Graphing Quadratic Equations

Lynn Marecek; MaryAnne Anthony-Smith

Quadratic Equations

85 Graphing Quadratic Equations

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

Before you get started, take this readiness quiz.

Graph the equation $y=3x-5$ by plotting points.
If you missed this problem, review (Figure).
Evaluate $2{x}^{2}+4x-1$ when $x=-3$ .
If you missed this problem, review (Figure).
Evaluate $-\frac{b}{2a}$ when $a=\frac{1}{3}$ and $b=\frac{5}{6}$ .
If you missed this problem, review (Figure).

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

$y=a{x}^{2}+bx+c$

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 $-2$ and 2 and find their $y$ values. See (Figure).

$y={x}^{2}$
$x$	$y$
0	0
1	1
$-1$	1
2	4
$-2$	4

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

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

The same thing happened when we let $x=2$ and $x=-2$ .

Now, we will plot the points to show the graph of $y={x}^{2}$ . See (Figure).

This figure shows an upward-opening u shaped curve graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The lowest point on the curve is at the point (0, 0). Other points on the curve are located at (-2, 4), (-1, 1), (1, 1) and (2, 4).

The graph is not a line. This figure is called a parabola. Every quadratic equation has a graph that looks like this.

In (Figure) you will practice graphing a parabola by plotting a few points.

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 $y={x}^{2}-1$ .

Graph $y=\text{−}{x}^{2}$ .

This figure shows a downward-opening u shaped curve graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The highest point on the curve is at the point (0, 0). Other points on the curve are located at (-2, -4), (-1, -1), (1, -1) and (2, -4).

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).

This figure shows two graphs side by side. The graph on the left side shows an upward-opening u shaped curve graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The lowest point on the curve is at the point (-2, -1). Other points on the curve are located at (-3, 0), and (-1, 0). Below the graph is the equation y equals a squared plus b x plus c. Below that is the equation of the graph, y equals x squared plus 4 x plus 3. Below that is the inequality a greater than 0 which means the parabola opens upwards. The graph on the right side shows a downward-opening u shaped curve graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The highest point on the curve is at the point (2, 7). Other points on the curve are located at (0, 3), and (4, 3). Below the graph is the equation y equals a squared plus b x plus c. Below that is the equation of the graph, y equals negative x squared plus 4 x plus 3. Below that is the inequality a less than 0 which means the parabola opens downwards.

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). 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:

The image shows two statements. The first statement reads “a greater than 0, the parabola opens upwards”. This statement is followed by the image of an upward opening parabola. The second statement reads “a less than 0, the parabola opens downward”. This statement is followed by the image of a downward opening parabola.

Determine whether each parabola opens upward or downward:

ⓐ $y=-3{x}^{2}+2x-4$ ⓑ $y=6{x}^{2}+7x-9$

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.

Determine whether each parabola opens upward or downward:

ⓐ $y=2{x}^{2}+5x-2$ ⓑ $y=-3{x}^{2}-4x+7$

ⓐ up ⓑ down

Determine whether each parabola opens upward or downward:

ⓐ $y=-2{x}^{2}-2x-3$ ⓑ $y=5{x}^{2}-2x-1$

ⓐ down ⓑ up

Find the Axis of Symmetry and Vertex of a Parabola

Look again at (Figure). 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 red. See (Figure).

This figure shows an two graphs side by side. The graph on the left side shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The lowest point on the curve is at the point (-2, -1). Other points on the curve are located at (-3, 0), and (-1, 0). Also on the graph is a dashed vertical line that goes through the center of the parabola at the point (-2, -1). Below the graph is the equation of the graph, y equals x squared plus 4 x plus 3. The graph on the right side shows an downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The highest point on the curve is at the point (2, 7). Other points on the curve are located at (0, 3), and (4, 3). Also on the graph is a dashed vertical line that goes through the center of the parabola at the point (2, 7). Below the graph is the equation of the graph, y equals negative x squared plus 4 x plus 3.

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). 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.

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.

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 x-coordinate will be $x=1$ .
Substitute $x=1$ into the equation and solve for y.
Simplify.
This is the y-coordinate.		$y=-1$ The vertex is $\left(1,\text{−}1\right).$

For the parabola $y=2{x}^{2}-8x+1$ find: ⓐ the axis of symmetry and ⓑ the vertex.

ⓐ $x=2$ ⓑ $\left(2,-7\right)$

For the parabola $y=2{x}^{2}-4x-3$ find: ⓐ the axis of symmetry and ⓑ the vertex.

ⓐ $x=1$ ⓑ $\left(1,-5\right)$

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).

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 $\left(\text{−}1,0\right)$ and $\left(\text{−}3,0\right).$

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.
$a=-1$ , $b=4$ , $c=3$
Simplify.
		The x intercepts are $\left(2+\sqrt{7},0\right)$ and $\left(2-\sqrt{7},0\right)$ .

We will use the decimal approximations of the x-intercepts, so that we can locate these points on the graph.

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

Do these results agree with our graphs? See (Figure).

Find the intercepts of a parabola.

To find the intercepts of a parabola with equation $y=a{x}^{2}+bx+c$ :

$\begin{array}{cccc}\hfill {\text{y}}\mathbf{\text{-intercept}}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}{\text{x}}\mathbf{\text{-intercepts}}\hfill \\ \hfill \text{Let}\phantom{\rule{0.2em}{0ex}}x=0\phantom{\rule{0.2em}{0ex}}\text{and solve for}\phantom{\rule{0.2em}{0ex}}y.\hfill & & & \hfill \phantom{\rule{4em}{0ex}}\text{Let}\phantom{\rule{0.2em}{0ex}}y=0\phantom{\rule{0.2em}{0ex}}\text{and solve for}\phantom{\rule{0.2em}{0ex}}x.\hfill \end{array}$

Find the intercepts of the parabola $y={x}^{2}-2x-8$ .

Solution


To find the y-intercept, let $x=0$ and solve for y.
		When $x=0$ , then $y=-8$ . The y-intercept is the point $\left(0,-8\right)$ .

To find the x-intercept, let $y=0$ and solve for x.
Solve by factoring.

When $y=0$ , then $x=4\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}x=-2$ . The x-intercepts are the points $\left(4,0\right)$ and $\left(-2,0\right)$ .

Find the intercepts of the parabola $y={x}^{2}+2x-8.$

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,-8\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\left(-4,0\right),\left(2,0\right)$

Find the intercepts of the parabola $y={x}^{2}-4x-12.$

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,-12\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\left(6,0\right),\left(-2,0\right)$

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:

$\begin{array}{cccc}\hfill \mathbf{\text{Quadratic equation}}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}\mathbf{\text{Quadratic equation in two variables}}\hfill \\ & & & \hfill \phantom{\rule{4em}{0ex}}y={x}^{2}-2x-15\hfill \\ \hfill \begin{array}{ccc}\hfill {x}^{2}-2x-15& =\hfill & 0\hfill \\ \hfill \left(x-5\right)\left(x+3\right)& =\hfill & 0\hfill \end{array}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}\text{let}\phantom{\rule{0.2em}{0ex}}y=0\phantom{\rule{1em}{0ex}}\begin{array}{c}0={x}^{2}-2x-15\hfill \\ 0=\left(x-5\right)\left(x+3\right)\hfill \end{array}\hfill \\ \hfill \begin{array}{cccccccc}\hfill x-5& =\hfill & 0\hfill & & & \hfill x+3& =\hfill & 0\hfill \\ \hfill x& =\hfill & 5\hfill & & & \hfill x& =\hfill & -3\hfill \end{array}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}\begin{array}{cccccccc}\hfill x-5& =\hfill & 0\hfill & & & \hfill x+3& =\hfill & 0\hfill \\ \hfill x& =\hfill & 5\hfill & & & \hfill x& =\hfill & -3\hfill \end{array}\hfill \\ & & & \hfill \phantom{\rule{4em}{0ex}}\left(5,0\right)\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\left(-3,0\right)\hfill \\ & & & \hfill \phantom{\rule{4em}{0ex}}x\text{-intercepts}\hfill \end{array}$

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.

Find the intercepts of the parabola $y=5{x}^{2}+x+4$ .

Solution


To find the y-intercept, let $x=0$ and solve for y.		When $x=0$ , then $y=4$ . The y-intercept is the point $\left(0,4\right)$ .

To find the x-intercept, let $y=0$ and solve for x.
Find the value of the discriminant to predict the number of solutions and so x-intercepts.		$\begin{array}{c}\begin{array}{ccc}\hfill {b}^{2}& -\hfill & 4ac\hfill \\ \hfill {1}^{2}& -\hfill & 4\cdot 5\cdot 4\hfill \\ 1\hfill & -\hfill & 80\hfill \end{array}\hfill \\ \hfill -79\phantom{\rule{0.8em}{0ex}}\hfill \end{array}$
Since the value of the discriminant is negative, there is no real solution to the equation.		There are no x-intercepts.

Find the intercepts of the parabola $y=3{x}^{2}+4x+4.$

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,4\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\text{none}$

Find the intercepts of the parabola $y={x}^{2}-4x-5.$

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,-5\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\left(5,0\right)\phantom{\rule{0.2em}{0ex}}\left(-1,0\right)$

Find the intercepts of the parabola $y=4{x}^{2}-12x+9$ .

Solution


To find the y-intercept, let $x=0$ and solve for y.
		When $x=0$ , then $y=9$ . The y-intercept is the point $\left(0,9\right)$ .

To find the x-intercept, let $y=0$ and solve for x.
Find the value of the discriminant to predict the number of solutions and so x-intercepts.		$\begin{array}{c}\begin{array}{ccc}\hfill {b}^{2}& -\hfill & 4ac\hfill \\ \hfill {2}^{2}& -\hfill & 4\cdot 4\cdot 9\hfill \\ \hfill 144& -\hfill & 144\hfill \end{array}\hfill \\ \hfill 0\phantom{\rule{1em}{0ex}}\hfill \end{array}$
		Since the value of the discriminant is 0, there is no real solution to the equation. So there is one x-intercept.
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 x-intercept is the point $\left(\frac{3}{2},0\right).$

Find the intercepts of the parabola $y=\text{−}{x}^{2}-12x-36.$

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,-36\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\left(-6,0\right)$

Find the intercepts of the parabola $y=9{x}^{2}+12x+4.$

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,4\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\left(-\frac{2}{3},0\right)$

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.

How To Graph a Quadratic Equation in Two Variables

Graph $y={x}^{2}-6x+8$ .

Solution

The image shows the steps to graph the quadratic equation y equals x squared minus 6 x plus 8. Step 1 is to write the quadratic equation with y on one side. This equation has y on one side already. The value of a is one, the value of b is -6 and the value of c is 8. Step 2 is to determine whether the parabola opens upward or downward. Since a is positive, the parabola opens upward. Step 3 is to find the axis of symmetry. The axis of symmetry is the line x equals negative b divided by the quantity 2 a. Plugging in the values of b and a the formula becomes x equals negative -6 divided by the quantity 2 times 1 which simplifies to x equals 3. The axis of symmetry is the line x equals 3. Step 4 is to find the vertex. The vertex is on the axis of symmetry. Substitute x equals 3 into the equation and solve for y. The equation is y equals x squared minus 6 x plus 8. Replacing x with 3 it becomes y equals 3 squared minus 6 times 3 plus 8 which simplifies to y equals -1. The vertex is (3, -1). Step 6 is to find the x-intercepts. We substitute y equals 0 into the equation. The equation becomes 0 equals x squared minus 6 x plus 8. We can solve this quadratic equation by factoring to get 0 equals the quantity x minus 2 times the quantity x minus 4. Solve each equation to get x equals 2 and x equals 4. The x-intercepts are (2, 0) and (4, 0).

Graph the parabola $y={x}^{2}+2x-8.$

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,-8\right)$ ; $x\text{:}\phantom{\rule{0.2em}{0ex}}\left(2,0\right),\left(-4,0\right)$ ;
axis: $x=-1$ ; vertex: $\left(-1,-9\right)$ ;

Graph the parabola $y={x}^{2}-8x+12.$

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,12\right);\phantom{\rule{0.2em}{0ex}}x\text{:}\phantom{\rule{0.2em}{0ex}}\left(2,0\right),\left(6,0\right);$
axis: $x=4;\phantom{\rule{0.2em}{0ex}}\text{vertex:}\phantom{\rule{0.2em}{0ex}}\left(4,-4\right)$ ;

Graph a quadratic equation in two variables.

Write the quadratic equation with $y$ on one side.
Determine whether the parabola opens upward or downward.
Find the axis of symmetry.
Find the vertex.
Find the y-intercept. Find the point symmetric to the y-intercept across the axis of symmetry.
Find the x-intercepts.
Graph the parabola.

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.

Graph $y=\text{−}{x}^{2}+6x-9$ .

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 $\left(3,0\right).$
The y-intercept occurs when $x=0.$ Substitute $x=0.$ Simplify. The point $\left(0,-9\right)$ is three units to the left of the line of symmetry. The point three units to the right of the line of symmetry is $\left(6,-9\right).$ Point symmetric to the y-intercept is $\left(6,-9\right)$		The y-intercept is $\left(0,-9\right).$
The x-intercept occurs when $y=0.$
Substitute $y=0.$
Factor the GCF.
Factor the trinomial.
Solve for x.
Connect the points to graph the parabola.

Graph the parabola $y=-3{x}^{2}+12x-12.$

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,-12\right);\phantom{\rule{0.2em}{0ex}}x\text{:}\phantom{\rule{0.2em}{0ex}}\left(2,0\right);$
axis: $x=2;\phantom{\rule{0.2em}{0ex}}\text{vertex:}\left(2,0\right)$ ;

The graph shows an downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -10 to 10. The y-axis of the plane runs from -1 to 10. The vertex is at the point (2, 0). One other point is plotted on the curve at (0, -12). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 2.

Graph the parabola $y=25{x}^{2}+10x+1.$

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,1\right);\phantom{\rule{0.2em}{0ex}}x\text{:}\phantom{\rule{0.2em}{0ex}}\left(-\frac{1}{5},0\right);$
axis: $x=-\frac{1}{5};\phantom{\rule{0.2em}{0ex}}\text{vertex:}\phantom{\rule{0.2em}{0ex}}\left(-\frac{1}{5},0\right)$ ;

The graph shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -5 to 5. The y-axis of the plane runs from -5 to 10. The vertex is at the point (-1 fifth, 0). One other point is plotted on the curve at (0, 1). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals -1 fifth.

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 $y={x}^{2}+4x+5$ ?

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 $x=-2.$
The vertex is on the line $x=-2.$
Find y when $x=-2.$		The vertex is $\left(-2,1\right).$
The y-intercept occurs when $x=0.$ Substitute $x=0.$ Simplify. The point $\left(0,5\right)$ is two units to the right of the line of symmetry. The point two units to the left of the line of symmetry is $\left(-4,5\right).$		The y-intercept is $\left(0,5\right).$ Point symmetric to the y- intercept is $\left(-4,5\right)$ .
The x– intercept occurs when $y=0.$
Substitute $y=0.$ Test the discriminant.
		${b}^{2}-4ac$ ${4}^{2}-4\cdot 15$ $16-20$ $\phantom{\rule{1em}{0ex}}-4$
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.

Graph the parabola $y=2{x}^{2}-6x+5.$

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,5\right);\phantom{\rule{0.2em}{0ex}}x\text{:}\phantom{\rule{0.2em}{0ex}}\text{none};$
axis: $x=\frac{3}{2};\phantom{\rule{0.2em}{0ex}}\text{vertex:}\phantom{\rule{0.2em}{0ex}}\left(\frac{3}{2},\frac{1}{2}\right)$ ;

The graph shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -5 to 5. The y-axis of the plane runs from -5 to 10. The vertex is at the point (3 halves, 1 half). One other point is plotted on the curve at (0, 5). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 3 halves.

Graph the parabola $y=-2{x}^{2}-1.$

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,-1\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\text{none};$
axis: $x=0;\phantom{\rule{0.2em}{0ex}}\text{vertex:}\phantom{\rule{0.2em}{0ex}}\left(0,-1\right)$ ;

The graph shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -10 to 10. The y-axis of the plane runs from -10 to 10. The vertex is at the point (0, -1). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 0.

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 (Figure). We will use the Quadratic Formula again in the next example.

Graph $y=2{x}^{2}-4x-3$ .

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 $\left(1,\text{−}5\right)$ .
The y-intercept occurs when $x=0.$
Substitute $x=0.$
Simplify.		The y-intercept is $\left(0,-3\right)$ .
The point $\left(0,-3\right)$ is one unit to the left of the line of symmetry. The point one unit to the right of the line of symmetry is $\left(2,-3\right)$		Point symmetric to the y-intercept is $\left(2,-3\right).$
The x-intercept 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 x-intercepts are $\left(2.5,0\right)$ and $\left(-0.6,0\right)$ .
Graph the parabola using the points found.

Graph the parabola $y=5{x}^{2}+10x+3.$

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,3\right);\phantom{\rule{0.2em}{0ex}}x\text{:}\phantom{\rule{0.2em}{0ex}}\left(-1.6,0\right),\left(-0.4,0\right);$
axis: $x=-1;\phantom{\rule{0.2em}{0ex}}\text{vertex:}\phantom{\rule{0.2em}{0ex}}\left(-1,-2\right)$ ;

Graph the parabola $y=-3{x}^{2}-6x+5.$

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,5\right);\phantom{\rule{0.2em}{0ex}}x\text{:}\phantom{\rule{0.2em}{0ex}}\left(0.6,0\right),\left(-2.6,0\right);$
axis: $x=-1;\phantom{\rule{0.2em}{0ex}}\text{vertex:}\phantom{\rule{0.2em}{0ex}}\left(-1,8\right)$ ;

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).

This figure shows two graphs side by side. The left graph shows an downward-opening parabola graphed on the x y-coordinate plane. The vertex of the parabola is in the upper right quadrant. The vertex is labeled “maximum”. The right graph shows an upward-opening parabola graphed on the x y-coordinate plane. The vertex of the parabola is in the lower right quadrant. The vertex is labeled “minimum”.

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.

Find the minimum value of the quadratic equation $y={x}^{2}+2x-8$ .

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 $x=-1$ .
The vertex is on the line $x=-1.$
Find y when $x=-1.$		The vertex is $\left(-1,-9\right)$ .
Since the parabola has a minimum, the y-coordinate of the vertex is the minimum y-value of the quadratic equation.
The minimum value of the quadratic is $-9$ and it occurs when $x=-1$ .
Show the graph to verify the result.

Find the maximum or minimum value of the quadratic equation $y={x}^{2}-8x+12$ .

The minimum value is $-4$ when $x=4$ .

Find the maximum or minimum value of the quadratic equation $y=-4{x}^{2}+16x-11$ .

The maximum value is 5 when $x=2$ .

We have used the formula

$h=-16{t}^{2}+{v}_{0}t+{h}_{0}$

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.

The quadratic equation $h=-16{t}^{2}+{v}_{0}t+{h}_{0}$ 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

$h=-16{t}^{2}+176t+4$

Since a is negative, the parabola opens downward.

The quadratic equation has a maximum.

ⓐ
$\begin{array}{cccc}\text{Find the axis of symmetry.}\hfill & & & \phantom{\rule{4em}{0ex}}\begin{array}{c}t=-\frac{b}{2a}\hfill \\ t=-\frac{176}{2\left(-16\right)}\hfill \\ t=5.5\hfill \end{array}\hfill \\ & & & \phantom{\rule{4em}{0ex}}\text{The axis of symmetry is}\phantom{\rule{0.2em}{0ex}}t=5.5.\hfill \\ \text{The vertex is on the line}\phantom{\rule{0.2em}{0ex}}t=5.5.\hfill & & & \phantom{\rule{4em}{0ex}}\text{The maximum occurs when}\phantom{\rule{0.2em}{0ex}}t=5.5\phantom{\rule{0.2em}{0ex}}\text{seconds.}\hfill \end{array}$

ⓑ

Find h when $t=5.5$ .
Use a calculator to simplify.
		The vertex is $\left(5.5,488\right)$ .
Since the parabola has a maximum, the h-coordinate of the vertex is the maximum y-value of the quadratic equation.		The maximum value of the quadratic is 488 feet and it occurs when $t=5.5$ seconds.

The quadratic equation $h=-16{t}^{2}+128t+32$ 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.

It will take 4 seconds to reach the maximum height of 288 feet.

A toy rocket shot upward from the ground at a rate of 208 ft/sec has the quadratic equation of $h=-16{t}^{2}+208t$ . When will the rocket reach its maximum height? What will be the maximum height? Round answers to the nearest tenth.

It will take 6.5 seconds to reach the maximum height of 676 feet.

Access these online resources for additional instruction and practice graphing quadratic equations:

Key Concepts

The graph of every quadratic equation is a parabola.
Parabola Orientation For the quadratic equation , 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 :
- 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 $y=a{x}^{2}+bx+c$ :
$\begin{array}{cccc}\hfill {\text{y}}\mathbf{\text{-intercept}}& & & \hfill {\text{x}}\mathbf{\text{-intercepts}}\\ \hfill \text{Let}\phantom{\rule{0.2em}{0ex}}x=0\phantom{\rule{0.2em}{0ex}}\text{and solve for}\phantom{\rule{0.2em}{0ex}}y.& & & \hfill \text{Let}\phantom{\rule{0.2em}{0ex}}y=0\phantom{\rule{0.2em}{0ex}}\text{and solve for}\phantom{\rule{0.2em}{0ex}}x.\end{array}$
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:

$y={x}^{2}+3$

This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -10 to 10. The y-axis of the plane runs from -10 to 10. The parabola has a vertex at (0, 3) and goes through the point (1, 4).

$y=\text{−}{x}^{2}+1$

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

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

down

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

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

up

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

Find the Axis of Symmetry and Vertex of a Parabola

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

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

ⓐ $x=-4$ ⓑ $\left(-4,-17\right)$

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

$y=\text{−}{x}^{2}+2x+5$

ⓐ $x=1$ ⓑ $\left(1,6\right)$

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

Find the Intercepts of a Parabola

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

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

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,6\right);\phantom{\rule{0.2em}{0ex}}x\text{:}\phantom{\rule{0.2em}{0ex}}\left(-1,0\right),\left(-6,0\right)$

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

$y=\text{−}{x}^{2}+8x-19$

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,19\right);\phantom{\rule{0.2em}{0ex}}x\text{:}\phantom{\rule{0.2em}{0ex}}\text{none}$

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

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

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,25\right);\phantom{\rule{0.2em}{0ex}}x\text{:}\phantom{\rule{0.2em}{0ex}}\left(\frac{5}{2},0\right)$

$y=\text{−}{x}^{2}-14x-49$

Graph Quadratic Equations in Two Variables

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

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

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,5\right);\phantom{\rule{0.2em}{0ex}}x\text{:}\phantom{\rule{0.2em}{0ex}}\left(-1,0\right),\left(-5,0\right);$
axis: $x=-3;\phantom{\rule{0.2em}{0ex}}\text{vertex}\text{:}\phantom{\rule{0.2em}{0ex}}\left(-3,-4\right)$

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

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

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,3\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\left(-1,0\right),\left(-3,0\right);$
axis: $x=-2;\phantom{\rule{0.2em}{0ex}}\text{vertex:}\phantom{\rule{0.2em}{0ex}}\left(-2,-1\right)$

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

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

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,4\right)\phantom{\rule{0.2em}{0ex}}x\text{:}\phantom{\rule{0.2em}{0ex}}\left(-\frac{2}{3},0\right);$
axis: $x=-\frac{2}{3};\phantom{\rule{0.2em}{0ex}}\text{vertex:}\phantom{\rule{0.2em}{0ex}}\left(-\frac{2}{3},0\right)$

This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -5 to 5. The y-axis of the plane runs from -5 to 5. The parabola has points plotted at the vertex (-2 thirds, 0) and the intercept (0, 4). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals -2 thirds.

$y=\text{−}{x}^{2}+8x-16$

$y=\text{−}{x}^{2}+2x-7$

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,-7\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\text{none};$
axis: $x=1;\text{vertex}\text{:}\phantom{\rule{0.2em}{0ex}}\left(1,-6\right)$

This figure shows a downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -10 to 10. The y-axis of the plane runs from -15 to 5. The parabola has points plotted at the vertex (1, -6) and the intercept (0, -7). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 1.

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

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

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,1\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\left(1.7,0\right),\left(0.3,0\right);$
axis: $x=1;\phantom{\rule{0.2em}{0ex}}\text{vertex}\text{:}\phantom{\rule{0.2em}{0ex}}\left(1,-1\right)$

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

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

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,2\right)\phantom{\rule{0.2em}{0ex}}x\text{:}\phantom{\rule{0.2em}{0ex}}\left(1,0\right);$
axis: $x=1;\phantom{\rule{0.2em}{0ex}}\text{vertex:}\phantom{\rule{0.2em}{0ex}}\left(1,0\right)$

This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -10 to 10. The y-axis of the plane runs from -10 to 10. The parabola has points plotted at the vertex (1, 0) and the intercept (0, 2). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 1.

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

$y=\text{−}{x}^{2}-4x+2$

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,2\right)\phantom{\rule{0.2em}{0ex}}x\text{:}\phantom{\rule{0.2em}{0ex}}\left(-4.4,0\right),\left(0.4,0\right);$
axis: $x=-2;\phantom{\rule{0.2em}{0ex}}\text{vertex:}\phantom{\rule{0.2em}{0ex}}\left(-2,6\right)$

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

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

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,8\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\text{none};$
axis: $x=1;\text{vertex}\text{:}\phantom{\rule{0.2em}{0ex}}\left(1,3\right)$

This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -10 to 10. The y-axis of the plane runs from -10 to 10. The parabola has points plotted at the vertex (1, 3) and the intercept(0, 8). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 1.

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

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

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,20\right)\phantom{\rule{0.2em}{0ex}}x\text{:}\phantom{\rule{0.2em}{0ex}}\left(-4.5,0\right),\left(-1.5,0\right);$
axis: $x=-3;\phantom{\rule{0.2em}{0ex}}\text{vertex:}\phantom{\rule{0.2em}{0ex}}\left(-3,-7\right)$

This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -10 to 10. The y-axis of the plane runs from -10 to 10. The parabola has points plotted at the vertex (-3, -7) and the intercepts (-4.5, 0) and (-1.5, 0). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals -3.

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

Solve Maximum and Minimum Applications

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

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

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

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

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

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

$y=\text{−}{x}^{2}+4x-5$

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

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

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

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 $h=-16{t}^{2}+168t+45$ 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 $h=-16{t}^{2}+160t+20$ 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.

20 computers will give the maximum of ?400 in receipts.

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 120 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.

Everyday Math

In the previous set of exercises, you worked with the quadratic equation $R=\text{−}{x}^{2}+40x$ 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 $R=\text{−}{x}^{2}+40x$ . ⓑ Find the values of the x-intercepts.

ⓐ
ⓑ $\left(0,0\right),\left(40,0\right)$

In the previous set of exercises, you worked with the quadratic equation $R=\text{−}{x}^{2}+100x$ 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 $R=\text{−}{x}^{2}+100x$ . ⓑ Find the values of the x-intercepts.

Writing Exercises

For the revenue model in (Figure) and (Figure), explain what the x-intercepts mean to the computer store owner.

Answers will vary.

For the revenue model in (Figure) and (Figure), explain what the x-intercepts 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?

Chapter 10 Review Exercises

10.1 Solve Quadratic Equations Using the Square Root Property

In the following exercises, solve using the Square Root Property.

${x}^{2}=100$

$x=±\phantom{\rule{0.2em}{0ex}}10$

${y}^{2}=144$

${m}^{2}-40=0$

$m=±\phantom{\rule{0.2em}{0ex}}2\sqrt{10}$

${n}^{2}-80=0$

$4{a}^{2}=100$

$a=±\phantom{\rule{0.2em}{0ex}}5$

$2{b}^{2}=72$

${r}^{2}+32=0$

no solution

${t}^{2}+18=0$

$\frac{4}{3}{v}^{2}+4=28$

$v=±\phantom{\rule{0.2em}{0ex}}3\sqrt{2}$

$\frac{2}{3}{w}^{2}-20=30$

$5{c}^{2}+3=19$

$c=±\phantom{\rule{0.2em}{0ex}}\frac{4\sqrt{5}}{5}$

$3{d}^{2}-6=43$

In the following exercises, solve using the Square Root Property.

${\left(p-5\right)}^{2}+3=19$

$p=1,9$

${\left(q+4\right)}^{2}=9$

${\left(u+1\right)}^{2}=45$

$u=-1±3\sqrt{5}$

${\left(z-5\right)}^{2}=50$

${\left(x-\frac{1}{4}\right)}^{2}=\frac{3}{16}$

$x=\frac{1}{4}±\frac{\sqrt{3}}{4}$

${\left(y-\frac{2}{3}\right)}^{2}=\frac{2}{9}$

${\left(m-7\right)}^{2}+6=30$

$m=7±2\sqrt{6}$

${\left(n-4\right)}^{2}-50=150$

${\left(5c+3\right)}^{2}=-20$

no solution

${\left(4c-1\right)}^{2}=-18$

${m}^{2}-6m+9=48$

$m=3±4\sqrt{3}$

${n}^{2}+10n+25=12$

$64{a}^{2}+48a+9=81$

$a=-\frac{3}{2},\frac{3}{4}$

$4{b}^{2}-28b+49=25$

10.2 Solve Quadratic Equations Using Completing the Square

In the following exercises, complete the square to make a perfect square trinomial. Then write the result as a binomial squared.

${x}^{2}+22x$

${\left(x+11\right)}^{2}$

${y}^{2}+6y$

${m}^{2}-8m$

${\left(m-4\right)}^{2}$

${n}^{2}-10n$

${a}^{2}-3a$

${\left(a-\frac{3}{2}\right)}^{2}$

${b}^{2}+13b$

${p}^{2}+\frac{4}{5}p$

${\left(p+\frac{2}{5}\right)}^{2}$

${q}^{2}-\frac{1}{3}q$

In the following exercises, solve by completing the square.

${c}^{2}+20c=21$

$c=1,-21$

${d}^{2}+14d=-13$

${x}^{2}-4x=32$

$x=-4,8$

${y}^{2}-16y=36$

${r}^{2}+6r=-100$

no solution

${t}^{2}-12t=-40$

${v}^{2}-14v=-31$

$v=7±3\sqrt{2}$

${w}^{2}-20w=100$

${m}^{2}+10m-4=-13$

$m=-9,-1$

${n}^{2}-6n+11=34$

${a}^{2}=3a+8$

$a=\frac{3}{2}±\frac{\sqrt{41}}{2}$

${b}^{2}=11b-5$

$\left(u+8\right)\left(u+4\right)=14$

$u=-6±2\sqrt{2}$

$\left(z-10\right)\left(z+2\right)=28$

$3{p}^{2}-18p+15=15$

$p=0,6$

$5{q}^{2}+70q+20=0$

$4{y}^{2}-6y=4$

$y=-\frac{1}{2},2$

$2{x}^{2}+2x=4$

$3{c}^{2}+2c=9$

$c=-\frac{1}{3}±\frac{2\sqrt{7}}{3}$

$4{d}^{2}-2d=8$

10.3 Solve Quadratic Equations Using the Quadratic Formula

In the following exercises, solve by using the Quadratic Formula.

$4{x}^{2}-5x+1=0$

$x=\frac{1}{4},1$

$7{y}^{2}+4y-3=0$

${r}^{2}-r-42=0$

$r=-6,7$

${t}^{2}+13t+22=0$

$4{v}^{2}+v-5=0$

$v=-\frac{5}{4},1$

$2{w}^{2}+9w+2=0$

$3{m}^{2}+8m+2=0$

$m=\frac{-4±\sqrt{10}}{3}$

$5{n}^{2}+2n-1=0$

$6{a}^{2}-5a+2=0$

no real solution

$4{b}^{2}-b+8=0$

$u\left(u-10\right)+3=0$

$u=5±\sqrt{22}$

$5z\left(z-2\right)=3$

$\frac{1}{8}{p}^{2}-\frac{1}{5}p=-\frac{1}{20}$

$p=\frac{4±\sqrt{6}}{5}$

$\frac{2}{5}{q}^{2}+\frac{3}{10}q=\frac{1}{10}$

$4{c}^{2}+4c+1=0$

$c=-\frac{1}{2}$

$9{d}^{2}-12d=-4$

In the following exercises, determine the number of solutions to each quadratic equation.

ⓐ $9{x}^{2}-6x+1=0$
ⓑ $3{y}^{2}-8y+1=0$
ⓓ $5{n}^{2}-n+1=0$

ⓐ $5{x}^{2}-7x-8=0$
ⓑ $7{x}^{2}-10x+5=0$
ⓓ $15{x}^{2}-8x+4=0$

In the following exercises, identify the most appropriate method (Factoring, Square Root, or Quadratic Formula) to use to solve each quadratic equation.

ⓐ $16{r}^{2}-8r+1=0$
ⓑ $5{t}^{2}-8t+3=9$

ⓐ $4{d}^{2}+10d-5=21$
ⓑ $25{x}^{2}-60x+36=0$

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 $-17$ and $-19.$

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 image shows a rectangular pathway with a right inlaid in the lower left corner. The right angle of the triangle overlays the lower left corner of the rectangle. The left leg of the right triangle overlays the left side of the rectangle and the hypotenuse of the right triangle runs from the upper left corner of the rectangle to a point on the bottom of the rectangle.

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 8.1 feet and its length is 30.8 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.

The image shows four rectangular tables placed side by side to create one large table.

A ball is thrown vertically in the air with a velocity of 160 ft/sec. Use the formula $h=-16{t}^{2}+{v}_{0}t$ 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 $h=-16{t}^{2}+{v}_{0}t$ 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.

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

down

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

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

up

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

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

$y=\text{−}{x}^{2}+6x+8$

ⓐ $x=3$ ⓑ $\left(3,17\right)$

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

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

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

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,5\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\left(5,0\right),\left(-1,0\right)$

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

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

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,10\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\text{none}$

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

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

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,1\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\left(\frac{1}{4},0\right)$

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

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

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

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,15\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\left(-3,0\right),\left(-5,0\right);$
axis: $x=-4;\text{vertex}\text{:}\phantom{\rule{0.2em}{0ex}}\left(-4,-1\right)$

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

$y=\text{−}{x}^{2}+8x-16$

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,-16\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\left(4,0\right);$
axis: $x=4;\text{vertex}\text{:}\phantom{\rule{0.2em}{0ex}}\left(4,0\right)$

This figure shows a downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -15 to 12. The y-axis of the plane runs from -20 to 2. The parabola has points plotted at the vertex (4, 0) and the intercept (0, -16). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 4.

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

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

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,13\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\text{none};$
axis: $x=-3;\text{vertex}\text{:}\phantom{\rule{0.2em}{0ex}}\left(-3,4\right)$

This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -10 to 10. The y-axis of the plane runs from -2 to 18. The parabola has points plotted at the vertex (-3, 4) and the intercept (0, 13). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals -3.

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

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

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,-11\right)\phantom{\rule{0.2em}{0ex}}x\text{:}\phantom{\rule{0.2em}{0ex}}\left(3.1,0\right),\left(0.9,0\right);$
axis: $x=2;\phantom{\rule{0.2em}{0ex}}\text{vertex:}\phantom{\rule{0.2em}{0ex}}\left(2,5\right)$

This figure shows a downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -10 to 10. The y-axis of the plane runs from -10 to 10. The parabola has points plotted at the vertex (2, 5) and the intercepts (3.1, 0) and (0.9, 0). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 2.

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

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

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

The minimum value is $-1$ when $x=-1$ .

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

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 $h=-16{t}^{2}+112t$ 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 $A=-2{x}^{2}+180x$ 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

Use the Square Root Property to solve the quadratic equation: $3{\left(w+5\right)}^{2}=27$ .

$w=-2,-8$

Use Completing the Square to solve the quadratic equation: ${a}^{2}-8a+7=23$ .

Use the Quadratic Formula to solve the quadratic equation: $2{m}^{2}-5m+3=0$ .

$m=1,\frac{3}{2}$

Solve the following quadratic equations. Use any method.

$8{v}^{2}+3=35$

$3{n}^{2}+8n+3=0$

$n=\frac{-4±\sqrt{7}}{3}$

$2{b}^{2}+6b-8=0$

$x\left(x+3\right)+12=0$

no real solution

$\frac{4}{3}{y}^{2}-4y+3=0$

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

$6{p}^{2}-13p+7=0$

2

$3{q}^{2}-10q+12=0$

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

Find two consecutive even numbers whose product is 360.

Two consecutive even number are $-20$ and $-18$ and 18 and 20.

The length of a diagonal of a rectangle is three more than the width. The length of the rectangle is three times the width. Find the length of the diagonal. (Round to the nearest tenth.)

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.

$y=3{x}^{2}+6x+8$

ⓐ up ⓑ $x=-1$ ⓒ $\left(-1,5\right)$ ⓓ $y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,8\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\text{none}$ ⓔ minimum value of 5 when $x=-1$

$y={x}^{2}-4$

$y={x}^{2}+10x+24$

ⓐ up ⓑ $x=-5$ ⓒ $\left(-5,-1\right)$ ⓓ $y;\left(0,24\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\left(-6,0\right),\left(-4,0\right)$ ⓔ minimum value of $-5$ when $x=-1$

$y=-3{x}^{2}+12x-8$

$y=\text{−}{x}^{2}-8x+16$

ⓐ down ⓑ $x=-4$
ⓒ $\left(-4,32\right)$ ⓓ $y;\left(0,16\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\left(-9.7,0\right),\left(1.7,0\right)$
ⓔ maximum value of $32$ when $x=-4$

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

$y=2{x}^{2}+6x+2$

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

$y\text{:}\phantom{\rule{0.2em}{0ex}}\left(0,9\right);x\text{:}\phantom{\rule{0.2em}{0ex}}\left(-\frac{3}{4},0\right);$
axis: $x=-\frac{3}{4};\text{vertex}\text{:}\phantom{\rule{0.2em}{0ex}}\left(-\frac{3}{4},0\right)$

This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from -10 to 10. The y-axis of the plane runs from -10 to 10. The parabola has points plotted at the vertex (3 fourths, 0) and the intercept (0, 9). Also on the graph is a dashed vertical line representing the axis of symmetry. The line goes through the vertex at x equals 3 fourths.

Solve.

A water balloon is launched upward at the rate of 86 ft/sec. Using the formula $h=-16{t}^{2}+86t$ , find how long it will take the balloon to reach the maximum height and then find the maximum height. Round to the nearest tenth.

Glossary

axis of symmetry: The axis of symmetry is the vertical line passing through the middle of the parabola $y=a{x}^{2}+bx+c.$

parabola: The graph of a quadratic equation in two variables is a parabola.

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 $y=a{x}^{2}+bx+c.$

vertex: The point on the parabola that is on the axis of symmetry is called the vertex of the parabola; it is the lowest or highest point on the parabola, depending on whether the parabola opens upwards or downwards.

x-intercepts of a parabola: The x-intercepts are the points on the parabola where $y=0.$

y-intercept of a parabola: The y-intercept is the point on the parabola where $x=0.$

License

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Elementary Algebra by OSCRiceUniversity is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

Learning Objectives

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

Key Concepts

Section Exercises

Practice Makes Perfect

Everyday Math

Writing Exercises

Self Check

Chapter 10 Review Exercises

10.1 Solve Quadratic Equations Using the Square Root Property

10.2 Solve Quadratic Equations Using Completing the Square

10.3 Solve Quadratic Equations Using the Quadratic Formula

10.4 Solve Applications Modeled by Quadratic Equations

10.5 Graphing Quadratic Equations in Two Variables

Practice Test

Glossary

License

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