Learning Objectives
By the end of this section, you will be able to:
 Recognize the graph of a quadratic equation in two variables
 Find the axis of symmetry and vertex of a parabola
 Find the intercepts of a parabola
 Graph quadratic equations in two variables
 Solve maximum and minimum applications
Be Prepared 10.13
Before you get started, take this readiness quiz.
Graph the equation $y=3x5$ by plotting points.
If you missed this problem, review Example 4.11.
Be Prepared 10.14
Evaluate $2{x}^{2}+4x1$ when $x=\mathrm{3}$.
If you missed this problem, review Example 1.57.
Be Prepared 10.15
Evaluate $\frac{b}{2a}$ when $a=\frac{1}{3}$ and $b=\frac{5}{6}$.
If you missed this problem, review Example 1.89.
Recognize the Graph of a Quadratic Equation in Two Variables
We have graphed equations of the form $Ax+By=C$. We called equations like this linear equations because their graphs are straight lines.
Now, we will graph equations of the form $y=a{x}^{2}+bx+c$. We call this kind of equation a quadratic equation in two variables.
Quadratic Equation in Two Variables
A quadratic equation in two variables, where $a,b,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}c$ are real numbers and $a\ne 0$, is an equation of the form
Just like we started graphing linear equations by plotting points, we will do the same for quadratic equations.
Let’s look first at graphing the quadratic equation $y={x}^{2}$. We will choose integer values of $x$ between $\mathrm{2}$ and 2 and find their $y$ values. See Table 10.1.
$y={x}^{2}$  
$x$  $y$ 
0  0 
1  1 
$\mathrm{1}$  1 
2  4 
$\mathrm{2}$  4 
Notice when we let $x=1$ and $x=\mathrm{1}$, we got the same value for $y$.
The same thing happened when we let $x=2$ and $x=\mathrm{2}$.
Now, we will plot the points to show the graph of $y={x}^{2}$. See Figure 10.2.
The graph is not a line. This figure is called a parabola. Every quadratic equation has a graph that looks like this.
In Example 10.43 you will practice graphing a parabola by plotting a few points.
Example 10.43
Graph $y={x}^{2}1$.
Solution
We will graph the equation by plotting points.
Choose integers values for x, substitute them into the equation and solve for y.  
Record the values of the ordered pairs in the chart.  
Plot the points, and then connect them with a smooth curve. The result will be the graph of the equation$y={x}^{2}1$. 
Try It 10.85
Graph $y=\text{\u2212}{x}^{2}$.
Try It 10.86
Graph $y={x}^{2}+1$.
How do the equations $y={x}^{2}$ and $y={x}^{2}1$ differ? What is the difference between their graphs? How are their graphs the same?
All parabolas of the form $y=a{x}^{2}+bx+c$ open upwards or downwards. See Figure 10.3.
Notice that the only difference in the two equations is the negative sign before the ${x}^{2}$ in the equation of the second graph in Figure 10.3. When the ${x}^{2}$ term is positive, the parabola opens upward, and when the ${x}^{2}$ term is negative, the parabola opens downward.
Parabola Orientation
For the quadratic equation $y=a{x}^{2}+bx+c$, if:
Example 10.44
Determine whether each parabola opens upward or downward:
ⓐ $y=\mathrm{3}{x}^{2}+2x4$ ⓑ $y=6{x}^{2}+7x9$
Solution
ⓐ Find the value of "a". 
Since the “a” is negative, the parabola will open downward. 

ⓑ Find the value of "a". 
Since the “a” is positive, the parabola will open upward. 
Try It 10.87
Determine whether each parabola opens upward or downward:
ⓐ $y=2{x}^{2}+5x2$ ⓑ $y=\mathrm{3}{x}^{2}4x+7$
Try It 10.88
Determine whether each parabola opens upward or downward:
ⓐ $y=\mathrm{2}{x}^{2}2x3$ ⓑ $y=5{x}^{2}2x1$
Find the Axis of Symmetry and Vertex of a Parabola
Look again at Figure 10.3. Do you see that we could fold each parabola in half and that one side would lie on top of the other? The ‘fold line’ is a line of symmetry. We call it the axis of symmetry of the parabola.
We show the same two graphs again with the axis of symmetry in blue. See Figure 10.4.
The equation of the axis of symmetry can be derived by using the Quadratic Formula. We will omit the derivation here and proceed directly to using the result. The equation of the axis of symmetry of the graph of $y=a{x}^{2}+bx+c$ is $x=\frac{b}{2a}.$
So, to find the equation of symmetry of each of the parabolas we graphed above, we will substitute into the formula $x=\frac{b}{2a}$.
Look back at Figure 10.4. Are these the equations of the dashed red lines?
The point on the parabola that is on the axis of symmetry is the lowest or highest point on the parabola, depending on whether the parabola opens upwards or downwards. This point is called the vertex of the parabola.
We can easily find the coordinates of the vertex, because we know it is on the axis of symmetry. This means its xcoordinate is $\frac{b}{2a}$. To find the ycoordinate of the vertex, we substitute the value of the xcoordinate into the quadratic equation.
Axis of Symmetry and Vertex of a Parabola
For a parabola with equation $y=a{x}^{2}+bx+c$:
 The axis of symmetry of a parabola is the line $x=\frac{b}{2a}$.
 The vertex is on the axis of symmetry, so its xcoordinate is $\frac{b}{2a}$.
To find the ycoordinate of the vertex, we substitute $x=\frac{b}{2a}$ into the quadratic equation.
Example 10.45
For the parabola $y=3{x}^{2}6x+2$ find: ⓐ the axis of symmetry and ⓑ the vertex.
Solution
ⓐ  
The axis of symmetry is the line $x=\frac{b}{2a}$.  
Substitute the values of a, b into the equation.  
Simplify.  $x=1$  
The axis of symmetry is the line $x=1$.  
ⓑ  
The vertex is on the line of symmetry, so its xcoordinate will be $x=1$.  
Substitute$x=1$ into the equation and solve for y.  
Simplify.  
This is the ycoordinate.  $y=\mathrm{1}$ The vertex is $(1,\text{\u2212}1).$ 
Try It 10.89
For the parabola $y=2{x}^{2}8x+1$ find: ⓐ the axis of symmetry and ⓑ the vertex.
Try It 10.90
For the parabola $y=2{x}^{2}4x3$ find: ⓐ the axis of symmetry and ⓑ the vertex.
Find the Intercepts of a Parabola
When we graphed linear equations, we often used the x and yintercepts to help us graph the lines. Finding the coordinates of the intercepts will help us to graph parabolas, too.
Remember, at the yintercept the value of $x$ is zero. So, to find the yintercept, we substitute $x=0$ into the equation.
Let’s find the yintercepts of the two parabolas shown in the figure below.
At an xintercept, the value of $y$ is zero. To find an xintercept, we substitute $y=0$ into the equation. In other words, we will need to solve the equation $0=a{x}^{2}+bx+c$ for $x$.
But solving quadratic equations like this is exactly what we have done earlier in this chapter.
We can now find the xintercepts of the two parabolas shown in Figure 10.5.
First, we will find the xintercepts of a parabola with equation $y={x}^{2}+4x+3$.
Let $y=0$.  
Factor.  
Use the zero product property.  
Solve.  
The x intercepts are $(\text{\u2212}1,0)$ and $(\text{\u2212}3,0).$ 
Now, we will find the xintercepts of the parabola with equation $y={x}^{2}+4x+3$.
Let $y=0$.  
This quadratic does not factor, so we use the Quadratic Formula.  
$a=\mathrm{1}$, $b=4$, $c=3$  
Simplify.  
The x intercepts are $(2+\sqrt{7},0)$ and $(2\sqrt{7},0)$. 
We will use the decimal approximations of the xintercepts, so that we can locate these points on the graph.
Do these results agree with our graphs? See Figure 10.6.
How To
Find the intercepts of a parabola.
To find the intercepts of a parabola with equation $y=a{x}^{2}+bx+c$:
Example 10.46
Find the intercepts of the parabola $y={x}^{2}2x8$.
Solution
To find the yintercept, let $x=0$ and solve for y.  
When $x=0$, then $y=\mathrm{8}$. The yintercept is the point $(0,\mathrm{8})$. 

To find the xintercept, let $y=0$ and solve for x.  
Solve by factoring.  
When $y=0$, then $x=4\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}x=\mathrm{2}$. The xintercepts are the points $\left(4,0\right)$ and $\left(\mathrm{2},0\right)$.
Try It 10.91
Find the intercepts of the parabola $y={x}^{2}+2x8.$
Try It 10.92
Find the intercepts of the parabola $y={x}^{2}4x12.$
In this chapter, we have been solving quadratic equations of the form $a{x}^{2}+bx+c=0$. We solved for $x$ and the results were the solutions to the equation.
We are now looking at quadratic equations in two variables of the form $y=a{x}^{2}+bx+c$. The graphs of these equations are parabolas. The xintercepts of the parabolas occur where $y=0$.
For example:
The solutions of the quadratic equation are the $x$ values of the xintercepts.
Earlier, we saw that quadratic equations have 2, 1, or 0 solutions. The graphs below show examples of parabolas for these three cases. Since the solutions of the equations give the xintercepts of the graphs, the number of xintercepts is the same as the number of solutions.
Previously, we used the discriminant to determine the number of solutions of a quadratic equation of the form $a{x}^{2}+bx+c=0$. Now, we can use the discriminant to tell us how many xintercepts there are on the graph.
Before you start solving the quadratic equation to find the values of the xintercepts, you may want to evaluate the discriminant so you know how many solutions to expect.
Example 10.47
Find the intercepts of the parabola $y=5{x}^{2}+x+4$.
Solution
To find the yintercept, let $x=0$ and solve for y.  When $x=0$, then $y=4$. The yintercept is the point $(0,4)$. 

To find the xintercept, let $y=0$ and solve for x.  
Find the value of the discriminant to predict the number of solutions and so xintercepts.  $\begin{array}{c}\begin{array}{ccc}\hfill {b}^{2}& \hfill & 4ac\hfill \\ \hfill {1}^{2}& \hfill & 4\cdot 5\cdot 4\hfill \\ 1\hfill & \hfill & 80\hfill \end{array}\hfill \\ \hfill \mathrm{79}\phantom{\rule{0.8em}{0ex}}\hfill \end{array}$  
Since the value of the discriminant is negative, there is no real solution to the equation.  There are no xintercepts. 
Try It 10.93
Find the intercepts of the parabola $y=3{x}^{2}+4x+4.$
Try It 10.94
Find the intercepts of the parabola $y={x}^{2}4x5.$
Example 10.48
Find the intercepts of the parabola $y=4{x}^{2}12x+9$.
Solution
To find the yintercept, let $x=0$ and solve for y.  
When $x=0$, then $y=9$. The yintercept is the point $(0,9)$. 

To find the xintercept, let $y=0$ and solve for x.  
Find the value of the discriminant to predict the number of solutions and so xintercepts.  $\begin{array}{c}\begin{array}{ccc}\hfill {b}^{2}& \hfill & 4ac\hfill \\ \hfill {2}^{2}& \hfill & 4\cdot 4\cdot 9\hfill \\ \hfill 144& \hfill & 144\hfill \end{array}\hfill \\ \hfill 0\phantom{\rule{1em}{0ex}}\hfill \end{array}$  
Since the value of the discriminant is 0, there is only one real solution to the equation. Therefore, there is only one xintercept.  
Solve the equation by factoring the perfect square trinomial.  
Use the Zero Product Property.  
Solve for x.  
When $y=0$, then $\frac{3}{2}=x.$  
The xintercept is the point $(\frac{3}{2},0).$ 
Try It 10.95
Find the intercepts of the parabola $y=\text{\u2212}{x}^{2}12x36.$
Try It 10.96
Find the intercepts of the parabola $y=9{x}^{2}+12x+4.$
Graph Quadratic Equations in Two Variables
Now, we have all the pieces we need in order to graph a quadratic equation in two variables. We just need to put them together. In the next example, we will see how to do this.
Example 10.49
How To Graph a Quadratic Equation in Two Variables
Graph $y={x}^{2}6x+8$.
Solution
Try It 10.97
Graph the parabola $y={x}^{2}+2x8.$
Try It 10.98
Graph the parabola $y={x}^{2}8x+12.$
How To
Graph a quadratic equation in two variables.
 Step 1. Write the quadratic equation with $y$ on one side.
 Step 2. Determine whether the parabola opens upward or downward.
 Step 3. Find the axis of symmetry.
 Step 4. Find the vertex.
 Step 5. Find the yintercept. Find the point symmetric to the yintercept across the axis of symmetry.
 Step 6. Find the xintercepts.
 Step 7. Graph the parabola.
We were able to find the xintercepts in the last example by factoring. We find the xintercepts in the next example by factoring, too.
Example 10.50
Graph $y=\text{\u2212}{x}^{2}+6x9$.
Solution
The equation y has on one side.  
Since a is $1$, the parabola opens downward. To find the axis of symmetry, find $x=\frac{b}{2a}$. 
The axis of symmetry is $x=3.$ The vertex is on the line $x=3.$ 

Find y when $x=3.$  The vertex is $(3,0).$ 

The yintercept occurs when $x=0.$ Substitute $x=0.$ Simplify. The point $(0,\mathrm{9})$ is three units to the left of the line of symmetry. The point three units to the right of the line of symmetry is $(6,\mathrm{9}).$ Point symmetric to the yintercept is $(6,\mathrm{9})$ 
The yintercept is $(0,\mathrm{9}).$ 

The xintercept occurs when $y=0.$  
Substitute $y=0.$  
Factor the GCF.  
Factor the trinomial.  
Solve for x.  
Connect the points to graph the parabola. 
Try It 10.99
Graph the parabola $y=\mathrm{3}{x}^{2}+12x12.$
Try It 10.100
Graph the parabola $y=25{x}^{2}+10x+1.$
For the graph of $y={x}^{2}+6x9$, the vertex and the xintercept were the same point. Remember how the discriminant determines the number of solutions of a quadratic equation? The discriminant of the equation $0=\text{\u2212}{x}^{2}+6x9$ is 0, so there is only one solution. That means there is only one xintercept, and it is the vertex of the parabola.
How many xintercepts would you expect to see on the graph of $y={x}^{2}+4x+5$?
Example 10.51
Graph $y={x}^{2}+4x+5$.
Solution
The equation has y on one side.  
Since a is 1, the parabola opens upward.  
To find the axis of symmetry, find $x=\frac{b}{2a}.$  The axis of symmetry is $x=\mathrm{2}.$ 

The vertex is on the line $x=\mathrm{2}.$  
Find y when $x=\mathrm{2}.$  The vertex is $(\mathrm{2},1).$ 

The yintercept occurs when $x=0.$ Substitute $x=0.$ Simplify. The point $(0,5)$ is two units to the right of the line of symmetry. The point two units to the left of the line of symmetry is $(\mathrm{4},5).$ 
The yintercept is $(0,5).$ Point symmetric to the y intercept is $(\mathrm{4},5)$. 

The x intercept occurs when $y=0.$  
Substitute $y=0.$ Test the discriminant. 

${b}^{2}4ac$ ${4}^{2}4\cdot 15$ $1620$ $\phantom{\rule{1em}{0ex}}\mathrm{4}$ 

Since the value of the discriminant is negative, there is no solution and so no x intercept. Connect the points to graph the parabola. You may want to choose two more points for greater accuracy. 
Try It 10.101
Graph the parabola $y=2{x}^{2}6x+5.$
Try It 10.102
Graph the parabola $y=\mathrm{2}{x}^{2}1.$
Finding the yintercept by substituting $x=0$ into the equation is easy, isn’t it? But we needed to use the Quadratic Formula to find the xintercepts in Example 10.51. We will use the Quadratic Formula again in the next example.
Example 10.52
Graph $y=2{x}^{2}4x3$.
Solution
The equation y has one side. Since a is 2, the parabola opens upward. 

To find the axis of symmetry, find $x=\frac{b}{2a}$.  The axis of symmetry is $x=1$. 

The vertex on the line $x=1.$  
Find y when $x=1$.  The vertex is $(1,\text{\u2212}5)$. 

The yintercept occurs when $x=0.$  
Substitute $x=0.$  
Simplify.  The yintercept is $(0,\mathrm{3})$. 

The point $(0,\mathrm{3})$ is one unit to the left of the line of symmetry. The point one unit to the right of the line of symmetry is $(2,\mathrm{3})$ 
Point symmetric to the yintercept is $(2,\mathrm{3}).$  
The xintercept occurs when $y=0$.  
Substitute $y=0$.  
Use the Quadratic Formula.  
Substitute in the values of a, b, c.  
Simplify.  
Simplify inside the radical.  
Simplify the radical.  
Factor the GCF.  
Remove common factors.  
Write as two equations.  
Approximate the values.  
The approximate values of the xintercepts are $(2.5,0)$ and $(\mathrm{0.6},0)$.  
Graph the parabola using the points found. 
Try It 10.103
Graph the parabola $y=5{x}^{2}+10x+3.$
Try It 10.104
Graph the parabola $y=\mathrm{3}{x}^{2}6x+5.$
Solve Maximum and Minimum Applications
Knowing that the vertex of a parabola is the lowest or highest point of the parabola gives us an easy way to determine the minimum or maximum value of a quadratic equation. The ycoordinate of the vertex is the minimum yvalue of a parabola that opens upward. It is the maximum yvalue of a parabola that opens downward. See Figure 10.7.
Minimum or Maximum Values of a Quadratic Equation
The ycoordinate of the vertex of the graph of a quadratic equation is the
 minimum value of the quadratic equation if the parabola opens upward.
 maximum value of the quadratic equation if the parabola opens downward.
Example 10.53
Find the minimum value of the quadratic equation $y={x}^{2}+2x8$.
Solution
Since a is positive, the parabola opens upward.  
The quadratic equation has a minimum.  
Find the axis of symmetry.  The axis of symmetry is $x=\mathrm{1}$. 

The vertex is on the line $x=\mathrm{1}.$  
Find y when $x=\mathrm{1}.$  The vertex is $(\mathrm{1},\mathrm{9})$. 

Since the parabola has a minimum, the ycoordinate of the vertex is the minimum yvalue of the quadratic equation.  
The minimum value of the quadratic is $\mathrm{9}$ and it occurs when $x=\mathrm{1}$.  
Show the graph to verify the result. 
Try It 10.105
Find the maximum or minimum value of the quadratic equation $y={x}^{2}8x+12$.
Try It 10.106
Find the maximum or minimum value of the quadratic equation $y=\mathrm{4}{x}^{2}+16x11$.
We have used the formula
to calculate the height in feet, $h$, of an object shot upwards into the air with initial velocity, ${v}_{0}$, after $t$ seconds.
This formula is a quadratic equation in the variable $t$, so its graph is a parabola. By solving for the coordinates of the vertex, we can find how long it will take the object to reach its maximum height. Then, we can calculate the maximum height.
Example 10.54
The quadratic equation $h=\mathrm{16}{t}^{2}+{v}_{0}t+{h}_{0}$ models the height of a volleyball hit straight upwards with velocity 176 feet per second from a height of 4 feet.
 ⓐ How many seconds will it take the volleyball to reach its maximum height?
 ⓑ Find the maximum height of the volleyball.
Solution
$h=\mathrm{16}{t}^{2}+176t+4$
Since a is negative, the parabola opens downward.
The quadratic equation has a maximum.
 ⓐ
$\begin{array}{cccc}\text{Find the axis of symmetry.}\hfill & & & \phantom{\rule{4em}{0ex}}\begin{array}{c}t=\frac{b}{2a}\hfill \\ t=\frac{176}{2(\mathrm{16})}\hfill \\ t=5.5\hfill \end{array}\hfill \\ & & & \phantom{\rule{4em}{0ex}}\text{The axis of symmetry is}\phantom{\rule{0.2em}{0ex}}t=5.5.\hfill \\ \text{The vertex is on the line}\phantom{\rule{0.2em}{0ex}}t=5.5.\hfill & & & \phantom{\rule{4em}{0ex}}\text{The maximum occurs when}\phantom{\rule{0.2em}{0ex}}t=5.5\phantom{\rule{0.2em}{0ex}}\text{seconds.}\hfill \end{array}$  ⓑ
Find h when $t=5.5$. Use a calculator to simplify. The vertex is $(5.5,488)$. Since the parabola has a maximum, the hcoordinate of the vertex is the maximum yvalue of the quadratic equation. The maximum value of the quadratic is 488 feet and it occurs when $t=5.5$ seconds.
Try It 10.107
The quadratic equation $h=\mathrm{16}{t}^{2}+128t+32$ is used to find the height of a stone thrown upward from a height of 32 feet at a rate of 128 ft/sec. How long will it take for the stone to reach its maximum height? What is the maximum height? Round answers to the nearest tenth.
Try It 10.108
A toy rocket shot upward from the ground at a rate of 208 ft/sec has the quadratic equation of $h=\mathrm{16}{t}^{2}+208t$. When will the rocket reach its maximum height? What will be the maximum height? Round answers to the nearest tenth.
Media
Access these online resources for additional instruction and practice graphing quadratic equations:
Section 10.5 Exercises
Practice Makes Perfect
Recognize the Graph of a Quadratic Equation in Two Variables
In the following exercises, graph:
$y=\text{\u2212}{x}^{2}+1$
In the following exercises, determine if the parabola opens up or down.
$y=6{x}^{2}+2x+3$
$y=\mathrm{9}{x}^{2}24x16$
Find the Axis of Symmetry and Vertex of a Parabola
In the following exercises, find ⓐ the axis of symmetry and ⓑ the vertex.
$y={x}^{2}+10x+25$
$y=\mathrm{2}{x}^{2}8x3$
Find the Intercepts of a Parabola
In the following exercises, find the x and yintercepts.
$y={x}^{2}+10x11$
$y={x}^{2}+6x+13$
$y=\text{\u2212}{x}^{2}14x49$
Graph Quadratic Equations in Two Variables
In the following exercises, graph by using intercepts, the vertex, and the axis of symmetry.
$y={x}^{2}+4x12$
$y={x}^{2}6x+8$
$y=\text{\u2212}{x}^{2}+8x16$
$y=5{x}^{2}+2$
$y=3{x}^{2}6x1$
$y=\mathrm{4}{x}^{2}6x2$
$y={x}^{2}+6x+8$
$y=\mathrm{16}{x}^{2}+24x9$
$y=\mathrm{2}{x}^{2}+8x10$
Solve Maximum and Minimum Applications
In the following exercises, find the maximum or minimum value.
$y=\mathrm{4}{x}^{2}+12x5$
$y=\text{\u2212}{x}^{2}+4x5$
$y=4{x}^{2}49$
In the following exercises, solve. Round answers to the nearest tenth.
An arrow is shot vertically upward from a platform 45 feet high at a rate of 168 ft/sec. Use the quadratic equation $h=\mathrm{16}{t}^{2}+168t+45$ to find how long it will take the arrow to reach its maximum height, and then find the maximum height.
A stone is thrown vertically upward from a platform that is 20 feet high at a rate of 160 ft/sec. Use the quadratic equation $h=\mathrm{16}{t}^{2}+160t+20$ to find how long it will take the stone to reach its maximum height, and then find the maximum height.
A computer store owner estimates that by charging $x$ dollars each for a certain computer, he can sell $40x$ computers each week. The quadratic equation $R=\text{\u2212}{x}^{2}+40x$ is used to find the revenue, $R$, received when the selling price of a computer is $x$. Find the selling price that will give him the maximum revenue, and then find the amount of the maximum revenue.
A retailer who sells backpacks estimates that, by selling them for $x$ dollars each, he will be able to sell $100x$ backpacks a month. The quadratic equation $R=\text{\u2212}{x}^{2}+100x$ is used to find the $R$ received when the selling price of a backpack is $x$. Find the selling price that will give him the maximum revenue, and then find the amount of the maximum revenue.
A rancher is going to fence three sides of a corral next to a river. He needs to maximize the corral area using 240 feet of fencing. The quadratic equation $A=x\left(2402x\right)$ gives the area of the corral, $A$, for the length, $x,$ of the corral along the river. Find the length of the corral along the river that will give the maximum area, and then find the maximum area of the corral.
A veterinarian is enclosing a rectangular outdoor running area against his building for the dogs he cares for. He needs to maximize the area using 100 feet of fencing. The quadratic equation $A=x\left(1002x\right)$ gives the area, $A$, of the dog run for the length, $x$, of the building that will border the dog run. Find the length of the building that should border the dog run to give the maximum area, and then find the maximum area of the dog run.
Everyday Math
In the previous set of exercises, you worked with the quadratic equation $R=\text{\u2212}{x}^{2}+40x$ that modeled the revenue received from selling computers at a price of $x$ dollars. You found the selling price that would give the maximum revenue and calculated the maximum revenue. Now you will look at more characteristics of this model.
ⓐ Graph the equation $R=\text{\u2212}{x}^{2}+40x$. ⓑ Find the values of the xintercepts.
In the previous set of exercises, you worked with the quadratic equation $R=\text{\u2212}{x}^{2}+100x$ that modeled the revenue received from selling backpacks at a price of $x$ dollars. You found the selling price that would give the maximum revenue and calculated the maximum revenue. Now you will look at more characteristics of this model.
ⓐ Graph the equation $R=\text{\u2212}{x}^{2}+100x$. ⓑ Find the values of the xintercepts.
Writing Exercises
For the revenue model in Exercise 10.205 and Exercise 10.209, explain what the xintercepts mean to the computer store owner.
For the revenue model in Exercise 10.206 and Exercise 10.210, explain what the xintercepts mean to the backpack retailer.
Self Check
ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?