Find One Solution to the Quadratic Equation

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

Recognize the Graph of a Quadratic Equation in Two Variables

We have graphed equations of the form . 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 between and 2 and find their values. See (Figure).

$y={x}^{2}$

0	0
1	1
	1
2	4
	4

Notice when we let and , we got the same value for .

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

The same thing happened when we let and .

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

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, 1). Other points on the curve are located at (-2, 5), (-1, 2), (1, 2) and (2, 5).

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).

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

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).

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.

The figure shows the steps to find the vertex for two parabolas. On the left side is the given equation y equals x squared plus 4 x plus 3. Below the equation is the statement

Axis of Symmetry and Vertex of a Parabola

For a parabola with equation $y=a{x}^{2}+bx+c$ :

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.

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

ⓐ ⓑ $\left(2,-7\right)$

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

ⓐ ⓑ $\left(1,-5\right)$

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

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

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 .
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 .
This quadratic does not factor, so we use the Quadratic Formula.
, ,
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 $y={x}^{2}-2x-8$ .

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 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 .

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 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.

This figure shows three graphs side by side. The leftmost graph shows an upward-opening parabola graphed on the x y-coordinate plane. The vertex of the parabola is in the lower right quadrant. Below the graph is the inequality b squared minus 4 a c greater than 0. Below that is the statement

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

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

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

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

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

Graph a quadratic equation in two variables.

Write the quadratic equation with 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 $y=\text{−}{x}^{2}+6x-9$ .

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

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

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

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

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

Finding the y-intercept by substituting 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 $y=2{x}^{2}-4x-3$ .

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

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

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 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

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

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

The minimum value is when .

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

The maximum value is 5 when .

We have used the formula

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

to calculate the height in feet, , of an object shot upwards into the air with initial velocity, ${v}_{0}$ , after seconds.

This formula is a quadratic equation in the variable , 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 .
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 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.

Key Concepts

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

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

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

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

Find the Axis of Symmetry and Vertex of a Parabola

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

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

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

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.

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

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

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

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

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

$y=\text{−}{x}^{2}+8x-16$

$y=\text{−}{x}^{2}+2x-7$

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

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

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

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

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

$y=\text{−}{x}^{2}-4x+2$

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

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

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

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

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

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 .

$y=\text{−}{x}^{2}+4x-5$

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

The maximum value is 16 when .

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

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 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, , for the length, , 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.

Everyday Math

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.

This table has six rows and four columns. The first row is a header row and it labels each column. The first column is labeled "I can …", the second "Confidently", the third

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

Chapter 10 Review Exercises

10.3 Solve Quadratic Equations Using the Quadratic Formula

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

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

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

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

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

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

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

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

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

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

$m=\frac{-4±\sqrt{10}}{3}$

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

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

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.

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

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 and

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

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

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

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

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

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

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.

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

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

$y=\text{−}{x}^{2}+8x-16$

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

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

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

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

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

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

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

The minimum value is when .

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

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.

Practice Test

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

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.

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

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

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 and 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.

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

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

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

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

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

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.

Find One Solution to the Quadratic Equation

Source: https://opentextbc.ca/elementaryalgebraopenstax/chapter/graphing-quadratic-equations/

Froggatt Yourselly64

Find One Solution to the Quadratic Equation

85 Graphing Quadratic Equations

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

10.3 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

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