MA001 Study Guide

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