Determining the Maximum and Minimum Values of Quadratic Functions

The output of the quadratic function at the vertex is the maximum or minimum value of the function, depending on the orientation of the parabola. We can see the maximum and minimum values in Figure 9.

Figure 9

There are many real-world scenarios that involve finding the maximum or minimum value of a quadratic function, such as applications involving area and revenue.

Example 5

Finding the Maximum Value of a Quadratic Function

A backyard farmer wants to enclose a rectangular space for a new garden within her fenced backyard. She has purchased 80 feet of wire fencing to enclose three sides, and she will use a section of the backyard fence as the fourth side.

ⓐFind a formula for the area enclosed by the fence if the sides of fencing perpendicular to the existing fence have length $L$.

ⓑWhat dimensions should she make her garden to maximize the enclosed area?

Solution
Let's use a diagram such as Figure 10 to record the given information. It is also helpful to introduce a temporary variable, $W$, to represent the width of the garden and the length of the fence section parallel to the backyard fence.

Figure 10

ⓐWe know we have only 80 feet of fence available, and $L+W+L=80$, or more simply, $2L+W=80$. This allows us to represent the width, $W$, in terms of $L$.

$W=80−2L$

Now we are ready to write an equation for the area the fence encloses. We know the area of a rectangle is length multiplied by width, so

$AL=W=L(80−2L)$

$A(L)=80L−2L^2$

This formula represents the area of the fence in terms of the variable length $L$. The function, written in general form, is

$A(L)=−2L^2+80L$.

ⓑThe quadratic has a negative leading coefficient, so the graph will open downward, and the vertex will be the maximum value for the area. In finding the vertex, we must be careful because the equation is not written in standard polynomial form with decreasing powers. This is why we rewrote the function in general form above. Since $a$ is the coefficient of the squared term, $a=−2$, $b=80$, and $c=0$.

To find the vertex:

$\begin{aligned}&h=-\frac{b}{2 a} \quad k=A(20)\\&=-\frac{80}{2(-2)} \text { and }=80(20)-2(20)^{2}\\&=20=800\end{aligned}$

The maximum value of the function is an area of 800 square feet, which occurs when $L=20$ feet. When the shorter sides are 20 feet, there is 40 feet of fencing left for the longer side. To maximize the area, she should enclose the garden so the two shorter sides have length 20 feet and the longer side parallel to the existing fence has length 40 feet.

Analysis

This problem also could be solved by graphing the quadratic function. We can see where the maximum area occurs on a graph of the quadratic function in Figure 11.

Figure 11

How To

Given an application involving revenue, use a quadratic equation to find the maximum.

1. Write a quadratic equation for a revenue function.
2. Find the vertex of the quadratic equation.
3. Determine the y-value of the vertex.

Example 6

Finding Maximum Revenue
The unit price of an item affects its supply and demand. That is, if the unit price goes up, the demand for the item will usually decrease. For example, a local newspaper currently has 84,000 subscribers at a quarterly charge of $30. Market research has suggested that if the owners raise the price to $32, they would lose 5,000 subscribers. Assuming that subscriptions are linearly related to the price, what price should the newspaper charge for a quarterly subscription to maximize their revenue?

Solution
Revenue is the amount of money a company brings in. In this case, the revenue can be found by multiplying the price per subscription times the number of subscribers, or quantity. We can introduce variables, $p$ for price per subscription and $Q$ for quantity, giving us the equation $Revenue=pQ$.

Because the number of subscribers changes with the price, we need to find a relationship between the variables. We know that currently $p=30$ and $Q=84,000$. We also know that if the price rises to $32, the newspaper would lose 5,000 subscribers, giving a second pair of values, $p=32$ and $Q=79,000$. From this we can find a linear equation relating the two quantities. The slope will be

$m=\dfrac{79,000−84,000}{32−30}$

$=\dfrac{−5,000}{2}$

$=−2,500$

This tells us the paper will lose 2,500 subscribers for each dollar they raise the price. We can then solve for the y-intercept.

$\begin{aligned}Q &=-2500 p+b & & \text { Substitute in the point } Q=84,000 \text { and } p=30 \\84,000 &=-2500(30)+b & & \text { Solve for } b \\b &=159,000 & &\end{aligned}$

This gives us the linear equation $Q=−2,500p+159,000$ relating cost and subscribers. We now return to our revenue equation.

$Revenue = pQ$

$Revenue=p(−2,500p+159,000)$

$Revenue=−2,500p^2+159,000p$

We now have a quadratic function for revenue as a function of the subscription charge. To find the price that will maximize revenue for the newspaper, we can find the vertex.

$h=−\dfrac{159,000}{2(−2,500)}$

$=31.8$

The model tells us that the maximum revenue will occur if the newspaper charges $31.80 for a subscription. To find what the maximum revenue is, we evaluate the revenue function.

$\text{maximum revenue }=−2,500(31.8)^2+159,000(31.8)$

$=2,528,100$

Analysis

This could also be solved by graphing the quadratic as in Figure 12. We can see the maximum revenue on a graph of the quadratic function.

Figure 12

Finding the x- and y-Intercepts of a Quadratic Function

Much as we did in the application problems above, we also need to find intercepts of quadratic equations for graphing parabolas. Recall that we find the y- intercept of a quadratic by evaluating the function at an input of zero, and we find the x- intercepts at locations where the output is zero. Notice in Figure 13 that the number of x- intercepts can vary depending upon the location of the graph.

Figure 13 Number of x-intercepts of a parabola

How To

Given a quadratic function $f(x)$, find the y- and x-intercepts.

1. Evaluate $f(0)$ to find the y-intercept.
2. Solve the quadratic equation $f(x)=0$ to find the x-intercepts.

Example 7

Finding the y- and x-Intercepts of a Parabola

Find the y- and x-intercepts of the quadratic $f(x)=3x^2+5x−2$.

Solution

We find the y-intercept by evaluating $f(0)$.

$f(0)=3(0)^2+5(0)−2$

$=−2$

So the y-intercept is at (0,−2).

For the x-intercepts, we find all solutions of $f(x)=0$.

$0=3x^2+5x−2$

In this case, the quadratic can be factored easily, providing the simplest method for solution.

$0=(3x−1)(x+2)$

So the x-intercepts are at $(\frac{1}{3},0)$ and $(−2,0)$.

Analysis

By graphing the function, we can confirm that the graph crosses the y-axis at $(0,−2)$. We can also confirm that the graph crosses the x-axis at $(\frac{1}{3}, 0)$ and $(−2,0)$. See Figure 14

Figure 14

Rewriting Quadratics in Standard Form

In Example 7, the quadratic was easily solved by factoring. However, there are many quadratics that cannot be factored. We can solve these quadratics by first rewriting them in standard form.

How To

Given a quadratic function, find the x- intercepts by rewriting in standard form.

1. Substitute $a$ and $b$ into $h=−\frac{b}{2a}$.
2. Substitute $x=h$ into the general form of the quadratic function to find $k$.
3. Rewrite the quadratic in standard form using $h$ and $k$.
4. Solve for when the output of the function will be zero to find the x- intercepts.
   
   

Example 8

Finding the x-Intercepts of a Parabola

Find the x- intercepts of the quadratic function $f(x)=2x^2+4x−4$.

Solution
We begin by solving for when the output will be zero.

$0=2x^2+4x−4$

Because the quadratic is not easily factorable in this case, we solve for the intercepts by first rewriting the quadratic in standard form.

$f(x)=a(x−h)^2+k$

We know that $a=2$. Then we solve for $h$ and $k$.

$\begin{aligned}h &=-\frac{b}{2 a} & k &=f(-1) \\&=-\frac{4}{2(2)} & &=2(-1)^{2}+4(-1)-4 \\&=-1 & &=-6\end{aligned}$

So now we can rewrite in standard form.

$f(x)=2(x+1)^2−6$

We can now solve for when the output will be zero.

$\begin{aligned}&0=2(x+1)^{2}-6 \\&6=2(x+1)^{2} \\&3=(x+1)^{2} \\&x+1=\pm \sqrt{3} \\&x=-1 \pm \sqrt{3}\end{aligned}$

The graph has x-intercepts at $(−1−\sqrt{3},0)$ and $(−1+\sqrt{3}, 0)$.

We can check our work by graphing the given function on a graphing utility and observing the x- intercepts. See Figure 15.

Figure 15

Analysis

We could have achieved the same results using the quadratic formula. Identify $a=2$, $b=4$ and $c=−4$.

$\begin{aligned}x &=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\&=\frac{-4 \pm \sqrt{4^{2}-4(2)(-4)}}{2(2)} \\&=\frac{-4 \pm \sqrt{48}}{4} \\&=\frac{-4 \pm \sqrt{3(16)}}{4} \\&=-1 \pm \sqrt{3}\end{aligned}$

So the x-intercepts occur at $(−1−\sqrt{3},0)$ and $(−1+\sqrt{3}, 0)$.

Try It #4

In a Try It, we found the standard and general form for the function $g(x)=13+x^2−6x$. Now find the y- and x-intercepts (if any).

Example 9

Applying the Vertex and x-Intercepts of a Parabola

A ball is thrown upward from the top of a 40 foot high building at a speed of 80 feet per second. The ball's height above ground can be modeled by the equation $H(t)=−16t^2+80t+40$.

    ⓐWhen does the ball reach the maximum height?
    ⓑWhat is the maximum height of the ball?
    ⓒWhen does the ball hit the ground?

Solution

ⓐThe ball reaches the maximum height at the vertex of the parabola.

$h=−\dfrac{80}{2(−16)}$

$=\dfrac{80}{32}$

$=\dfrac{5}{2}$

$=2.5$

The ball reaches a maximum height after 2.5 seconds.

ⓑTo find the maximum height, find the y- coordinate of the vertex of the parabola.

$k=H(−\dfrac{b}{2a})$

$= H(2.5)$

$=−16(2.5)^2+80(2.5)+40$

$=140$

The ball reaches a maximum height of 140 feet.

ⓒTo find when the ball hits the ground, we need to determine when the height is zero, $H(t)=0$.

We use the quadratic formula.

$t=\dfrac{−80±\sqrt{80^2−4(−16)(40)}}{2(−16)}$

$= \dfrac{−80±\sqrt{8960}}{−32}$

Because the square root does not simplify nicely, we can use a calculator to approximate the values of the solutions.

$t=\dfrac{−80−\sqrt{8960}}{−32} ≈5.458$ or $t=\dfrac{−80+\sqrt{8960}}{−32}≈−0.458$

The second answer is outside the reasonable domain of our model, so we conclude the ball will hit the ground after about 5.458 seconds. See Figure 16.

Figure 16

Note that the graph does not represent the physical path of the ball upward and downward. Keep the quantities on each axis in mind while interpreting the graph.

Try It #5

A rock is thrown upward from the top of a 112-foot high cliff overlooking the ocean at a speed of 96 feet per second. The rock's height above ocean can be modeled by the equation $H(t)=−16t^2+96t+112$.
    ⓐWhen does the rock reach the maximum height?
    ⓑWhat is the maximum height of the rock?
    ⓒWhen does the rock hit the ocean?

Source: Rice University, https://openstax.org/books/college-algebra/pages/5-1-quadratic-functions
This work is licensed under a Creative Commons Attribution 4.0 License.