Subsection$x$-Intercepts of a Parabola

So far we have graphed quadratic equations by plotting points. But we can use some algebraic techniques to make the process easier.

In Lesson 6.2 we saw that the solutions of the equation

\begin{equation*} 40p=p^2=0 \end{equation*}

namely, $p=0$ and $p=40\text{,}$ are also the horizontal intercepts of the graph of

\begin{equation*} R=40p-p^2 \end{equation*}

This is the same connection we saw between the $x$-intercepts of a line and the solutions of its equation. Recall that the $x$-intercept of the line $y=mx+b$ is the point where the graph crosses the $x$-axis. To find the $x$-intercepts of a line, we set $y=0$ and solve for $x\text{.}$ The same strategy applies to quadratic equations.

$x$-Intercepts of a Parabola.

To find the $x$-intercepts of the graph of

\begin{equation*} y=ax^2+bx+c \end{equation*}

we set $y=0$ and solve the equation

\begin{equation*} ax^2+bx+c=0 \end{equation*}
Example6.20.
1. Find the intercepts of the graph of $y=4x-x^2\text{.}$
2. Sketch the graph.
Solution
1. We set $y=0$ and solve the equation
\begin{align*} 4x-x^2 \amp =0 \amp \amp \blert{\text{Factor the left side.}}\\ x(4-x) \amp =0 \amp \amp \blert{\text{Set each factor equal to 0.}}\\ x=0~~~~4-x \amp = 0 \amp \amp \blert{\text{Solve each equation.}}\\ x = 0~~~~x \amp = 4 \end{align*}
The $x$-intercepts are the points $(0,0)$ and $(4,0)\text{.}$
2. We make a table of values that includes the $x$-intercepts.

 $x$ $-1$ $0$ $1$ $2$ $3$ $4$ $5$ $y$ $-5$ $0$ $3$ $4$ $3$ $0$ $-5$

We plot the points and connect them with a parabola, as shown in the figure.

1.

How do we find the $x$-intercepts of the graph of $y=ax^2+bx+c\text{?}$

SubsectionThe Vertex

The graph of a quadratic equation $y=ax^2+bx+c$ is a smooth curve, called a parabola, that bends upwards or downwards.

The high or low point of a parabola is called its vertex.

Look again at the graph of $y=4x-x^2$ from Example 6.20, shown at right. Notice that the graph is symmetric about a vertical line (called the axis of symmetry) that passes through the vertex. All the parabolas we'll study have a vertical axis of symmetry.

Look Closer.

Because of this symmetry, the $x$-intercepts are located at equal distances on either side of the axis of symmetry. Or we can say that the $x$-coordinate of the vertex is exactly halfway between the two $x$-intercepts.

We can locate the vertex of a parabola by taking the average of its $x$-intercepts.

2.

What is the vertex of a parabola?

Example6.21.

Find the vertex of the graph of $y=4x-x^2.$

Solution

In Example 1 we found that the $x$-intercepts of the graph are $x=0$ and $x=4\text{.}$ The $x$-coordinate of the vertex is the average of these two numbers.

\begin{equation*} x=\dfrac{0+4}{2} \end{equation*}

The $x$-coordinate of the vertex is $x=2\text{.}$ To find the $y$-coordinate of the vertex, we substitute $x=\alert{2}$ into the equation of the parabola.

The coordinates of the vertex are $(2,4)\text{,}$ as you can see in the graph above.

The Vertex of a Parabola.
1. The $x$-coordinate of the vertex is the average of the $x$-intercepts.
2. To find the $y$-coordinate of the vertex, substitute its $x$-coordinate into the equation of the parabola.

3.

How can we find the $x$-coordinate of the vertex of a parabola?

4.

How can we find the $y$-coordinate of the vertex of a parabola?

SubsectionGraphing Parabolas

By locating the $x$-intercepts and the vertex of the graph, we can make a quick sketch of a parabola.

Example6.22.

Sketch a graph of $~y=x^2-8x+7\text{.}$

Solution

First we find the $x$-intercepts: we substitute $y=0$ into the equation, and solve for $x\text{.}$

\begin{align*} x^2-8x+7 \amp =0 \amp \amp \blert{\text{Factor the left side.}}\\ (x-7)(x-1) \amp =0 \amp \amp \blert{\text{Set each factor equal to 0.}}\\ x-7=0~~~~x-1 \amp = 0 \amp \amp \blert{\text{Solve each equation.}}\\ x = 7~~~~x \amp = 1 \end{align*}

The $x$-intercepts are the points $(7,0)$ and $(1,0)\text{.}$ Next, we locate the vertex of the graph. The $x$-coordinate of the vertex is the average of the $x$-intercepts, so

\begin{equation*} x=\dfrac{1+7}{2}=4 \end{equation*}

You can check that $x=4$ is halfway between the two $x$-intercepts of the graph shown in the figure. To find the $y$-coordinate of the vertex, we evaluate the formula for the parabola at$x=\alert{4}\text{.}$

The vertex of the parabola is the point $(4,-9)\text{.}$ We can also find the $y$-intercept of the graph by substituting $x=\alert{0}$ into the equation.

The $y$-intercept is the point $(0,7)\text{.}$ We plot the vertex and the intercepts and draw a smooth curve through them. The completed graph is shown at right.

3.

How can we find the $y$-intercept of the graph of $y=ax^2+bx+c\text{?}$

By combining the techniques we studied in this Lesson, we write the following guidelines for sketching an accurate graph of a quadratic equation.

1. Find the $x$-intercepts: set $y=0$ and solve for $x\text{.}$
2. Find the vertex: the $x$-coordinate is the average of the $x$-intercepts. Find the $y$-coordinate by substituting the $x$-coordinate into the equation of the parabola.
3. Find the $y$-intercept: set $x=0$ and solve for $y\text{.}$
4. Draw a parabola through the points. The graph is symmetric about a vertical line through the vertex.

SubsectionSkills Warm-Up

ExercisesExercises

1. Solve the equation.
2. Write the equation in the form $ax+b=0\text{.}$
3. Graph the equation $y=ax+b\text{.}$
4. Find the $x$-intercept of your graph. Compare with your answer to part (a).
1.
$2x+5=11$
2.
$2x-3=5x+9$
3.
$0.7x+0.2(100-x) = 0.3(100)$
4.
$4(7-x)=-2(6x-5)-6$

ExercisesHomework 6.4

For Problems 1–3, without graphing, find the $x$-intercepts of the graph of the equation.

1.
$y=(x-3)(2x+5)$
2.
$y=2x^2-6x$
3.
$y=8x-3x^2$

For Problems 4–5, find the intercepts and graph the two parabolas on the same grid.

4.
1. $y=x^2+1$
2. $y=x^2-4$
5.
1. $y=-x^2-3$
2. $y=1-x^2$
6.
1. For each graph in Problems 4 and 5, give the coordinates of the vertex. Then use your answers to part (a) to help you answer the following questions. In these questions, $k$ is a positive constant.
2. What is the vertex of the graph of $y=ax^2+k\text{?}$
3. What is the vertex of the graph of $y=ax^2-k\text{?}$

For Problems 7–8,

1. Make a table of values and sketch the graph.
2. What is the vertex of the graph?
7.

$y=(x+2)^2$

8.

$y=(x-3)^2$

For Problems 9–12,

1. Find the $x$-intercepts and the $y$-intercept of the graph.
2. Find the vertex of the graph.
3. Sketch the graph.
9.

$y=x^2+2x$

10.

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

11.

$y=9-x^2$

12.

$y=-2x^2+12x-10$

For Problems 13–14,

1. Find the $x$- and $y$-intercepts of the parabola.
2. Find the vertex of the parabola.
3. Graph the pair of parabolas on the same grid and compare.
13.
1. $y=x^2-x-2$
2. $y=-x^2+x+2$
14.
1. $y=x^2-2x-15$
2. $y=x^2+2x-15$
15.

The bridge over the Rushing River at Marionville is 48 feet high. Francine stands on the bridge and tosses a rock into the air off the edge of the bridge. The height of the rock above the water $t$ seconds later is given in feet by

\begin{equation*} h=48+32t-16t^2 \end{equation*}
1. Complete the table of values.

 $t$ $0$ $0.5$ $1$ $1.5$ $2$ $2.5$ $3$ $h$ $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$
2. Sketch a graph of the equation on the grid.

3. Use the graph to estimate the height of the rock after 1.75 seconds. Verify your answer algebraically by substituting $t=1.75$ into the equation for $h\text{.}$
16.
1. Use your graph from Problem 15 to answer the question: When is the rock about 40 feet above the water?
2. Write an equation that you could solve to answer the question in part (a).
17.

How long is the rock in Problem 15 more than 60 feet high?

18.

Refer to Problem 15.

1. What is the highest point the rock reaches?
2. After reaching its highest point, how long is the rock falling before it hits the water?