6.3 Graphing Parabolas

Introduction

The graph of a quadratic function $f (x) = a x^{2} + b x + c$ is called a parabola. Some parabolas are shown below.

All these parabolas share certain features.

The graph has either a highest point (if the parabola opens downward, as in figure (a) or a lowest point (if the parabola opens upward, as in figure (b). This high or low point is called the vertex of the graph.
The parabola is symmetric about a vertical line, called the axis of symmetry, that runs through the vertex.
The $y$ -intercept is the point where the parabola intersects the $y$ -axis. The graph of a quadratic function always has exactly one $y$ -intercept.
However, the graph may cross the $x$ -axis at one point, at two points, or not at all. Points where the parabola intersects the $x$ -axis are called the $x$ -intercepts. If there are two $x$ -intercepts, they are equidistant from the axis of symmetry.
The values of the constants $a$ , $b$ , and $c$ determine the location and orientation of the parabola. We will begin by considering each of these constants separately.

Which point on a parabola always lies on the axis of symmetry?

_____

The vertex

Which point on a parabola always lies on the axis of symmetry?

The $x$ -intercept
The $y$ -intercept
The vertex
The origin

The Graph of $y = a x^{2}$

In Modeling with Functions, we saw that the graph of $y = a f (x)$ is a transformation of the graph of $y = f (x)$ . The scale factor, $a$ , stretches or compresses the graph vertically, and if $a$ is negative, the graph is reflected about the $x$ -axis.

Example 1

Sketch a graph of each quadratic function by hand.

$y equals 2 x squared$
$y = - \frac{1}{2} x^{2}$

Both functions are of the form $y = a x^{2}$ . The graph of $y equals 2 x squared$ opens upward because $a equals 2 greater than 0$ , and the graph of $y = - \frac{1}{2} x^{2}$ opens downward because $a = - \frac{1}{2} < 0$ .

To make a reasonable sketch by hand, it is enough to plot a few guidepoints; the points with $x$ -coordinates $1$ and $minus 1$ are easy to compute.

$x$	$y equals 2 x squared$	$y = - \frac{1}{2} x^{2}$
$minus 1$	$2$	$- \frac{1}{2}$
$0$	$0$	$0$
$1$	$2$	$- \frac{1}{2}$

We sketch parabolas through each set of guidepoints, as shown at left.

What does the value of $a$ tell us about the graph of $y = a x^{2} + b x + c$ ?

_____

The width of the parabola

What does the value of $a$ tell us about the graph of $y = a x^{2} + b x + c$ ?

The $y$ -intercept
The number of $x$ -intercepts
The $x$ -coordinate of the vertex
The width of the parabola

basic parabola and four vertically scaled parabolas

Match each parabola in the figure above with its equation. The basic parabola is shown in black.

$y = - \frac{3}{4} x^{2}$ _____
$y = \frac{1}{4} x^{2}$ _____
$y = \frac{5}{2} x^{2}$ _____
$y = - \frac{5}{4} x^{2}$ _____

Match each parabola in the figure above with its equation. The basic parabola is shown in black.

$y = - \frac{3}{4} x^{2}$
$y = \frac{1}{4} x^{2}$
$y = \frac{5}{2} x^{2}$
$y = - \frac{5}{4} x^{2}$

The Graph of $y = x^{2} + c$

Next, we consider the effect of the constant term, $c$ , on the graph. Adding a constant $c$ to the formula for $y = f (x)$ causes a vertical translation of the graph.

Example 2

Sketch graphs for the following quadratic functions.

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

The graph of $y = x^{2} - 2$ is shifted downward by two units, compared to the basic parabola. The vertex is the point $open parenthesis 0, minus 2 close parenthesis$ and the $x$ -intercepts are the solutions of the equation
$0 = x^{2} - 2$
or $the square root of 2$ and $- \sqrt{2}$ . The graph is shown below.
The graph of $y = - x^{2} + 4$ opens downward and is shifted $4$ units up, compared to the basic parabola. Its vertex is the point $open parenthesis 0, 4 close parenthesis$ . Its $x$ -intercepts are the solutions of the equation
$0 = - x^{2} + 4$
or $2$ and $minus 2$ . You can verify both graphs with your graphing calculator.

Which of these is the equation of a parabola that has no $x$ -intercepts?

_____

$y = x^{2} + 4$

Which of these is the equation of a parabola that has no $x$ -intercepts?

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

Find an equation for the parabola shown above.
$y =$ _____
Give the $x$ - and $y$ -intercepts of the graph.
$x$ -intercepts: _____ Note: Use a comma to separate different points.
$y$ -intercept: _____

$y = x^{2} - 5$
$x$ -intercepts: $(- \sqrt{5}, 0), (\sqrt{5}, 0)$ ; $y$ -intercept: $open parenthesis 0, minus 5 close parenthesis$

Find an equation for the parabola shown at right.
Give the $x$ - and $y$ -intercepts of the graph.

$y = x^{2} - 5$
$x$ -intercepts: $(- \sqrt{5}, 0), (\sqrt{5}, 0)$ ; $y$ -intercept: $open parenthesis 0, minus 5 close parenthesis$

Describe what the parameters $a$ and $c$ tell you about the graph of $y = a x^{2} + c$ .

_____

Describe what the parameters $a$ and $c$ tell you about the graph of $y = a x^{2} + c$ .

The Graph of $y = a x^{2} + b x$

How does the linear term, $b x$ , affect the graph? Let us begin by considering an example. Graph the function

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

on your calculator. The graph is shown at right.

Note that $a equals 2$ and that $2 greater than 0$ , so the parabola opens upward. We can find the $x$ -intercepts of the graph by setting $y$ equal to zero:

$\begin{aligned} 0 & = 2 x^{2} + 8 x \\ = 2 x (x + 4) \end{aligned}$

The solutions of this equation are $0$ and $minus 4$ , so the $x$ -intercepts are the points $open parenthesis 0, 0 close parenthesis$ and $open parenthesis minus 4, 0 close parenthesis$ .

Recall that the parabola is symmetric about a vertical line through its vertex. (We will prove that this is true in the Homework problems.) The two $x$ -intercepts are equidistant from this line of symmetry, so the $x$ -coordinate of the vertex lies exactly halfway between the $x$ -intercepts. We can average their values to find

$x = \frac{1}{2} [0 + (- 4)] = - 2$

To find the $y$ -coordinate of the vertex, substitute $x = - 2$ into the equation for the parabola:

$\begin{aligned} y & = 2 (- 2)^{2} + 8 (- 2) \\ = 8 - 16 = - 8 \end{aligned}$

Thus, the vertex is the point $(- 2, - 8)$ .

Find the $x$ -intercepts and the vertex of the parabola $y = 6 x - x^{2}$ .
$x$ -intercepts: _____ Note: Use a comma to separate different points.
Vertex: _____
Verify your answers by graphing the function in the window
$\begin{aligned} Xmin & = - 9.4 & Xmax = 9.4 \\ Ymin & = - 10 & Ymax = 10 \end{aligned}$

$x$ -intercepts: $open parenthesis 0, 0 close parenthesis$ and $open parenthesis 6, 0 close parenthesis$ ; vertex: $open parenthesis 3, 9 close parenthesis$

Find the $x$ -intercepts and the vertex of the parabola $y = 6 x - x^{2}$ .
Verify your answers by graphing the function in the window
$\begin{aligned} Xmin & = - 9.4 & Xmax = 9.4 \\ Ymin & = - 10 & Ymax = 10 \end{aligned}$

$x$ -intercepts: $open parenthesis 0, 0 close parenthesis$ and $open parenthesis 6, 0 close parenthesis$ ; vertex: $open parenthesis 3, 9 close parenthesis$

Finding the Vertex

We can use the same method to find a formula for the vertex of any parabola of the form

$y = a x^{2} + b x$

We proceed as we did in the previous example.

First, find the $x$ -intercepts of the graph by setting $y$ equal to zero and solving for $x$ .

$\begin{aligned} 0 & = a x^{2} + b x & Factor. \\ = x (a x + b) \end{aligned}$

Thus,

$\begin{aligned} x & = 0 & or & a x + b & = 0 \\ x & = 0 & or & x & = \frac{- b}{a} \end{aligned}$

parabola with vertex and line of symmetry

The $x$ -intercepts are the points $open parenthesis 0, 0 close parenthesis$ and $(\frac{- b}{a}, 0)$ .

Next, we find the $x$ -coordinate of the vertex by taking the average of the two $x$ -intercepts found above:

$x = \frac{1}{2} [0 + (\frac{- b}{a})] = \frac{- b}{2 a}$

This gives us a formula for the $x$ -coordinate of the vertex.

Also, the axis of symmetry is the vertical line $x = \frac{- b}{2 a}$ as shown in the figure above. Finally, we find the $y$ -coordinate of the vertex by substituting its $x$ -coordinate into the equation for the parabola.

Example 3

Find the vertex of the graph of $f (x) = - 1.8 x^{2} - 16.2 x$ .
Find the $x$ -intercepts of the graph.

The $x$ -coordinate of the vertex is
$x_{v} = \frac{- b}{2 a} = \frac{- (- 16.2)}{2 (- 1.8)} = - 4.5$
To find the $y$ -coordinate of the vertex, evaluate $f (x)$ at $x = - 4.5$ .
$y_{v} = - 1.8 (- 4.5)^{2} - 16.2 (- 4.5) = 36.45$
The vertex is $open parenthesis minus 4.5, 36.45 close parenthesis$ .
To find the $x$ -intercepts of the graph, set $f open parenthesis x close parenthesis equals 0$ and solve.
$\begin{aligned} - 1.8 x^{2} - 16.2 x & = 0 & Factor. \\ - x (1.8 x + 16.2) & = 0 & Set each factor equal to zero. \\ - x = 0 1.8 x + 16.2 & = 0 & Solve each equation. \\ x = 0 x & = - 9 \end{aligned}$
The $x$ -intercepts of the graph are $open parenthesis 0, 0 close parenthesis$ and $open parenthesis minus 9, 0 close parenthesis$ . The graph is shown below.

Explain why the $x$ -coordinate of the vertex is the average of the $x$ -intercepts of the graph.

_____

Explain why the $x$ -coordinate of the vertex is the average of the $x$ -intercepts of the graph.

The Graph of $y = a x^{2} + b x + c$

Now we will see that the vertex formula holds for any parabola. Consider the function

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

Adding $6$ to $2 x squared plus 8 x$ shifts each point on the graph $6$ units upward, as shown at right. The $x$ -coordinate of the vertex will not be affected by an upward shift. Thus, the formula

$x_{v} = \frac{- b}{2 a}$

for the $x$ -coordinate of the vertex still holds. We have

$x_{v} = \frac{- 8}{2 (2)} = - 2$

We find the $y$ -coordinate of the vertex by substituting $x_{v} = - 2$ into the equation for the parabola.

Two upward-opening parabolas on a coordinate grid. The dashed blue graph of y = 2x squared + 8x has its vertex below the x-axis; the solid red graph of y = 2x squared + 8x + 6 is that parabola shifted up 6 units, passing through the points (negative 4, 6) and (0, 6).

$\begin{aligned} y_{v} & = 2 (- 2)^{2} + 8 (- 2) + 6 \\ = 8 - 16 + 6 = - 2 \end{aligned}$

So the vertex is the point $(- 2, - 2)$ . (Notice that this point is shifted $6$ units upward from the vertex of $y = 2 x^{2} + 8 x$ .)

We find the $x$ -intercepts of the graph by setting $y$ equal to zero.

$\begin{aligned} 0 & = 2 x^{2} + 8 x + 6 & Factor the right side. \\ = 2 (x + 1) (x + 3) & Set each factor equal to zero. \\ x + 1 & = 0 or x + 3 = 0 \\ x & = - 1 x = - 3 \end{aligned}$

The $x$ -intercepts are the points $open parenthesis minus 1, 0 close parenthesis$ and $open parenthesis minus 3, 0 close parenthesis$ .

The $y$ -intercept of the graph is found by setting $x$ equal to zero:

$y = 2 (0) 2 + 8 (0) + 6 = 6$

You can see that the $y$ -intercept, $6$ , is just the constant term of the quadratic equation. The completed graph is shown above.

The $y$ -intercept of $y = a x^{2} + b x + c$ is the same as the value of which parameter?

_____

$c$

The $y$ -intercept of $y = a x^{2} + b x + c$ is the same as the value of which parameter?

$a$
$b$
$c$
None of these

Example 4

Find the vertex of the graph of $f (x) = - 2 x^{2} + x + 1$ .

For this function, $a = - 2$ , $b equals 1$ , and $c equals 1$ . The $x$ -coordinate of the vertex is given by

$x_{v} = \frac{- b}{2 a} = \frac{- 1}{2 (- 2)} = \frac{1}{4}$

To find the $y$ -coordinate of the vertex, we substitute $x = \frac{1}{4}$ into the equation. We can do this by hand to find

$\begin{aligned} y_{v} & = - 2 {(\frac{1}{4})}^{2} + \frac{1}{4} + 1 \\ = - 2 (\frac{1}{16}) + \frac{4}{16} + \frac{16}{16} = \frac{18}{16} = \frac{9}{8} \end{aligned}$

So the coordinates of the vertex are $(\frac{1}{4}, \frac{9}{8})$ . Alternatively, we can use the calculator to evaluate $- 2 x^{2} + x + 1$ for $x equals 0.25$ . The calculator returns the $y$ -value $1.125$ . Thus, the vertex is the point $open parenthesis 0.25, 1.125 close parenthesis$ , which is the decimal equivalent of $(\frac{1}{4}, \frac{9}{8})$ .

Find the vertex of the graph of $f (x) = 3 x^{2} - 6 x + 4$ .

_____

Decide whether the vertex is a maximum point or a minimum point of the graph._____

$open parenthesis 1, 1 close parenthesis$ , minimum

Find the vertex of the graph of $f (x) = 3 x^{2} - 6 x + 4$ . Decide whether the vertex is a maximum point or a minimum point of the graph.

$open parenthesis 1, 1 close parenthesis$ , minimum

Number of $x$ -Intercepts

The graph of the quadratic function

$f (x) = a x^{2} + b x + c$

may have two, one, or no $x$ -intercepts, according to the number of distinct real-valued solutions of the equation $a x^{2} + b x + c = 0$ . Consider the three functions graphed below.

parabolas with zero, one, or two x-intercepts

The graph of
$f (x) = x^{2} - 4 x + 3$
has two $x$ -intercepts, because the equation
$x^{2} - 4 x + 3 = 0$
has two real-valued solutions, $x equals 1$ and $x equals 3$ .
The graph of
$g (x) = x^{2} - 4 x + 4$
has only one $x$ -intercept, because the equation
$x^{2} - 4 x + 4 = 0$
has only one (repeated) real-valued solution, $x equals 2$ .
The graph of
$h (x) = x^{2} - 4 x + 6$
has no $x$ -intercepts, because the equation
$x^{2} - 4 x + 6 = 0$
has no real-valued solutions.

A closer look at the quadratic formula reveals useful information about the solutions of quadratic equations. For the three functions above, we have the following:

The expression $b squared minus 4 a c$ , which appears under the radical in the quadratic formula, is called the discriminant, $D$ , of the equation. The value of the discriminant determines the nature of the solutions of the equation. In particular, if the discriminant is negative, the solutions of the quadratic equation are complex numbers. (We will study complex numbers in Complex Numbers.)

Note

We can also use the discriminant to decide whether a quadratic equation can be solved by factoring. First, clear the equation of fractions. If the discriminant is a perfect square, that is, the square of an integer, the solutions are rational numbers. This in turn means that the equation can be solved by factoring.

If the discriminant is not a perfect square, the solutions will be irrational. Irrational solutions always occur in conjugate pairs,

$\frac{- b + \sqrt{b^{2} - 4 a c}}{2 a} and \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a}$

The only difference between the two solutions is the sign between the terms. For example, if we know that one solution of a particular quadratic equation is $3 + \sqrt{2}$ , then the other solution must be $3 - \sqrt{2}$ .

Example 5

Use the discriminant to determine the nature of the solutions of each equation. Can the equation be solved by factoring?

$x^{2} - x - 3 = 0$
$2 x^{2} + x + 1 = 0$
$x^{2} - 2 x - 3 = 0$

$D = b^{2} - 4 a c = (- 1)^{2} - 4 (1) (- 3) = 13 > 0$ .
The equation has two real, unequal solutions. Because $13$ is not a perfect square, the solutions will be irrational numbers, so the equation cannot be solved by factoring.
$D = b^{2} - 4 a c = 1^{2} - 4 (2) (1) = - 7 < 0$ .
The equation has two complex solutions, which cannot be found by factoring.
$D = b^{2} - 4 a c = (- 2)^{2} - 4 (1) (- 3) = 16 > 0$ .
The equation has two real, unequal solutions. Because $16 equals 4 squared$ , the solutions are rational numbers and can be found by factoring.

(You can verify the conclusions above by solving each equation.)

Use the discriminant to discover how many $x$ -intercepts the graph of each function has.

$y = x^{2} + 5 x + 7$ _____
$y = - \frac{1}{2} x^{2} + 4 x - 8$ _____

None: the discriminant is negative.
One: the discriminant is 0.

Use the discriminant to discover how many $x$ -intercepts the graph of each function has.

$y = x^{2} + 5 x + 7$
$y = - \frac{1}{2} x^{2} + 4 x - 8$

None: the discriminant is negative.
One: the discriminant is 0.

In Practice 5, you can check that the single $x$ -intercept is also the vertex of the parabola.

Explain what the discriminant tells us about a quadratic equation.

_____

Explain what the discriminant tells us about a quadratic equation.

Sketching a Parabola

Once we have located the vertex of the parabola, the $x$ -intercepts, and the $y$ -intercept, we can sketch a reasonably accurate graph. Recall that the graph should be symmetric about a vertical line through the vertex. We summarize the procedure as follows.

Function graph showing y = a*(x - h)^2 + k, the parametric curve (h + 0*t, t) for t in [-6, 6] and the parametric curve (h + 0.12*cos(t), k + 0.12*sin(t)) for t in [0, 6.28]. Adjustable parameters: Opens/width a (a) = 1, Vertex x: h (h) = 1, Vertex y: k (k) = -2. Viewing window: x from -8.64 to 8.64, y from -5.34 to 5.34. — The sketching recipe's two landmarks, live: y = a(x − h)² + k with the vertex marked in red at (h, k) and the axis of symmetry dashed at x = h. Slide h and k and the whole parabola follows its vertex; the graph is always the mirror image of itself across the dashed line, which is why finding the vertex is most of the sketch. The slider a controls the rest: its sign says whether the parabola opens upward (a > 0, vertex a minimum) or downward (a < 0, vertex a maximum), and its size says how narrow the bowl is. At a = 0 the parabola flattens into the horizontal line through its vertex — the boundary between the two cases.

Example 6

Sketch a graph of the equation $f (x) = x^{2} + 3 x + 1$ , showing the significant points.

We follow the steps outlined above.

Because $a equals 1 greater than 0$ , we know that the parabola opens upward.
We compute the coordinates of the vertex:
$\begin{aligned} x_{v} & = \frac{- b}{2 a} = \frac{- 3}{2 (1)} = - 1.5 \\ y_{v} & = (- 1.5) 2 + 3 (- 1.5) + 1 = - 1.25 \end{aligned}$
The vertex is the point $(- 1.5, - 1.25)$ .
We set $y$ equal to zero to find the $x$ -intercepts.
$\begin{aligned} 0 & = x^{2} + 3 x + 1 & Use the quadratic formula. \\ x & = \frac{- 3 \pm \sqrt{3^{2} - 4 (1) (1)}}{2 (1)} \\ = \frac{- 3 \pm \sqrt{5}}{2} \end{aligned}$
Rounding to the nearest tenth, we find that the $x$ -intercepts are approximately $open parenthesis minus 2.6, 0 close parenthesis$ and $open parenthesis minus 0.4, 0 close parenthesis$ .
We substitute $x equals 0$ to find the $y$ -intercept, $open parenthesis 0, 1 close parenthesis$ .
The axis of symmetry is the vertical line $x = - 1.5$ , so the $y$ -intercept lies $1.5$ units to the right of the axis of symmetry.
There must be another point on the parabola with the same $y$ -coordinate as the $y$ -intercept but $1.5$ units to the left of the axis of symmetry. The coordinates of this point are $open parenthesis minus 3, 1 close parenthesis$ .
Finally, plot the $x$ -intercepts, the vertex, and the $y$ -intercept and its symmetric point, and draw a parabola through them. The finished graph is shown below.

Find the intercepts and the vertex of the graph of $f (x) = x^{2} - 5 x + 4$ .
Intercepts: _____ Note: Use a comma to separate different points.
Vertex: _____
Sketch the graph by hand.
Use your calculator to verify your graph.

$open parenthesis 0, 4 close parenthesis$ ; $open parenthesis 1, 0 close parenthesis$ , $open parenthesis 4, 0 close parenthesis$ ; vertex $(\frac{5}{2}, \frac{- 9}{4})$
A graph is below.

A graph for part (b):

Find the intercepts and the vertex of the graph of $f (x) = x^{2} - 5 x + 4$ .
Sketch the graph by hand.
Use your calculator to verify your graph.

$open parenthesis 0, 4 close parenthesis$ ; $open parenthesis 1, 0 close parenthesis$ , $open parenthesis 4, 0 close parenthesis$ ; vertex $(\frac{5}{2}, \frac{- 9}{4})$

Which points on a parabola should you find to help you sketch its graph? Describe how to find each of these points.

_____

Which points on a parabola should you find to help you sketch its graph? Describe how to find each of these points.

Section Summary

Vocabulary

Look up the definitions of new terms in the Glossary.

Vertex
Conjugate pair
Axis of symmetry

CONCEPTS

The graph of a quadratic function $f (x) = a x^{2} + b x + c$ is called a parabola. The values of the constants $a$ , $b$ , and $c$ determine the location and orientation of the parabola.
For the graph of $f (x) = a x^{2} + b x + c$ , the $x$ -coordinate of the vertex is $x_{v} = \frac{- b}{2 a}$ . To find the $y$ -coordinate of the vertex, we substitute $x_{v}$ into the formula for the parabola.
The graph of the quadratic function $f (x) = a x^{2} + b x + c$ may have two, one, or no $x$ -intercepts, according to the number of distinct real-valued solutions of the equation $a x^{2} + b x + c = 0$ .
Note
The Discriminant
The discriminant of a quadratic equation is $D = b^{2} - 4 a c$ .
1. If $D greater than 0$ , there are two unequal real solutions.
2. If $D equals 0$ , there is one real solution of multiplicity two.
3. If $D less than 0$ , there are two complex solutions.
Note
To Graph the Quadratic Function $f (x) = a x^{2} + b x + c$ :
1. Determine whether the parabola opens upward (if $a greater than 0$ ) or downward (if $a less than 0$ ).
2. Locate the vertex of the parabola.
  1. The $x$ -coordinate of the vertex is $x_{v} = \frac{- b}{2 a}$ .
  2. Find the $y$ -coordinate of the vertex by substituting $x_{v}$ into the equation of the parabola.
3. Locate the $x$ -intercepts (if any) by setting $y equals 0$ and solving for $x$ .
4. Locate the $y$ -intercept by evaluating $y$ for $x equals 0$ .
5. Locate the point symmetric to the $y$ -intercept across the axis of symmetry.

STUDY QUESTIONS

Sketch a parabola that opens downward. Show the location of the $x$ -intercepts, the $y$ -intercept, the vertex, and the axis of symmetry.
Describe how the value of $a$ in $y = a x^{2}$ alters the graph of the basic parabola.
Describe how the value of $c$ in $y = x^{2} + c$ alters the graph of the basic parabola.
Suppose you know that the $x$ -intercepts of a parabola are $open parenthesis minus 8, 0 close parenthesis$ and $open parenthesis 2, 0 close parenthesis$ . What is the equation of the parabola's axis of symmetry?
State a formula for the $x$ -coordinate of the vertex of a parabola. How can you find the $y$ -coordinate of the vertex?
Suppose that a given parabola has only one $x$ -intercept. What can you say about the vertex of the parabola?
Explain why a quadratic equation has one (repeated) solution if its discriminant is zero, and none if the discriminant is negative.

SKILLS

Practice each skill in the Homework problems listed.

Graph transformations of the basic parabola: #1 and 2, 7, and 8
Locate the $x$ -intercepts of a parabola: #3–6
Locate the vertex of a parabola: #3–6, 13, and 14
Sketch the graph of a quadratic function: #15–24, 41, and 42
Use the discriminant to describe the solutions of a quadratic equation: #25–40

Homework 6.3

For Problems 1–2, describe what each graph will look like compared to the basic parabola. Then sketch a graph by hand and label the coordinates of three points on the graph.

$y equals 2 x squared$
$y = 2 + x^{2}$
$y = (x + 2)^{2}$
$y = x^{2} - 2$

The parabola opens up, twice as steep as the standard parabola.
The parabola is the standard parabola shifted 2 units up.
The parabola is the standard parabola shifted 2 units left.
The parabola is the standard parabola shifted 2 units down.

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

For problems 3–6, find the vertex and the $x$ -intercepts (if there are any) of the graph. Then sketch the graph by hand.

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

Vertex $open parenthesis 0, minus 16 close parenthesis$ ; $x$ -intercepts $(\pm 4, 0)$
Vertex $open parenthesis 0, 16 close parenthesis$ ; $x$ -intercepts $(\pm 4, 0)$
Vertex $open parenthesis 8, 64 close parenthesis$ ; $x$ -intercepts $open parenthesis 0, 0 close parenthesis$ and $open parenthesis 16, 0 close parenthesis$
Vertex $open parenthesis 8, minus 64 close parenthesis$ ; $x$ -intercepts $open parenthesis 0, 0 close parenthesis$ and $open parenthesis 16, 0 close parenthesis$

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

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

Vertex $open parenthesis 1, minus 3 close parenthesis$ ; $x$ -intercepts $open parenthesis 0, 0 close parenthesis$ and $open parenthesis minus 2, 0 close parenthesis$
Vertex $open parenthesis 1, minus 3 close parenthesis$ ; $x$ -intercepts $open parenthesis 0, 0 close parenthesis$ and $open parenthesis 2, 0 close parenthesis$
Vertex $open parenthesis 0, 6 close parenthesis$ ; no $x$ -intercepts
Vertex $open parenthesis 0, minus 6 close parenthesis$ ; $x$ -intercepts $(\pm \sqrt{2}, 0)$

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

Match each function with its graph. In each equation, $a greater than 0$ .

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

Match each function with its graph. In each equation, $b greater than 0$ .

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

Commercial fishermen rely on a steady supply of fish in their area. To avoid overfishing, they adjust their harvest to the size of the population. The function

$g (x) = 0.4 x - 0.0001 x^{2}$

gives the annual rate of growth, in tons per year, of a fish population of biomass $x$ tons.

Find the vertex of the graph. What does it tell us about the fish population?
Sketch the graph for $0 less than or equal to x less than or equal to 5000$ .
For what values of $x$ does the fish population decrease rather than increase? Suggest a reason why the population might decrease.

$open parenthesis 2000, 400 close parenthesis$ ; The largest annual increase in biomass, $400$ tons, occurs when the biomass is $2000$ tons.
$4000 less than x less than or equal to 5000$ ; When there are too many fish, there will not be enough food to support all of them.

The annual increase, $I$ , in the deer population in a national park depends on the size, $x$ , of the population that year, according to the function

$I = f (x) = 1.2 x - 0.0002 x^{2}$

Find the vertex of the graph. What does it tell us about the deer population?
Sketch the graph for $0 less than or equal to x less than or equal to 7000$ .
For what values of $x$ does the deer population decrease rather than increase? Suggest a reason why the population might decrease.

Many animals live in groups. A species of marmot found in Colorado lives in harems composed of a single adult male and several females with their young. The number of offspring each female can raise depends on the number of females in the harem. On average, if there are $x$ females in the harem, each female can raise $y = 2 - 0.4 x$ young marmots each year.

Complete the table of values for the average number of offspring per female, and the total number of young marmots, $A$ , produced by the entire harem in one year.
$x$ $1$ $2$ $3$ $4$ $5$
$y$
$A$
Write a formula for $A$ in terms of $x$ .
Graph $A$ as a function of $x$ .
What is the maximum number of young marmots a harem can produce (on average)? What is the optimal number of female marmots per harem?

$x$	$1$	$2$	$3$	$4$	$5$
$y$	$1.6$	$1.2$	$0.8$	$0.4$	$0$
$A$	$1.6$	$2.4$	$2.4$	$1.6$	$0$

$A = x (2 - 0.4 x)$ or $A = 2 x - 0.4 x^{2}$
The maximum number of young marmots, on average, is $2.5$ ; the optimal number of female marmots is $2.5$ .

Greenshield's model for traffic flow assumes that the average speed, $u$ , of cars on a highway is a linear function of the traffic density, $k$ , in vehicles per mile, given by

$u = u_{f} (1 - \frac{k}{k_{j}})$

where $u_{f}$ is the free-flow speed and $k_{j}$ is the maximum density (the point when traffic jams). Then the traffic flow, $q$ , in vehicles per hour, is given by $q = u k$ .

Write a formula for $q$ as a function of $k$ .
If the free-flow speed is $70$ mph and the maximum density is $240$ vehicles per mile, graph $q$ as a function of $k$ .
What value of $k$ gives the maximum traffic flow? What is the average speed of vehicles at that density?

After touchdown, the distance the space shuttle travels is given by

$d = v T + \frac{v^{2}}{2 a}$

where $v$ is the shuttle's velocity in ft/sec at touchdown, $T$ is the pilot's reaction time before the brakes are applied, and $a$ is the shuttle's deceleration.

Graph $d = f (v)$ for $T equals 0.5$ seconds and $a equals 12 ft/sec squared$ . Find the coordinates of the vertex and the horizontal intercepts. Explain their meaning, if any, in this context.
The runway at Edwards Air Force base is $15, 000$ feet long. What is the maximum velocity the shuttle can have at touchdown and still stop on the runway?

Vertex: $(- 6, - 1.5)$ ; Horizontal intercepts $open parenthesis minus 12, 0 close parenthesis$ and $open parenthesis 0, 0 close parenthesis$ . The point $open parenthesis 0, 0 close parenthesis$ means that no distance is required to stop a plane that is not moving.
$594$ ft/sec

When setting the pump pressure at the engine, firefighters must take into account the pressure loss due to friction inside the fire hose. For every $100$ feet of hoseline, a hose of diameter $2.5$ inches loses pressure according to the formula

$L = {\begin{cases} 2 Q^{2} + Q, & Q \geq 1 \\ 2 Q^{2} + \frac{1}{2} Q, & Q < 1 \end{cases}$

where $Q$ is the water flow in hundreds of gallos per minute. The friction loss, $L$ , is measured in pounds per square inch (psi). (Source: www.hcc.hawaii.edu/~jkemmer)

Graph $L = g (Q)$ on the domain $[ 0, 5 ]$ .
The firefighters have unrolled $600$ feet of $2.5$ -inch-diameter hose, and they would like to deliver water at a rate of $200$ gallons per minute, with nozzle pressure at $100$ psi. They must add the friction loss to the nozzle pressure to calculate the engine pressure required. What should the engine pressure be?

For Problems 15–16, find the coordinates of the vertex. Decide whether the vertex is a maximum point or a minimum point on the graph.

$y = 2 + 3 x - x^{2}$
$y = \frac{1}{2} x^{2} - \frac{2}{3} x + \frac{1}{3}$
$y = 2.3 - 7.2 x - 0.8 x^{2}$

$(\frac{3}{2}, \frac{17}{4})$ , maximum
$(\frac{2}{3}, \frac{1}{9})$ , minimum
$open parenthesis minus 4.5, 18.5 close parenthesis$ , maximum

$y = 3 - 5 x + x^{2}$
$y = \frac{- 3}{4} x^{2} + \frac{1}{2} x - \frac{1}{4}$
$y = 5.1 - 0.2 x + 4.6 x^{2}$

In Problems 17–26,

Find the coordinates of the intercepts and the vertex.
Sketch the graph by hand.
Use your calculator to verify your graph.

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

$x$ -intercepts: $(\frac{- 1}{2}, 0)$ and $open parenthesis 4, 0 close parenthesis$ ; $y$ -intercept: $open parenthesis 0, 4 close parenthesis$ ; vertex: $(\frac{7}{4}, \frac{81}{8})$

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

$y = 0.6 x^{2} + 0.6 x - 1.2$

$x$ -intercepts: $open parenthesis minus 2, 0 close parenthesis$ and $open parenthesis 1, 0 close parenthesis$ ; $y$ -intercept: $open parenthesis 0, minus 1.2 close parenthesis$ ; vertex: $(- 0.5, - 1.35)$

$y = 0.5 x^{2} - 0.25 x - 0.75$

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

No $x$ -intercepts; $y$ -intercept: $open parenthesis 0, 7 close parenthesis$ ; vertex: $open parenthesis minus 2, 3 close parenthesis$

$y = x^{2} - 6 x + 10$

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

$x$ -intercepts: $(- 1 \pm \sqrt{2}, 0)$ ; $y$ -intercept: $open parenthesis 0, minus 1 close parenthesis$ ; vertex: $(- 1, - 2)$

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

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

$x$ -intercepts: $(\frac{3 \pm \sqrt{3}}{2}, 0)$ ; $y$ -intercept: $open parenthesis 0, minus 3 close parenthesis$ ; vertex: $(\frac{3}{2}, \frac{3}{2})$

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

Graph the three functions
$\begin{aligned} f (x) & = x^{2} - 6 x + 5 \\ g (x) & = x^{2} - 6 x + 9 \\ h (x) & = x^{2} - 6 x + 12 \end{aligned}$
in the window
$\begin{aligned} Xmin & = - 2 & Xmax = 7.4 \\ Ymin & = - 5 & Ymax = 15 \end{aligned}$
Use the Trace to locate the $x$ -intercepts of each graph.
Set $y equals 0$ for each of the equations in part (a) and calculate the discriminant. What does the discriminant tell you about the solutions of the equation? How does your answer relate to the graphs in part (a)?

$f (x) = x^{2} - 6 x + 5$ : $x$ -intercepts $open parenthesis 1, 0 close parenthesis$ and $open parenthesis 5, 0 close parenthesis$ ; $g (x) = x^{2} - 6 x + 9$ : $x$ -intercept $open parenthesis 3, 0 close parenthesis$ ; $h (x) = x^{2} - 6 x + 12$ : No $x$ -intercept.
$16, 0, minus 12$ : $D equals 16$ means that there are two rational $x$ -intercepts, $D equals 0$ means that there is exactly one $x$ -intercept, $D = - 12$ means that there is no $x$ -intercept.

Graph the three functions
$\begin{aligned} F (x) & = 3 - 2 x - x^{2} \\ G (x) & = - 1 - 2 x - x^{2} \\ H (x) & = - 4 - 2 x - x^{2} \end{aligned}$
in the window
$\begin{aligned} Xmin & = - 6.4 & Xmax = 3 \\ Ymin & = - 10 & Ymax = 5 \end{aligned}$
Use the Trace to locate the $x$ -intercepts of each graph.
Set $y equals 0$ for each of the equations in part (a) and calculate the discriminant. What does the discriminant tell you about the solutions of the equation? How does your answer relate to the graphs in part (a)?

For Problems 29–34, use the discriminant to determine the nature of the solutions of each equation.

$3 x^{2} + 26 = 17 x$

Two complex solutions

$4 x^{2} + 23 x = 19$

$16 x^{2} - 712 x + 7921 = 0$

One repeated rational solution

$121 x^{2} + 1254 x + 3249 = 0$

$65.2 x = 13.2 x^{2} + 41.7$

Two distinct real solutions

$0.03 x^{2} = 0.05 x - 0.12$

For problems 35–38, use the discriminant to decide if we can solve the equation by factoring.

$3 x^{2} - 7 x + 6 = 0$

$6 x^{2} - 11 x - 7 = 0$

$15 x^{2} - 52 x - 32 = 0$

Yes

$17 x^{2} + 65 x - 12 = 0$

For Problems 39–42,

Given one solution of a quadratic equation with rational coefficients, find the other solution.
Write a quadratic equation that has those solutions.

$2 + \sqrt{5}$

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

$3 - \sqrt{2}$

$4 - 3 \sqrt{2}$

$4 + 3 \sqrt{2}$
$x^{2} - 8 x - 2 = 0$

$5 + 2 \sqrt{3}$

For Problems 43 and 44, match each equation with one of the eight graphs shown.

$y = 1 - x^{2}$
$y = (x + 2)^{2}$
$y equals 2 x squared$
$y = (x - 4) (x + 2)$

$y = - 2 - (x - 2)^{2}$
$y = x - x^{2}$
$y = x^{2} - 4$
$y = - 0.5 x^{2}$

Write an equation for a parabola that has $x$ -intercepts at $open parenthesis 2, 0 close parenthesis$ and $open parenthesis minus 3, 0 close parenthesis$ . What is the equation of the parabola's axis of symmetry?
Write an equation for another parabola that has the same $x$ -intercepts. What is the equation of the parabola's axis of symmetry?

$y = x^{2} + x - 6$ ; $x = \frac{- 1}{2}$
$y = 2 x^{2} + 2 x - 12$ ; $x = \frac{- 1}{2}$

Write an equation for a parabola that opens upward and has $x$ -intercepts at $open parenthesis minus 1, 0 close parenthesis$ and $open parenthesis 4, 0 close parenthesis$ . What is the equation of the parabola's axis of symmetry?
Write an equation for a parabola that opens downward and has $x$ -intercepts $open parenthesis minus 1, 0 close parenthesis$ and $open parenthesis 4, 0 close parenthesis$ . What is the equation of its axis of symmetry?

Graph the functions in the same window on your calculator:
$\begin{matrix} f (x) = x^{2} + 2 x, g (x) = x^{2} + 4 x, \\ h (x) = x^{2} + 6 x, j (x) = x^{2} + 8 x \end{matrix}$
Find the vertex of each graph in part (a) and plot the points.
Find the equation of the curve in part (b).
Show that the vertex of $y = x^{2} + 2 k x$ lies on the curve for any value of $k$ .

$(- 1, - 1), (- 2, - 4), (- 3, - 9), (- 4, - 16)$
$y = - x^{2}$
The vertex of $y = x^{2} + 2 k x$ is $(- k, - k^{2})$

Graph the functions in the same window on your calculator:
$\begin{matrix} F (x) = x - \frac{1}{2} x^{2}, G (x) = 3 x - \frac{1}{2} x^{2}, \\ H (x) = 5 x - \frac{1}{2} x^{2}, J (x) = 7 x - \frac{1}{2} x^{2} \end{matrix}$
Find the vertex of each graph in part (a) and plot the points.
Find the equation of the curve in part (b).
Show that the vertex of $y = k x - \frac{1}{2} x^{2}$ lies on the curve for any value of $k$ .

Because of air resistance, the path of a kicked soccer ball is not actually parabolic. However, both the horizontal and vertical coordinates of points on its trajectory can be approximated by quadratic functions. For a soccer ball kicked from the ground, these functions are

$\begin{matrix} x = f (t) = 12.8 t - 1.3 t^{2} \\ y = g (t) = 17.28 t - 4.8 t^{2} \end{matrix}$

where $x$ and $y$ are given in meters and $t$ is the number of seconds since the ball was kicked.

Fill in the table.
$t$ $0$ $0.5$ $1.0$ $1.5$ $2.0$ $2.5$ $3.0$ $3.5$
$x$
$y$
Plot the points $(x, y)$ from your table and connect them with a smooth curve to represent the path of the ball.
Use your graph to estimate the maximum height of the ball.
Estimate the horizontal distance traveled by the ball before it strikes the ground.
Using the formula given for $y$ , determine how long the ball is in the air.
Use your answer from part (e) and the formula for $x$ to find the horizontal distance traveled by the ball before it strikes the ground.
Use the formula given for $y$ to find the maximum height for the ball.

$t$	$0$	$0.5$	$1.0$	$1.5$	$2.0$	$2.5$	$3.0$	$3.5$
$x$	$0$	$6.075$	$11.5$	$16.275$	$20.4$	$23.875$	$26.7$	$28.875$
$y$	$0$	$7.44$	$12.48$	$15.12$	$15.36$	$13.2$	$8.64$	$1.68$

$y approximately equals 15.4$ m
$x approximately equals 30$ m
$3.6$ sec
$x approximately equals 29.2$ m
$y approximately equals 15.55$ m

How far can you throw a baseball? The distance depends on the initial speed of the ball, $v$ , and on the angle at which you throw it. For maximum range, you should throw the ball at $45 degrees$ .

If there were no air resistance, the height, $x$ , of the ball t seconds after its release would be given in meters by the function
$h = f (t) = \frac{v t}{\sqrt{2}} - \frac{g t^{2}}{2}$
where $g$ is the acceleration due to gravity. Find an expression for the total time the ball is in the air. (Hint: Set $h equals 0$ and solve for $t$ in terms of the other variables.)
At time $t$ , the ball has traveled a horizontal distance $d$ given by
$d = \frac{v t}{\sqrt{2}}$
Find an expression for the range of the ball in terms of its velocity, $v$ . (Hint: In part (a), you found an expression for $t$ when $h equals 0$ . Use that value of $t$ to calculate $d$ when $h equals 0$ .)
The fastest baseball pitch on record was $45$ meters per second, or about $100$ miles per hour. Use your formula from part (b) to calculate the theoretical range of such a pitch. The value of $g$ is $9.8$ .
The maximum distance a baseball has actually been thrown is $136$ meters. Can you explain the discrepancy between this figure and your answer to part (c)?

Modeling, Functions, and Graphs by Katherine Yoshiwara (yoshiwarabooks.org), GNU Free Documentation License 1.2 or later. Adapted for the XYZ HTML edition with the authors' permission (recorded 2026-07-04). License: GFDL-1.2-or-later.

6.3 Graphing Parabolas

Introduction

The Graph of y = a x 2

The Graph of y = x 2 + c

The Graph of y = a x 2 + b x

Finding the Vertex

The Graph of y = a x 2 + b x + c

Number of x -Intercepts

Sketching a Parabola

Section Summary

Vocabulary

CONCEPTS

STUDY QUESTIONS

SKILLS

Homework 6.3

The Graph of $y = a x^{2}$

The Graph of $y = x^{2} + c$

The Graph of $y = a x^{2} + b x$

The Graph of $y = a x^{2} + b x + c$

Number of $x$ -Intercepts