11.2 Parabolas

Graph: y = −3 x 2 + 12 x − 12 . y = −3 x 2 + 12 x − 12 .
If you missed this problem, review Example 9.47.

Be Prepared 11.5

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.

Be Prepared 11.6

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.

y equals ax squared plus bx plus c and standard form is y equals a open parentheses x minus h close parentheses squared plus k. Row one: orientation: general form is a greater than 0, up and a less than 0 down. Standard form is the same. Row 2: Axis of symmetry: general form is x equals minus b upon 2a and standard form is x equals h. Row 3: vertex: general form, substitute x equals minus b upon 2a and solve for y; standard form is point h, k. Row 4: y intercept: general and standard forms, let x be 0. Row 5: x intercept: general and standard forms, let y be 0." data-label="">
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 = − b 2 a x = − b 2 a x = h x = h
Vertex Substitute x = − b 2 a x = − 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.

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 or f ( x ) = a ( x − h ) 2 + k ) using properties.

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

Solution

y equals ax squared plus bx plus c. Since a is minus 1, the parabola opens downward. To find the axis of symmetry, find x equals minus b upon 2a. Substituting values of b and a, we get x equals 3. This is the axis of symmetry. The vertex is on the line x equals 3. Substituting this value in the equation, we get y equals 1. The vertex is the point 3, 1. The y intercept occurs when x equals 0. Substituting in the equation and simplifying, we get y equals minus 8. The point 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, negative 8. The x intercept occurs when y equals 0. We substitute this in the original equation and factor the trinomial. We get x intercepts 4, 0 and 2, 0. Graph the parabola." data-label="">
Since a is −1 , −1 , the parabola opens downward.
To find the axis of symmetry, find x = − b 2 a . x = − 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.

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 = 3 , a = 3 , 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 · 0 2 − 6 · 0 + 5 y = 3 · 0 2 − 6 · 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. y = 3 ( x − 1 ) 2 + 2 0 = 3 ( x − 1 ) 2 + 2 − 2 = 3 ( x − 1 ) 2 − 2 3 = ( x − 1 ) 2 ± − 2 3 = x − 1 y = 3 ( x − 1 ) 2 + 2 0 = 3 ( x − 1 ) 2 + 2 − 2 = 3 ( x − 1 ) 2 − 2 3 = ( x − 1 ) 2 ± − 2 3 = x − 1
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 = − b 2 a y = − b 2 a y = k y = k
Vertex Substitute y = − b 2 a y = − 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

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.

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 or x = a ( y − k ) 2 + h ) using properties.

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

Example 11.14

Graph x = 2 y 2 x = 2 y 2 by using properties.

Solution

This is the axis of symmetry. The vertex is on this line. Let y be 0. Substituting in equation, we get x equals 0. The vertex is (0, 0). Since the vertex is (0, 0) both the x- and y-intercepts are the point (0, 0). To graph the parabola we need more points. In this case it is easiest to choose values of y." data-label="">
Since a = 2 , a = 2 , the parabola opens to the right.
To find the axis of symmetry, find y = − b 2 a . y = − 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.


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.

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.

Solution

Since a = −1 , a = −1 , the parabola opens to the left.
To find the axis of symmetry, find y = − b 2 a . y = − 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.

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 . x = 2 ( y − 2 ) 2 + 1 x = 2 ( 0 − 2 ) 2 + 1 x = 9 x = 2 ( y − 2 ) 2 + 1 x = 2 ( 0 − 2 ) 2 + 1 x = 9
The x-intercept is ( 9 , 0 ) . ( 9 , 0 ) .
Find the point symmetric to ( 9 , 0 ) ( 9 , 0 ) across the
axis of symmetry.		 ( 9 , 4 ) ( 9 , 4 )
Find the y-intercepts. Let x = 0 . x = 0 . x = 2 ( y − 2 ) 2 + 1 0 = 2 ( y − 2 ) 2 + 1 −1 = 2 ( y − 2 ) 2 x = 2 ( y − 2 ) 2 + 1 0 = 2 ( y − 2 ) 2 + 1 −1 = 2 ( y − 2 ) 2
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.

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 . x = −4 ( y + 1 ) 2 + 4 x = −4 ( 0 + 1 ) 2 + 4 x = 0 x = −4 ( y + 1 ) 2 + 4 x = −4 ( 0 + 1 ) 2 + 4 x = 0
The x-intercept is ( 0 , 0 ) . ( 0 , 0 ) .
Find the point symmetric to ( 0 , 0 ) ( 0 , 0 ) across the
axis of symmetry.		 ( 0 , −2 ) ( 0 , −2 )
Find the y-intercepts. x = −4 ( y + 1 ) 2 + 4 x = −4 ( y + 1 ) 2 + 4
Let x = 0 . x = 0 . 0 = −4 ( y + 1 ) 2 + 4 −4 = −4 ( y + 1 ) 2 1 = ( y + 1 ) 2 y + 1 = ± 1 0 = −4 ( y + 1 ) 2 + 4 −4 = −4 ( y + 1 ) 2 1 = ( y + 1 ) 2 y + 1 = ± 1
y = −1 + 1 y = −1 − 1 y = −1 + 1 y = −1 − 1
y = 0 y = −2 y = 0 y = −2
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.

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 , h = −1 , k = −3
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 .		 x = 2 ( y + 3 ) 2 − 1 x = 2 ( 0 + 3 ) 2 − 1 x = 17 x = 2 ( y + 3 ) 2 − 1 x = 2 ( 0 + 3 ) 2 − 1 x = 17
The x-intercept is ( 17 , 0 ) . ( 17 , 0 ) .
Find the point symmetric to ( 17 , 0 ) ( 17 , 0 )
across the axis of symmetry.		 ( 17 , −6 ) ( 17 , −6 )
Find the y-intercepts.

Let x = 0 . x = 0 .		 x = 2 ( y + 3 ) 2 − 1 0 = 2 ( y + 3 ) 2 − 1 1 = 2 ( y + 3 ) 2 1 2 = ( y + 3 ) 2 y + 3 = ± 1 2 y = −3 ± 2 2 x = 2 ( y + 3 ) 2 − 1 0 = 2 ( y + 3 ) 2 − 1 1 = 2 ( y + 3 ) 2 1 2 = ( y + 3 ) 2 y + 3 = ± 1 2 y = −3 ± 2 2
y = −3 + 2 2 y = −3 − 2 2 y = −3 + 2 2 y = −3 − 2 2
y ≈ − 2.3 y ≈ − 3.7 y ≈ − 2.3 y ≈ − 3.7
The y-intercepts are ( 0 , −3 + 2 2 ) , ( 0 , −3 − 2 2 ) . ( 0 , −3 + 2 2 ) , ( 0 , −3 − 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.

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 .

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) and the rightmost point (20, 0). From this we know that the x coordinate of the vertex will also be 10. The vertex is 10, 10. So h is 10 and k is 10. Substitute the values into the standard form y equals a open parentheses x minus h close parentheses squared plus k. The value of a is still unknown. To find the value of a use one of the other points on the parabola, point (0, 0). Substituting the values into the equation, we get a equal to minus 1 by 10. Substitute the value for a into the equation. We get y equals minus 1 upon 10 open parentheses x minus 10 close parentheses squared plus 10." data-label="">
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.		 y = a ( x − h ) 2 + k y = a ( x − 10 ) 2 + 10 ( x , y ) = ( 0 , 0 ) y = a ( x − h ) 2 + k y = a ( x − 10 ) 2 + 10 ( x , y ) = ( 0 , 0 )
Substitute the values of the other point
into the equation.		 y = a ( x − 10 ) 2 + 10 0 = a ( 0 − 10 ) 2 + 10 y = a ( x − 10 ) 2 + 10 0 = a ( 0 − 10 ) 2 + 10
Solve for a. 0 = a ( 0 − 10 ) 2 + 10 −10 = a ( −10 ) 2 −10 = 100 a −10 100 = a a = − 1 10 0 = a ( 0 − 10 ) 2 + 10 −10 = a ( −10 ) 2 −10 = 100 a −10 100 = a a = − 1 10
y = a ( x − 10 ) 2 + 10 y = a ( x − 10 ) 2 + 10
Substitute the value for a into the
equation.		 y = − 1 10 ( x − 10 ) 2 + 10 y = − 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.

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.

Media

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

Section 11.2 Exercises

Practice Makes Perfect

Graph Vertical Parabolas

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

y = − x 2 + 4 x − 3 y = − x 2 + 4 x − 3

y = − x 2 + 8 x − 15 y = − x 2 + 8 x − 15

y = 6 x 2 + 2 x − 1 y = 6 x 2 + 2 x − 1

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.

y = − x 2 + 2 x − 4 y = − x 2 + 2 x − 4

y = 2 x 2 + 4 x + 6 y = 2 x 2 + 4 x + 6

y = −2 x 2 − 4 x − 5 y = −2 x 2 − 4 x − 5

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.

x = −2 y 2 x = −2 y 2

x = 3 y 2 x = 3 y 2

x = 4 y 2 x = 4 y 2

x = −4 y 2 x = −4 y 2

x = − y 2 − 2 y + 3 x = − y 2 − 2 y + 3

x = − y 2 − 4 y + 5 x = − y 2 − 4 y + 5

x = y 2 + 6 y + 8 x = y 2 + 6 y + 8

x = y 2 − 4 y − 12 x = y 2 − 4 y − 12

x = ( y − 2 ) 2 + 3 x = ( y − 2 ) 2 + 3

x = ( y − 1 ) 2 + 4 x = ( y − 1 ) 2 + 4

x = − ( y − 1 ) 2 + 2 x = − ( y − 1 ) 2 + 2

x = − ( y − 4 ) 2 + 3 x = − ( y − 4 ) 2 + 3

x = ( y + 2 ) 2 + 1 x = ( y + 2 ) 2 + 1

x = ( y + 1 ) 2 + 2 x = ( y + 1 ) 2 + 2

x = − ( y + 3 ) 2 + 2 x = − ( y + 3 ) 2 + 2

x = − ( y + 4 ) 2 + 3 x = − ( y + 4 ) 2 + 3

x = −3 ( y − 2 ) 2 + 3 x = −3 ( y − 2 ) 2 + 3

x = −2 ( y − 1 ) 2 + 2 x = −2 ( y − 1 ) 2 + 2

x = 4 ( y + 1 ) 2 − 4 x = 4 ( y + 1 ) 2 − 4

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.

x = y 2 + 4 y − 5 x = y 2 + 4 y − 5

x = y 2 + 2 y − 3 x = y 2 + 2 y − 3

x = −2 y 2 − 12 y − 16 x = −2 y 2 − 12 y − 16

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: ⓐ x 2 + y 2 = 64 ⓑ x 2 + y 2 = 49
ⓒ (x + 5) 2 + (y + 2) 2 = 4 ⓓ (x − 2) 2 + (y − 3) 2 = 9 ⓔ y = −x 2 + 8x − 15 ⓕ y = 6x 2 + 2x − 1

Solve Applications with Parabolas

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

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

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

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

Writing Exercises

In your own words, define a parabola.

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.

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.

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.

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

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