Contents

Preface
1Foundations
- Introduction
- 1.1Use the Language of Algebra
- 1.2Integers
- 1.3Fractions
- 1.4Decimals
- 1.5Properties of Real Numbers
- Chapter Review
  Key Terms
  Key Concepts
- Exercises
  Review Exercises
  Practice Test
2Solving Linear Equations
3Graphs and Functions
4Systems of Linear Equations
5Polynomials and Polynomial Functions
6Factoring
7Rational Expressions and Functions
8Roots and Radicals
9Quadratic Equations and Functions
10Exponential and Logarithmic Functions
11Conics
12Sequences, Series and Binomial Theorem
- Introduction
- 12.1Sequences
- 12.2Arithmetic Sequences
- 12.3Geometric Sequences and Series
- 12.4Binomial Theorem
- Chapter Review
  Key Terms
Answer Key
Index

Parabolas

Learning Objectives

By the end of this section, you will be able to:

Graph vertical parabolas
Graph horizontal parabolas
Solve applications with parabolas

Be Prepared 11.2

Before you get started, take this readiness quiz.

Graph: $y = −3 x^{2} + 12 x - 12 . y = −3 x^{2} + 12 x - 12 .$
If you missed this problem, review Example 9.47.
Solve by completing the square: $x^{2} - 6 x + 6 = 0 . x^{2} - 6 x + 6 = 0 .$
If you missed this problem, review Example 9.12.
Write in standard form: $y = 3 x^{2} - 6 x + 5 . y = 3 x^{2} - 6 x + 5 .$
If you missed this problem, review Example 9.59.

Graph Vertical Parabolas

The next conic section we will look at is a parabola. We define a parabola as all points in a plane that are the same distance from a fixed point and a fixed line. The fixed point is called the focus, and the fixed line is called the directrix of the parabola.

Parabola

A parabola is all points in a plane that are the same distance from a fixed point and a fixed line. The fixed point is called the focus, and the fixed line is called the directrix of the parabola.

Previously, we learned to graph vertical parabolas from the general form or the standard form using properties. Those methods will also work here. We will summarize the properties here.

Vertical Parabolas
	General form $y = a x^{2} + b x + c y = a x^{2} + b x + c$	Standard form $y = a {(x - h)}^{2} + k y = a {(x - h)}^{2} + k$
Orientation	$a > 0 a > 0$ up; $a < 0 a < 0$ down	$a > 0 a > 0$ up; $a < 0 a < 0$ down
Axis of symmetry	$x = - \frac{b}{2 a} x = - \frac{b}{2 a}$	$x = h x = h$
Vertex	Substitute $x = - \frac{b}{2 a} x = - \frac{b}{2 a}$ and solve for y.	$(h, k) (h, k)$
y-intercept	Let $x = 0 x = 0$	Let $x = 0 x = 0$
x-intercepts	Let $y = 0 y = 0$	Let $y = 0 y = 0$

The graphs show what the parabolas look like when they open up or down. Their position in relation to the x- or y-axis is merely an example.

This figure shows two parabolas with axis x equals h and vertex h, k. The one on the left opens up and A is greater than 0. The one on the right opens down. Here A is less than 0.

To graph a parabola from these forms, we used the following steps.

How To

Graph vertical parabolas $(y = a x^{2} + b x + c or f (x) = a {(x - h)}^{2} + k) (y = a x^{2} + b x + c$ using properties.

Step 1. Determine whether the parabola opens upward or downward.
Step 2. Find the axis of symmetry.
Step 3. Find the vertex.
Step 4. Find the y-intercept. Find the point symmetric to the y-intercept across the axis of symmetry.
Step 5. Find the x-intercepts.
Step 6. Graph the parabola.

The next example reviews the method of graphing a parabola from the general form of its equation.

Example 11.12

Graph $y = − x^{2} + 6 x - 8 y = − x^{2} + 6 x - 8$ by using properties.

Show solution

Solution


Since a is $−1, −1,$ the parabola opens downward.

To find the axis of symmetry, find $x = - \frac{b}{2 a} . x = - \frac{b}{2 a} .$


	The axis of symmetry is $x = 3 . x = 3 .$

The vertex is on the line $x = 3 . x = 3 .$
Let $x = 3 . x = 3 .$


	The vertex is $(3, 1) . (3, 1) .$

The y-intercept occurs when $x = 0 . x = 0 .$
Substitute $x = 0 . x = 0 .$
Simplify.
	The y-intercept is $(0, −8) . (0, −8) .$
The point $(0, −8) (0, −8)$ is three units to the left of the line of symmetry. The point three units to the right of the line of symmetry is $(6, −8) . (6, −8) .$	Point symmetric to the y-intercept is $(6, −8) . (6, −8) .$

The x-intercept occurs when $y = 0 . y = 0 .$
Let $y = 0 . y = 0 .$
Factor the GCF.
Factor the trinomial.
Solve for x.
	The x-intercepts are $(4, 0), (2, 0) . (4, 0), (2, 0) .$
Graph the parabola.

Try It 11.23

Graph $y = − x^{2} + 5 x - 6 y = − x^{2} + 5 x - 6$ by using properties.

Try It 11.24

Graph $y = − x^{2} + 8 x - 12 y = − x^{2} + 8 x - 12$ by using properties.

The next example reviews the method of graphing a parabola from the standard form of its equation, $y = a {(x - h)}^{2} + k . y = a {(x - h)}^{2} + k .$

Example 11.13

Write $y = 3 x^{2} - 6 x + 5 y = 3 x^{2} - 6 x + 5$ in standard form and then use properties of standard form to graph the equation.

Show solution

Solution

Rewrite the function in $y = a {(x - h)}^{2} + k y = a {(x - h)}^{2} + k$ form by completing the square.	$y = 3 x^{2} - 6 x + 5 y = 3 x^{2} - 6 x + 5$
	$y = 3 (x^{2} - 2 x) + 5 y = 3 (x^{2} - 2 x + 5$
	$y = 3 (x^{2} - 2 x + 1) + 5 - 3 y = 3 (x^{2} - 2 x + 1) + 5 - 3$
	$y = 3 {(x - 1)}^{2} + 2 y = 3 {(x - 1)}^{2} + 2$
Identify the constants a, h, k.	$a = 3 a = 3$ , $h = 1 h = 1$ , $k = 2 k = 2$
Since $a = 2, a = 2,$ the parabola opens upward.

The axis of symmetry is $x = h . x = h .$	The axis of symmetry is $x = 1 . x = 1 .$
The vertex is $(h, k) . (h, k) .$	The vertex is $(1, 2) . (1, 2) .$
Find the y-intercept by substituting $x = 0 . x = 0 .$	$y = 3 {(x - 1)}^{2} + 2 y = 3 {(x - 1)}^{2} + 2$ $y = 3 \cdot 0^{2} - 6 \cdot 0 + 5 y = 3 \cdot 0^{2} - 6 \cdot 0 + 5$
	$y = 5 y = 5$
	y-intercept $(0, 5) (0, 5)$
Find the point symmetric to $(0, 5) (0, 5)$ across the axis of symmetry.	$(2, 5) (2, 5)$
Find the x-intercepts.	$\begin{matrix} y & = & 3 {(x - 1)}^{2} + 2 \\ 0 & = & 3 {(x - 1)}^{2} + 2 \\ - 2 & = & 3 {(x - 1)}^{2} \\ - \frac{2}{3} & = & {(x - 1)}^{2} \\ \pm \sqrt{- \frac{2}{3}} & = & x - 1 \end{matrix} \begin{matrix} y & = & 3 {(x - 1)}^{2} + 2 \\ 0 & = & 3 {(x - 1)}^{2} + 2 \\ - 2 & = & 3 {(x - 1)}^{2} \\ - \frac{2}{3} & = & {(x - 1)}^{2} \\ \pm \sqrt{- \frac{2}{3}} & = & x - 1 \end{matrix}$
	The square root of a negative number tells us the solutions are complex numbers. So there are no x-intercepts.
Graph the parabola.

Try It 11.25

ⓐ Write $y = 2 x^{2} + 4 x + 5 y = 2 x^{2} + 4 x + 5$ in standard form and ⓑ use properties of standard form to graph the equation.

Try It 11.26

ⓐ Write $y = −2 x^{2} + 8 x - 7 y = −2 x^{2} + 8 x - 7$ in standard form and ⓑ use properties of standard form to graph the equation.

Graph Horizontal Parabolas

Our work so far has only dealt with parabolas that open up or down. We are now going to look at horizontal parabolas. These parabolas open either to the left or to the right. If we interchange the x and y in our previous equations for parabolas, we get the equations for the parabolas that open to the left or to the right.

Horizontal Parabolas
	General form $x = a y^{2} + b y + c x = a y^{2} + b y + c$	Standard form $x = a {(y - k)}^{2} + h x = a {(y - k)}^{2} + h$
Orientation	$a > 0 a > 0$ right; $a < 0 a < 0$ left	$a > 0 a > 0$ right; $a < 0 a < 0$ left
Axis of symmetry	$y = - \frac{b}{2 a} y = - \frac{b}{2 a}$	$y = k y = k$
Vertex	Substitute $y = - \frac{b}{2 a} y = - \frac{b}{2 a}$ and solve for x.	$(h, k) (h, k)$
y-intercepts	Let $x = 0 x = 0$	Let $x = 0 x = 0$
x-intercept	Let $y = 0 y = 0$	Let $y = 0 y = 0$

Table 11.1

The graphs show what the parabolas look like when they to the left or to the right. Their position in relation to the x- or y-axis is merely an example.

This figure shows two parabolas with axis of symmetry y equals k,) and vertex (h, k. The one on the left is labeled a greater than 0 and opens to the right. The other parabola opens to the left.

Looking at these parabolas, do their graphs represent a function? Since both graphs would fail the vertical line test, they do not represent a function.

To graph a parabola that opens to the left or to the right is basically the same as what we did for parabolas that open up or down, with the reversal of the x and y variables.

How To

Graph horizontal parabolas $(x = a y^{2} + b y + c or x = a {(y - k)}^{2} + h) (x = a y^{2} + b y + c$ using properties.

Step 1. Determine whether the parabola opens to the left or to the right.
Step 2. Find the axis of symmetry.
Step 3. Find the vertex.
Step 4. Find the x-intercept. Find the point symmetric to the x-intercept across the axis of symmetry.
Step 5. Find the y-intercepts.
Step 6. Graph the parabola.

Example 11.14

Graph $x = 2 y^{2} x = 2 y^{2}$ by using properties.

Show solution

Solution


Since $a = 2, a = 2,$ the parabola opens to the right.

To find the axis of symmetry, find $y = - \frac{b}{2 a} . y = - \frac{b}{2 a} .$


	The axis of symmetry is $y = 0 . y = 0 .$
The vertex is on the line $y = 0 . y = 0 .$
Let $y = 0 . y = 0 .$

	The vertex is $(0, 0) . (0, 0) .$

Since the vertex is $(0, 0), (0, 0),$ both the x- and y-intercepts are the point $(0, 0) . (0, 0) .$ To graph the parabola we need more points. In this case it is easiest to choose values of y.

In the equation x equals 2 y squared, when y is 1, x is 2 and when y is 2, x is 8. The points are (2, 1) and (8, 2).

We also plot the points symmetric to $(2, 1) (2, 1)$ and $(8, 2) (8, 2)$ across the y-axis, the points $(2, −1), (2, −1),$ $(8, −2) . (8, −2) .$

Graph the parabola.

This graph shows right opening parabola with vertex (0, 0). Four points are marked on it: point (2, 1), point (2, negative 1), point (8, 2) and point (8 minus 2).

Try It 11.27

Graph $x = y^{2} x = y^{2}$ by using properties.

Try It 11.28

Graph $x = − y^{2} x = − y^{2}$ by using properties.

In the next example, the vertex is not the origin.

Example 11.15

Graph $x = − y^{2} + 2 y + 8 x = − y^{2} + 2 y + 8$ by using properties.

Show solution

Solution


Since $a = −1, a = −1,$ the parabola opens to the left.

To find the axis of symmetry, find $y = - \frac{b}{2 a} . y = - \frac{b}{2 a} .$


	The axis of symmetry is $y = 1 . y = 1 .$
The vertex is on the line $y = 1 . y = 1 .$
Let $y = 1 . y = 1 .$

	The vertex is $(9, 1) . (9, 1) .$
The x-intercept occurs when $y = 0 . y = 0 .$


	The x-intercept is $(8, 0) . (8, 0) .$
The point $(8, 0) (8, 0)$ is one unit below the line of symmetry. The symmetric point one unit above the line of symmetry is $(8, 2) (8, 2)$	Symmetric point is $(8, 2) . (8, 2) .$
The y-intercept occurs when $x = 0 . x = 0 .$
Substitute $x = 0 . x = 0 .$
Solve.


	The y-intercepts are $(0, 4) (0, 4)$ and $(0, −2) . (0, −2) .$
Connect the points to graph the parabola.

Try It 11.29

Graph $x = − y^{2} - 4 y + 12 x = − y^{2} - 4 y + 12$ by using properties.

Try It 11.30

Graph $x = − y^{2} + 2 y - 3 x = − y^{2} + 2 y - 3$ by using properties.

In Table 11.1, we see the relationship between the equation in standard form and the properties of the parabola. The How To box lists the steps for graphing a parabola in the standard form $x = a {(y - k)}^{2} + h . x = a {(y - k)}^{2} + h .$ We will use this procedure in the next example.

Example 11.16

Graph $x = 2 {(y - 2)}^{2} + 1 x = 2 {(y - 2)}^{2} + 1$ using properties.

Show solution

Solution


Identify the constants a, h, k.	$a = 2, a = 2,$ $h = 1, h = 1,$ $k = 2 k = 2$
Since $a = 2, a = 2,$ the parabola opens to the right.

The axis of symmetry is $y = k . y = k .$	The axis of symmetry is $y = 2 . y = 2 .$
The vertex is $(h, k) . (h, k) .$	The vertex is $(1, 2) . (1, 2) .$
Find the x-intercept by substituting $y = 0 . y = 0 .$	$\begin{matrix} x & = & 2 {(y - 2)}^{2} + 1 \\ x & = & 2 {(0 - 2)}^{2} + 1 \\ x & = & 9 \end{matrix} \begin{matrix} x & = & 2 {(y - 2)}^{2} + 1 \\ x & = & 2 {(0 - 2)}^{2} + 1 \\ x & = & 9 \end{matrix}$
	The x-intercept is $(9, 0) . (9, 0) .$
Find the point symmetric to $(9, 0) (9, 0)$ across the axis of symmetry.	$(9, 4)$
Find the y-intercepts. Let $x = 0 . x = 0 .$	$\begin{matrix} x & = & 2 {(y - 2)}^{2} + 1 \\ 0 & = & 2 {(y - 2)}^{2} + 1 \\ −1 & = & 2 {(y - 2)}^{2} \end{matrix} \begin{matrix} x & = & 2 {(y - 2)}^{2} + 1 \\ 0 & = & 2 {(y - 2)}^{2} + 1 \\ −1 & = & 2 {(y - 2)}^{2} \end{matrix}$
	A square cannot be negative, so there is no real solution. So there are no y-intercepts.
Graph the parabola.

Try It 11.31

Graph $x = 3 {(y - 1)}^{2} + 2 x = 3 {(y - 1)}^{2} + 2$ using properties.

Try It 11.32

Graph $x = 2 {(y - 3)}^{2} + 2 x = 2 {(y - 3)}^{2} + 2$ using properties.

In the next example, we notice the a is negative and so the parabola opens to the left.

Example 11.17

Graph $x = −4 {(y + 1)}^{2} + 4 x = −4 {(y + 1)}^{2} + 4$ using properties.

Show solution

Solution


Identify the constants a, h, k.	$a = −4, a = −4,$ $h = 4, h = 4,$ $k = −1 k = −1$
Since $a = −4, a = −4,$ the parabola opens to the left.

The axis of symmetry is $y = k . y = k .$	The axis of symmetry is $y = −1 . y = −1 .$
The vertex is $(h, k) . (h, k) .$	The vertex is $(4, −1) . (4, −1) .$
Find the x-intercept by substituting $y = 0 . y = 0 .$	$\begin{matrix} x & = & −4 {(y + 1)}^{2} + 4 \\ x & = & −4 {(0 + 1)}^{2} + 4 \\ x & = & 0 \end{matrix}$
	The x-intercept is $(0, 0) . (0, 0) .$
Find the point symmetric to $(0, 0) (0, 0)$ across the axis of symmetry.	$(0, −2)$
Find the y-intercepts.	$x = −4 {(y + 1)}^{2} + 4$
Let $x = 0 . x = 0 .$	$\begin{matrix} 0 & = & −4 {(y + 1)}^{2} + 4 \\ −4 & = & −4 {(y + 1)}^{2} \\ 1 & = & {(y + 1)}^{2} \\ y + 1 & = & \pm 1 \end{matrix} \begin{matrix} 0 & = & −4 {(y + 1)}^{2} + 4 \\ −4 & = & −4 {(y + 1)}^{2} \\ 1 & = & {(y + 1)}^{2} \\ y + 1 & = & \pm 1 \end{matrix}$
	$y = −1 + 1 y = −1 - 1 y = −1 + 1$
	$y = 0 y = −2 y = 0$
	The y-intercepts are $(0, 0) (0, 0)$ and $(0, −2) . (0, −2) .$
Graph the parabola.

Try It 11.33

Graph $x = −4 {(y + 2)}^{2} + 4 x = −4 {(y + 2)}^{2} + 4$ using properties.

Try It 11.34

Graph $x = −2 {(y + 3)}^{2} + 2 x = −2 {(y + 3)}^{2} + 2$ using properties.

The next example requires that we first put the equation in standard form and then use the properties.

Example 11.18

Write $x = 2 y^{2} + 12 y + 17 x = 2 y^{2} + 12 y + 17$ in standard form and then use the properties of the standard form to graph the equation.

Show solution

Solution


Rewrite the function in $x = a {(y - k)}^{2} + h x = a {(y - k)}^{2} + h$ form by completing the square.



Identify the constants a, h, k.	$a = 2, h = −1, k = −3 a = 2,$
Since $a = 2, a = 2,$ the parabola opens to the right.

The axis of symmetry is $y = k . y = k .$	The axis of symmetry is $y = −3 . y = −3 .$
The vertex is $(h, k) . (h, k) .$	The vertex is $(−1, −3) . (−1, −3) .$
Find the x-intercept by substituting $y = 0 . y = 0 .$	$\begin{matrix} x & = & 2 {(y + 3)}^{2} - 1 \\ x & = & 2 {(0 + 3)}^{2} - 1 \\ x & = & 17 \end{matrix}$
	The x-intercept is $(17, 0) . (17, 0) .$
Find the point symmetric to $(17, 0) (17, 0)$ across the axis of symmetry.	$(17, −6)$
Find the y-intercepts. Let $x = 0 . x = 0 .$	$\begin{matrix} x & = & 2 {(y + 3)}^{2} - 1 \\ 0 & = & 2 {(y + 3)}^{2} - 1 \\ 1 & = & 2 {(y + 3)}^{2} \\ \frac{1}{2} & = & {(y + 3)}^{2} \\ y + 3 & = & \pm \sqrt{\frac{1}{2}} \\ y & = & −3 \pm \frac{\sqrt{2}}{2} \end{matrix} \begin{matrix} x & = & 2 {(y + 3)}^{2} - 1 \\ 0 & = & 2 {(y + 3)}^{2} - 1 \\ 1 & = & 2 {(y + 3)}^{2} \\ \frac{1}{2} & = & {(y + 3)}^{2} \\ y + 3 & = & \pm \sqrt{\frac{1}{2}} \\ y & = & −3 \pm \frac{\sqrt{2}}{2} \end{matrix}$
	$y = −3 + \frac{\sqrt{2}}{2} y = −3 - \frac{\sqrt{2}}{2} y = −3 + \frac{\sqrt{2}}{2}$
	$y \approx - 2.3 y \approx - 3.7 y \approx - 2.3$
	The y-intercepts are $(0, −3 + \frac{\sqrt{2}}{2}), (0, −3 - \frac{\sqrt{2}}{2}) . (0, −3 + \frac{\sqrt{2}}{2}), (0, −3 - \frac{\sqrt{2}}{2}) .$
Graph the parabola.

Try It 11.35

ⓐ Write $x = 3 y^{2} + 6 y + 7 x = 3 y^{2} + 6 y + 7$ in standard form and ⓑ use properties of the standard form to graph the equation.

Try It 11.36

ⓐ Write $x = −4 y^{2} - 16 y - 12 x = −4 y^{2} - 16 y - 12$ in standard form and ⓑ use properties of the standard form to graph the equation.

Solve Applications with Parabolas

Many architectural designs incorporate parabolas. It is not uncommon for bridges to be constructed using parabolas as we will see in the next example.

Example 11.19

Find the equation of the parabolic arch formed in the foundation of the bridge shown. Write the equation in standard form.

This figure shows a parabolic arch formed in the foundation of a bridge. It is 10 feet high and 20 feet wide at the base.

Show solution

Solution

We will first set up a coordinate system and draw the parabola. The graph will give us the information we need to write the equation of the graph in the standard form $y = a {(x - h)}^{2} + k . y = a {(x - h)}^{2} + k .$

Let the lower left side of the bridge be the origin of the coordinate grid at the point $(0, 0) . (0, 0) .$ Since the base is 20 feet wide the point $(20, 0) (20, 0)$ represents the lower right side. The bridge is 10 feet high at the highest point. The highest point is the vertex of the parabola so the y-coordinate of the vertex will be 10. Since the bridge is symmetric, the vertex must fall halfway between the left most point, $(0, 0), (0, 0),$ and the rightmost point $(20, 0) . (20, 0) .$ From this we know that the x-coordinate of the vertex will also be 10.
Identify the vertex, $(h, k) . (h, k) .$	$(h, k) = (10, 10) (h, k) = (10, 10)$
	$h = 10, k = 10 h = 10, k = 10$
Substitute the values into the standard form. The value of a is still unknown. To find the value of a use one of the other points on the parabola.	$\begin{matrix} y & = & a {(x - h)}^{2} + k \\ y & = & a {(x - 10)}^{2} + 10 \\ (x, y) & = & (0, 0) \end{matrix}$
Substitute the values of the other point into the equation.	$\begin{matrix} y & = & a {(x - 10)}^{2} + 10 \\ 0 & = & a {(0 - 10)}^{2} + 10 \end{matrix}$
Solve for a.	$\begin{matrix} 0 & = & a {(0 - 10)}^{2} + 10 \\ −10 & = & a {(−10)}^{2} \\ −10 & = & 100 a \\ \frac{−10}{100} & = & a \\ a & = & - \frac{1}{10} \end{matrix}$
	$y = a {(x - 10)}^{2} + 10 y = a {(x - 10)}^{2} + 10$
Substitute the value for a into the equation.	$y = - \frac{1}{10} {(x - 10)}^{2} + 10$

Try It 11.37

Find the equation of the parabolic arch formed in the foundation of the bridge shown. Write the equation in standard form.

This figure shows a parabolic arch formed in the foundation of a bridge. It is 20 feet high and 40 feet wide at the base.

Try It 11.38

Find the equation of the parabolic arch formed in the foundation of the bridge shown. Write the equation in standard form.

This figure shows a parabolic arch formed in the foundation of a bridge. It is 5 feet high and 10 feet wide at the base.

Media

Access these online resources for additional instructions and practice with quadratic functions and parabolas.

Practice Makes Perfect

Graph Vertical Parabolas

In the following exercises, graph each equation by using properties.

53.

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

54.

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

55.

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

56.

$y = 8 x^{2} - 10 x + 3 y = 8 x^{2} - 10 x + 3$

In the following exercises, ⓐ write the equation in standard form and ⓑ use properties of the standard form to graph the equation.

57.

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

58.

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

59.

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

60.

$y = 3 x^{2} - 12 x + 7 y = 3 x^{2} - 12 x + 7$

Graph Horizontal Parabolas

In the following exercises, graph each equation by using properties.

61.

$x = −2 y^{2} x = −2 y^{2}$

62.

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

63.

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

64.

$x = −4 y^{2} x = −4 y^{2}$

65.

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

66.

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

67.

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

68.

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

69.

$x = {(y - 2)}^{2} + 3 x = {(y - 2)}^{2} + 3$

70.

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

71.

$x = − {(y - 1)}^{2} + 2 x = − {(y - 1)}^{2} + 2$

72.

$x = − {(y - 4)}^{2} + 3 x = − {(y - 4)}^{2} + 3$

73.

$x = {(y + 2)}^{2} + 1 x = {(y + 2)}^{2} + 1$

74.

$x = {(y + 1)}^{2} + 2 x = {(y + 1)}^{2} + 2$

75.

$x = − {(y + 3)}^{2} + 2 x = − {(y + 3)}^{2} + 2$

76.

$x = − {(y + 4)}^{2} + 3 x = − {(y + 4)}^{2} + 3$

77.

$x = −3 {(y - 2)}^{2} + 3 x = −3 {(y - 2)}^{2} + 3$

78.

$x = −2 {(y - 1)}^{2} + 2 x = −2 {(y - 1)}^{2} + 2$

79.

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

80.

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

In the following exercises, ⓐ write the equation in standard form and ⓑ use properties of the standard form to graph the equation.

81.

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

82.

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

83.

$x = −2 y^{2} - 12 y - 16 x = −2 y^{2} - 12 y - 16$

84.

$x = −3 y^{2} - 6 y - 5 x = −3 y^{2} - 6 y - 5$

Mixed Practice

In the following exercises, match each graph to one of the following equations: ⓐ x² + y² = 64 ⓑ x² + y² = 49
ⓒ (x + 5)² + (y + 2)² = 4 ⓓ (x − 2)² + (y − 3)² = 9 ⓔ y = −x² + 8x − 15 ⓕ y = 6x² + 2x − 1

85.

This graph shows circle with center (0, 0) and radius 8 units.

86.

This graph shows a parabola opening upwards. Its vertex has an x value of slightly less than 0 and a y value of slightly less than negative 1. A point on it is close to (negative 1, 3).

87.

This graph shows circle with center (0, 0) and radius 7 units.

88.

This graph shows a parabola opening downwards with vertex (4, 1) and x intercepts (3, 0) and (5, 0).

89.

This graph shows circle with center (2, 3) and radius 3 units.

90.

This graph shows circle with center (negative 5, negative 2) and radius 2 units.

Solve Applications with Parabolas

91.

Write the equation in standard form of the parabolic arch formed in the foundation of the bridge shown. Use the lower left side of the bridge as the origin (0, 0).

92.

Write the equation in standard form of the parabolic arch formed in the foundation of the bridge shown. Use the lower left side of the bridge as the origin (0, 0).

This figure shows a parabolic arch formed in the foundation of a bridge. It is 50 feet high and 100 feet wide at the base.

93.

Write the equation in standard form of the parabolic arch formed in the foundation of the bridge shown. Use the lower left side of the bridge as the origin (0, 0).

This figure shows a parabolic arch formed in the foundation of a bridge. It is 90 feet high and 60 feet wide at the base.

94.

Write the equation in standard form of the parabolic arch formed in the foundation of the bridge shown. Use the lower left side of the bridge as the origin (0, 0).

This figure shows a parabolic arch formed in the foundation of a bridge. It is 45 feet high and 30 feet wide at the base.

Writing Exercises

95.

In your own words, define a parabola.

96.

Is the parabola $y = x^{2} y = x^{2}$ a function? Is the parabola $x = y^{2} x = y^{2}$ a function? Explain why or why not.

97.

Write the equation of a parabola that opens up or down in standard form and the equation of a parabola that opens left or right in standard form. Provide a sketch of the parabola for each one, label the vertex and axis of symmetry.

98.

Explain in your own words, how you can tell from its equation whether a parabola opens up, down, left or right.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has four columns, 3 rows and a header row. The header row labels each column I can, confidently, with some help and no, I don’t get it. The first column has the following statements: graph vertical parabolas, graph horizontal parabolas, solve applications with parabolas. The remaining columns are blank.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

Learning Objectives

Be Prepared 11.2

Graph Vertical Parabolas

Parabola

How To

Graph vertical parabolas (y=ax2+bx+corf(x)=a(x−h)2+k)(y=ax2+bx+corf(x)=a(x−h)2+k) using properties.

Example 11.12

Solution

Try It 11.23

Try It 11.24

Example 11.13

Solution

Try It 11.25

Try It 11.26

Graph Horizontal Parabolas

How To

Graph horizontal parabolas (x=ay2+by+corx=a(y−k)2+h)(x=ay2+by+corx=a(y−k)2+h) using properties.

Example 11.14

Solution

Try It 11.27

Try It 11.28

Example 11.15

Solution

Try It 11.29

Try It 11.30

Example 11.16

Solution

Try It 11.31

Try It 11.32

Example 11.17

Solution

Try It 11.33

Try It 11.34

Example 11.18

Solution

Try It 11.35

Try It 11.36

Solve Applications with Parabolas

Example 11.19

Solution

Try It 11.37

Try It 11.38

Media

Practice Makes Perfect

Writing Exercises

Self Check

Graph vertical parabolas $(y = a x^{2} + b x + c or f (x) = a {(x - h)}^{2} + k) (y = a x^{2} + b x + c$ using properties.

Graph horizontal parabolas $(x = a y^{2} + b y + c or x = a {(y - k)}^{2} + h) (x = a y^{2} + b y + c$ using properties.