terms you will need to know to successfully complete the final exam.

• axis of symmetry
• center of ellipse
• conjugate axis
• co-vertex of ellipse
• co-vertex of hyperbola
• directrix
• ellipse
• focus
• hyperbola
• latus rectum
• major axis
• minor axis
• parabola
• transverse axis
• vertex

11a. write the terms of a sequence given a formula

• Write a few terms of a sequence given an explicit formula for the sequence.
• Write a few terms of a sequence given a recursive formula for the sequence.

A sequence is an ordered list of numbers; each number in the list is a term of the sequence. The notation $a_{n}$ denotes the $n$th term of the sequence; that is, the subscript $n$ indicates the position of the term, so $a_{3}$ represents the third term of a sequence.

Listing all the terms of a sequence is often not feasible, particularly if the sequence continues without end. There are typically two ways to express a formula for a sequence. An explicit formula provides a way to find the $n$th term of the sequence; terms of the sequence can be found by evaluating the formula for $n=1,2,3, \ldots$ or a specific value of $n$. Evaluating a sequence expressed with an explicit formula is equivalent to evaluating a rule for a function. In fact, a sequence is a function whose domain is the set of positive integers.

Given a sequence defined by the formula $a_{n}=2 \cdot n^{3}+4$. To find the first five terms of the sequence, evaluate this formula for $n=1,2,3,4,5$. Thus, the first five terms of the sequence are $6,20,58,132,254$.

A second way to describe a sequence is with a recursive formula. A recursive formula consists of two parts: the initial term; and a formula to obtain $a_{n}$ from previous terms. For instance, given a sequence described by the following recursive formula.

$\begin{aligned} &a_{1}=12 \\ &a_{n}=5 \cdot a_{n-1}-6 \end{aligned}$

The first term is 12. Then $a_{2}=5 \cdot a_{1}-6=5 \cdot 12-6=54.$ Next, $a_{3}=5 \cdot a_{2}-6=5 \cdot 54-6=264$. Continuing, $a_{4}=5 \cdot a_{3}-6=5 \cdot 264-6=1314$ and $a_{5}=5 \cdot a_{4}-6=5 \cdot 1314-6=6564$. So, the first five terms of the sequence are $12,54,264,1314,6564$. Notice that the first term was given. Each subsequent term is found by multiplying the previous term by 5 and then subtracting 6.

Review the material in Sequences Defined by an Explicit Formula.

11b. write the formula for a sequence given the first few terms, or a recursive relationship

• Given the beginning terms of a sequence, write an explicit or recursive formula for the sequence.
• Replace a recursive formula for a sequence with an explicit formula for the sequence.
• Compare arithmetic and geometric sequences. What are the differences between the formulas for arithmetic and geometric sequences?

At times, the first few terms of a sequence are given but the rule for the sequence is not stated. In order to determine values of the sequence beyond those that are listed, it is helpful to find a formula to describe the sequence. This is the reverse of the skills used in objective 11a.

To find an explicit formula for a sequence, it is important to look for patterns among the terms. If the terms of the sequence are rational numbers, the patterns for the numerator and denominator might be different. Knowing the powers of the first few positive integers, such as 2, 3, 4, or 5, is often useful to find patterns among the terms. The pattern needs to be written in terms of $n$, where $n$ is the position of the term in the sequence. Be careful! It is important to check any proposed formula for multiple terms in the sequence to ensure that the pattern holds for more than just the first or second terms of the sequence.

For instance, consider the sequence whose first few terms are $2,5,10,17,26,37, \ldots$ To find a rule, start by checking to see if there is a constant difference between the terms or perhaps a constant ratio between the terms. Neither is visible for this sequence. There might not seem to be a pattern. However, notice that the terms are 1 more than a perfect square. That is $2=1^{2}+1,5=2^{2}+1,10=3^{2}+1,17=4^{2}+1,26=5^{2}+1$, and $37=6^{2}+1$. So, a formula for the sequence seems to be $a_{n}=n^{2}+1$.

A small adjustment in the sequence can lead to a major adjustment in the formula. Suppose the previous sequence had been $5, \, 10, \, 17, \, 26, \, 37, \ldots$; that is, the sequence has the same terms except the sequence starts at 5 rather than at 2. Each term of the sequence is still 1 more than a perfect square. Now, however, the number being squared does not correspond to the position of the term but to 1 more than the position of the term. So, for this sequence, the explicit formula is $a_{n}=(n+1)^{2}+1$.

There are two special types of sequences that regularly appear. An arithmetic sequence is a sequence in which the terms differ by a constant amount, called the common difference and generally denoted $d$. An arithmetic sequence can be described recursively:

$\begin{aligned} &a_{1} \\ &a_{n}=a_{n-1}+d \end{aligned} \text { for } n \geq 2$

This recursive formula gives rise to the following terms of the sequence:

$a_{1}, a_{1}+d, a_{1}+2 d, a_{1}+3 d, a_{1}+4 d, \ldots$

So, an explicit formula for this arithmetic sequence is $a_{n}=a_{1}+(n-1) d$

Given terms of a sequence that appear to differ by a constant amount, a formula can be written either recursively or explicitly using these two relationships. For instance, given the sequence $12, \, 5, \, -2,  \, -9,  \,-16,  \, \ldots$ The difference between successive terms is $-7$ and the first term is 12. So, the sequence can be described with one of the two formulas below:}}+(n-1) d\).

Although either formula can be used to write terms of the sequence, the explicit formula would typically be used to find a specific term. For example, the 100 th term is $a_{100}=12-7(100-1)=-681$.

A second type of special sequence is a geometric sequence, which is a sequence in which terms differ by a common factor, called the common ratio and generally denoted $r$. Just as an arithmetic sequence can be written either recursively or explicitly, a geometric sequence can be described either way.

Notice that the two relationships for a geometric sequence are analogous to those for an arithmetic sequence if addition is replaced by multiplication and the common difference is replaced by the common ratio.

Given a sequence that is known to be geometric, with $g_{2}=100$ and $g_{5}=12.5$. Writing recursive and explicit formulas for this sequence requires finding the first term and the common ratio. The two given terms can be substituted in the explicit formula to obtain two equations.

$100=g_{1} \cdot r \quad$ and $\quad 12.5=g_{1} \cdot r^{4}$

Solving these two equations simultaneously yields $r=\frac{1}{2}$ so that $g_{1}=200$. Then the sequence can be described as shown below.

Review the material in Sequences Defined by an Explicit Formula, Write the Terms of an Arithmetic Sequence, and Write the Terms of a Geometric Sequence.

11c. apply the formulas for arithmetic and geometric series

• Compare the formula for the sum of the first n terms of an arithmetic series to the formula for the first n terms of a geometric series.
• Under what conditions can the sum of an infinite geometric series be found? Identify an appropriate formula that can be used to find the sum of an infinite geometric series when it exists.

A series is the sum of the terms of a sequence. Although it is often not possible to find the sum if the sequence is infinite, it is generally possible to find the sum of a specific number of terms. For the special cases of arithmetic and geometric series, that is, the sums of terms in an arithmetic and geometric sequence, respectively, formulas exist to facilitate finding the sum of the first $n$ terms.

An arithmetic series is the sum of the terms of an arithmetic sequence. The sum of the first $n$ terms, denoted $S_{n}$, is given by $S_{n}=\frac{n\left(a_{1}+a_{n}\right)}{2}$. To find the sum of an arithmetic series, only three values are needed: the first term; the $n$th term; and the number of terms to be summed.

In contrast, a geometric series is the sum of the terms of a geometric sequence. The sum of the first $n$ terms, again denoted $S_{n}$, is given by $S_{n}=\frac{g_{1}\left(1-r^{n}\right)}{1-r}$ for $r \neq 1$. Notice that to find the sum of a geometric series, again only three values are needed: the first term; the common ratio; and the number of terms to be summed.

Given the sequence: $8, \, 11, \, 14, \, 17, \, 20, \ldots$. To find the sum of the first 100 terms of the related series, first notice that the terms of the sequence differ by 3, so this would be an arithmetic series. The 100th term is $a_{100}=8+99(2)=206$. Using the formula for the sum of an arithmetic series yields $S_{100}=\frac{100(8+206)}{2}= 10,700$.

A sequence that continues indefinitely is an infinite sequence; the sum of the terms of such a sequence is an infinite series. If the infinite series is geometric, then it can be summed under some conditions. In the formula for the sum of a geometric series, the behavior of $r^{n}$ determines whether a sum can be found. If $-1$, then $r^{n} \rightarrow 0$ as $n \rightarrow \infty$; so $S_{n} \rightarrow \frac{g_{1}}{1-r}$. If $r$ or $r > 1$, then $r^{n} \rightarrow \infty$ as $n \rightarrow \infty$, so $S_{n} \rightarrow \infty$ and no sum can be found. For instance, consider the geometric series $81+27+9+3+\ldots$ This is a geometric series with $r=\frac{1}{3}$; as the series continues, the terms get smaller and smaller in value. The sum of this infinite geometric series is $S=\frac{81}{1-\frac{1}{3}}=121.5$.

Review material in Use the Formula for an Arithmetic Series.

Unit 11 Vocabulary

This vocabulary list includes terms you will need to know to successfully complete the final exam.

• arithmetic sequence
• arithmetic series
• common difference of an arithmetic sequence
• common ratio of a geometric sequence
• explicit formula for a sequence
• geometric sequence
• geometric series
• infinite series
• recursive formula for a sequence
• sequence
• series
• term of a sequence