10.5 Graphing Quadratic Equations in Two Variables

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.

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.

Notice when we let $x=1$ and $x=\mathrm{-1}$, we got the same value for $y$.

$$\begin{array}{cccc}y=x^2\hfill & & & y=x^2\hfill \\ y=1^2\hfill & & & y={\left(\mathrm{-1}\right)}^2\hfill \\ y=1\hfill & & & y=1\hfill \end{array}$$

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 1 you will practice graphing a parabola by plotting a few points.

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.

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 x-coordinate is $-\frac{b}{2a}$. To find the y-coordinate of the vertex, we substitute the value of the x-coordinate into the quadratic equation.

Find the Intercepts of a Parabola

When we graphed linear equations, we often used the x- and y-intercepts to help us graph the lines. Finding the coordinates of the intercepts will help us to graph parabolas, too.

Remember, at the y-intercept the value of $x$ is zero. So, to find the y-intercept, we substitute $x=0$ into the equation.

Let’s find the y-intercepts of the two parabolas shown in the figure below.

At an x-intercept, the value of $y$ is zero. To find an x-intercept, we substitute $y=0$ into the equation. In other words, we will need to solve the equation $0=a{x}^2+bx+c$ for $x$.

$$\begin{array}{}\\ \\ y=a{x}^2+bx+c\hfill \\ 0=a{x}^2+bx+c\hfill \end{array}$$

But solving quadratic equations like this is exactly what we have done earlier in this chapter.

We can now find the x-intercepts of the two parabolas shown in Figure 10.5.

First, we will find the x-intercepts of a parabola with equation $y=x^2+4x+3$.

Let $y=0$.
Factor.
Use the zero product property.
Solve.
	The x intercepts are $(\text{−}1,0)$ and $(\text{−}3,0).$

Now, we will find the x-intercepts 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 x-intercepts, so that we can locate these points on the graph.

$$\begin{array}{cccc}\left(2+\sqrt{7},0\right)\approx \left(4.6,0\right)\hfill & & & \left(2-\sqrt{7},0\right)\approx \left(\mathrm{-0.6},0\right)\hfill \end{array}$$

Do these results agree with our graphs? See Figure 10.6.

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 x-intercepts of the parabolas occur where $y=0$.

For example:

$$\begin{array}{cccc}\hfill \mathbf{\text{Quadratic equation}}\hfill & & & \hfill \mathbf{\text{Quadratic equation in two variables}}\hfill \\ & & & \hfill y=x^2-2x-15\hfill \\ \hfill \begin{array}{ccc}\hfill x^2-2x-15& =\hfill & 0\hfill \\ \hfill \left(x-5\right)\left(x+3\right)& =\hfill & 0\hfill \end{array}\hfill & & & \hfill \text{let}y=0\begin{array}{c}0=x^2-2x-15\hfill \\ 0=\left(x-5\right)\left(x+3\right)\hfill \end{array}\hfill \\ \hfill \begin{array}{cccccccc}\hfill x-5& =\hfill & 0\hfill & & & \hfill x+3& =\hfill & 0\hfill \\ \hfill x& =\hfill & 5\hfill & & & \hfill x& =\hfill & \mathrm{-3}\hfill \end{array}\hfill & & & \hfill \begin{array}{cccccccc}\hfill x-5& =\hfill & 0\hfill & & & \hfill x+3& =\hfill & 0\hfill \\ \hfill x& =\hfill & 5\hfill & & & \hfill x& =\hfill & \mathrm{-3}\hfill \end{array}\hfill \\ & & & \hfill \left(5,0\right)\text{and}\left(\mathrm{-3},0\right)\hfill \\ & & & \hfill x\text{-intercepts}\hfill \end{array}$$

The solutions of the quadratic equation are the $x$ values of the x-intercepts.

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 x-intercepts of the graphs, the number of x-intercepts 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 x-intercepts there are on the graph.

Before you start solving the quadratic equation to find the values of the x-intercepts, you may want to evaluate the discriminant so you know how many solutions to expect.

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.

We were able to find the x-intercepts in the last example by factoring. We find the x-intercepts in the next example by factoring, too.

For the graph of $y=-x^2+6x-9$, the vertex and the x-intercept were the same point. Remember how the discriminant determines the number of solutions of a quadratic equation? The discriminant of the equation $0=\text{−}{x}^2+6x-9$ is 0, so there is only one solution. That means there is only one x-intercept, and it is the vertex of the parabola.

How many x-intercepts would you expect to see on the graph of $y=x^2+4x+5$?

Finding the y-intercept by substituting $x=0$ into the equation is easy, isn’t it? But we needed to use the Quadratic Formula to find the x-intercepts in Example 9. We will use the Quadratic Formula again in the next example.

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 y-coordinate of the vertex is the minimum y-value of a parabola that opens upward. It is the maximum y-value of a parabola that opens downward. See Figure 10.7.

We have used the formula

$$h=\mathrm{-16}{t}^2+v_{0}t+h_0$$

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.

Key Concepts

• The graph of every quadratic equation is a parabola.
• Parabola Orientation For the quadratic equation $y=a{x}^2+bx+c$, if
  • $a>0$, the parabola opens upward.
  • $a<0$, the parabola opens downward.
• 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 x-coordinate is $-\frac{b}{2a}$.
  • To find the y-coordinate of the vertex we substitute $x=-\frac{b}{2a}$ into the quadratic equation.
• Find the Intercepts of a Parabola To find the intercepts of a parabola with equation $y=a{x}^2+bx+c$:
  $\begin{array}{cccc}\hfill \mathit{\text{y}}\mathbf{\text{-intercept}}& & & \hfill \mathit{\text{x}}\mathbf{\text{-intercepts}}\\ \hfill \text{Let}x=0\text{and solve for}y.& & & \hfill \text{Let}y=0\text{and solve for}x.\end{array}$
• To Graph a Quadratic Equation in Two Variables
  1. Write the quadratic equation with $y$ on one side.
  2. Determine whether the parabola opens upward or downward.
  3. Find the axis of symmetry.
  4. Find the vertex.
  5. Find the y-intercept. Find the point symmetric to the y-intercept across the axis of symmetry.
  6. Find the x-intercepts.
  7. Graph the parabola.
• Minimum or Maximum Values of a Quadratic Equation
  • The y-coordinate 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.

Section Exercises

Practice Makes Perfect

Recognize the Graph of a Quadratic Equation in Two Variables

In the following exercises, graph:

$$y=x^2+3$$

$$y=\text{−}{x}^2+1$$

In the following exercises, determine if the parabola opens up or down.

$$y=\mathrm{-2}{x}^2-6x-7$$

down

$$y=6x^2+2x+3$$

$$y=4x^2+x-4$$

up

$$y=\mathrm{-9}{x}^2-24x-16$$

Find the Axis of Symmetry and Vertex of a Parabola

In the following exercises, find ⓐ the axis of symmetry and ⓑ the vertex.

$$y=x^2+8x-1$$

ⓐ $x=\mathrm{-4}$ ⓑ $\left(\mathrm{-4},\mathrm{-17}\right)$

$$y=x^2+10x+25$$

$$y=\text{−}{x}^2+2x+5$$

ⓐ $x=1$ ⓑ $\left(1,6\right)$

$$y=\mathrm{-2}{x}^2-8x-3$$

Find the Intercepts of a Parabola

In the following exercises, find the x- and y-intercepts.

$$y=x^2+7x+6$$

$$y\text{:}(0,6);x\text{:}(\mathrm{-1},0),(\mathrm{-6},0)$$

$$y=x^2+10x-11$$

$$y=\text{−}{x}^2+8x-19$$

$$y\text{:}(0,-19);x\text{:}\text{none}$$

$$y=x^2+6x+13$$

$$y=4x^2-20x+25$$

$$y\text{:}(0,25);x\text{:}(\frac{5}{2},0)$$

$$y=\text{−}{x}^2-14x-49$$

Graph Quadratic Equations in Two Variables

In the following exercises, graph by using intercepts, the vertex, and the axis of symmetry.

$$y=x^2+6x+5$$

$y\text{:}(0,5);x\text{:}(\mathrm{-1},0),(\mathrm{-5},0);$
axis: $x=\mathrm{-3};\text{vertex}\text{:}(\mathrm{-3},\mathrm{-4})$

$$y=x^2+4x-12$$

$$y=x^2+4x+3$$

$y\text{:}(0,3);x\text{:}(\mathrm{-1},0),(\mathrm{-3},0);$
axis: $x=\mathrm{-2};\text{vertex:}(\mathrm{-2},\mathrm{-1})$

$$y=x^2-6x+8$$

$$y=9x^2+12x+4$$

$y\text{:}(0,4)x\text{:}(-\frac{2}{3},0);$
axis: $x=-\frac{2}{3};\text{vertex:}(-\frac{2}{3},0)$

$$y=\text{−}{x}^2+8x-16$$

$$y=\text{−}{x}^2+2x-7$$

$y\text{:}(0,\mathrm{-7});x\text{:}\text{none};$
axis: $x=1;\text{vertex}\text{:}(1,\mathrm{-6})$

$$y=5x^2+2$$

$$y=2x^2-4x+1$$

$y\text{:}(0,1);x\text{:}(1.7,0),(0.3,0);$
axis: $x=1;\text{vertex}\text{:}(1,\mathrm{-1})$

$$y=3x^2-6x-1$$

$$y=2x^2-4x+2$$

$y\text{:}(0,2)x\text{:}(1,0);$
axis: $x=1;\text{vertex:}(1,0)$

$$y=\mathrm{-4}{x}^2-6x-2$$

$$y=\text{−}{x}^2-4x+2$$

$y\text{:}(0,2)x\text{:}(\mathrm{-4.4},0),(0.4,0);$
axis: $x=\mathrm{-2};\text{vertex:}(\mathrm{-2},6)$

$$y=x^2+6x+8$$

$$y=5x^2-10x+8$$

$y\text{:}(0,8);x\text{:}\text{none};$
axis: $x=1;\text{vertex}\text{:}(1,3)$

$$y=\mathrm{-16}{x}^2+24x-9$$

$$y=3x^2+18x+20$$

$y\text{:}(0,20)x\text{:}(\mathrm{-4.5},0),(\mathrm{-1.5},0);$
axis: $x=\mathrm{-3};\text{vertex:}(\mathrm{-3},\mathrm{-7})$

$$y=\mathrm{-2}{x}^2+8x-10$$

Solve Maximum and Minimum Applications

In the following exercises, find the maximum or minimum value.

$$y=2x^2+x-1$$

The minimum value is $-\frac{9}{8}$ when $x=-\frac{1}{4}$.

$$y=\mathrm{-4}{x}^2+12x-5$$

$$y=x^2-6x+15$$

The minimum value is 6 when $x=3$.

$$y=\text{−}{x}^2+4x-5$$

$$y=\mathrm{-9}{x}^2+16$$

The maximum value is 16 when $x=0$.

$$y=4x^2-49$$

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.

In 5.3 sec the arrow will reach maximum height of 486 ft.

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 $40-x$ computers each week. The quadratic equation $R=\text{−}{x}^2+40x$ 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.

Charging $20 for each computer will give the maximum revenue of $400.

A retailer who sells backpacks estimates that, by selling them for $x$ dollars each, he will be able to sell $100-x$ backpacks a month. The quadratic equation $R=\text{−}{x}^2+100x$ 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 $A=x\left(240-2x\right)$ 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.

The length of the side along the river of the corral is 60 feet and the maximum area is 7,200 sq ft.

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 $A=x\left(100-2x\right)$ 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.

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?

Chapter 10 Review Exercises

10.4 Solve Applications Modeled by Quadratic Equations

In the following exercises, solve by using methods of factoring, the square root principle, or the quadratic formula.

Find two consecutive odd numbers whose product is 323.

Two consecutive odd numbers whose product is 323 are 17 and 19, and $\mathrm{-17}$ and $\mathrm{-19}.$

Find two consecutive even numbers whose product is 624.

A triangular banner has an area of 351 square centimeters. The length of the base is two centimeters longer than four times the height. Find the height and length of the base.

The height of the banner is 13 cm and the length of the side is 54 cm.

Julius built a triangular display case for his coin collection. The height of the display case is six inches less than twice the width of the base. The area of the of the back of the case is 70 square inches. Find the height and width of the case.

A tile mosaic in the shape of a right triangle is used as the corner of a rectangular pathway. The hypotenuse of the mosaic is 5 feet. One side of the mosaic is twice as long as the other side. What are the lengths of the sides? Round to the nearest tenth.

The lengths of the sides of the mosaic are 2.2 and 4.4 feet.

A rectangular piece of plywood has a diagonal which measures two feet more than the width. The length of the plywood is twice the width. What is the length of the plywood’s diagonal? Round to the nearest tenth.

The front walk from the street to Pam’s house has an area of 250 square feet. Its length is two less than four times its width. Find the length and width of the sidewalk. Round to the nearest tenth.

The width of the front walk is 8.2 feet and its length is 30.6 feet.

For Sophia’s graduation party, several tables of the same width will be arranged end to end to give a serving table with a total area of 75 square feet. The total length of the tables will be two more than three times the width. Find the length and width of the serving table so Sophia can purchase the correct size tablecloth. Round answer to the nearest tenth.

A ball is thrown vertically in the air with a velocity of 160 ft/sec. Use the formula $h=\mathrm{-16}{t}^2+v_{0}t$ to determine when the ball will be 384 feet from the ground. Round to the nearest tenth.

The ball will reach 384 feet on its way up in 4 seconds and on the way down in 6 seconds.

A bullet is fired straight up from the ground at a velocity of 320 ft/sec. Use the formula $h=\mathrm{-16}{t}^2+v_{0}t$ to determine when the bullet will reach 800 feet. Round to the nearest tenth.

10.5 Graphing Quadratic Equations in Two Variables

In the following exercises, graph by plotting point.

Graph $y=x^2-2$

Graph $y=\text{−}{x}^2+3$

In the following exercises, determine if the following parabolas open up or down.

$$y=\mathrm{-3}{x}^2+3x-1$$

down

$$y=5x^2+6x+3$$

$$y=x^2+8x-1$$

up

$$y=\mathrm{-4}{x}^2-7x+1$$

In the following exercises, find ⓐ the axis of symmetry and ⓑ the vertex.

$$y=\text{−}{x}^2+6x+8$$

ⓐ $x=3$ ⓑ $\left(3,17\right)$

$$y=2x^2-8x+1$$

In the following exercises, find the x- and y-intercepts.

$$y=x^2-4x-5$$

$$y\text{:}(0,5);x\text{:}(5,0),(\mathrm{-1},0)$$

$$y=x^2-8x+15$$

$$y=x^2-4x+10$$

$$y\text{:}\left(0,10\right);x\text{:}\text{none}$$

$$y=\mathrm{-5}{x}^2-30x-46$$

$$y=16x^2-8x+1$$

$$y\text{:}(0,1);x\text{:}\left(\frac{1}{4},0\right)$$

$$y=x^2+16x+64$$

In the following exercises, graph by using intercepts, the vertex, and the axis of symmetry.

$$y=x^2+8x+15$$

$y\text{:}(0,15);x\text{:}(\mathrm{-3},0),(\mathrm{-5},0);$
axis: $x=\mathrm{-4};\text{vertex}\text{:}(\mathrm{-4},\mathrm{-1})$

$$y=x^2-2x-3$$

$$y=\text{−}{x}^2+8x-16$$

$y\text{:}(0,\mathrm{-16});x\text{:}(4,0);$
axis: $x=4;\text{vertex}\text{:}(4,0)$

$$y=4x^2-4x+1$$

$$y=x^2+6x+13$$

$y\text{:}(0,13);x\text{:}\text{none};$
axis: $x=\mathrm{-3};\text{vertex}\text{:}(\mathrm{-3},4)$

$$y=\mathrm{-2}{x}^2-8x-12$$

$$y=\mathrm{-4}{x}^2+16x-11$$

$y\text{:}(0,\mathrm{-11})x\text{:}(3.1,0),(0.9,0);$
axis: $x=2;\text{vertex:}(2,5)$

$$y=x^2+8x+10$$

In the following exercises, find the minimum or maximum value.

$$y=7x^2+14x+6$$

The minimum value is $\mathrm{-1}$ when $x=\mathrm{-1}$.

$$y=\mathrm{-3}{x}^2+12x-10$$

In the following exercises, solve. Rounding answers to the nearest tenth.

A ball is thrown upward from the ground with an initial velocity of 112 ft/sec. Use the quadratic equation $h=\mathrm{-16}{t}^2+112t$ to find how long it will take the ball to reach maximum height, and then find the maximum height.

In 3.5 seconds the ball is at its maximum height of 196 feet.

A daycare facility is enclosing a rectangular area along the side of their building for the children to play outdoors. They need to maximize the area using 180 feet of fencing on three sides of the yard. The quadratic equation $A=\mathrm{-2}{x}^2+180x$ gives the area, $A$, of the yard for the length, $x$, of the building that will border the yard. Find the length of the building that should border the yard to maximize the area, and then find the maximum area.

Practice Test

Solve the following quadratic equations. Use any method.

Use the discriminant to determine the number of solutions of each quadratic equation.

Solve by factoring, the Square Root Property, or the Quadratic Formula.

For each parabola, find ⓐ which ways it opens, ⓑ the axis of symmetry, ⓒ the vertex, ⓓ the x- and y-intercepts, and ⓔ the maximum or minimum value.

Graph the following parabolas by using intercepts, the vertex, and the axis of symmetry.

$y\text{:}(0,9);x\text{:}\left(-\frac{3}{4},0\right);$
axis: $x=-\frac{3}{4};\text{vertex}\text{:}\left(-\frac{3}{4},0\right)$

Solve.