Quadratic Functions and Graphs Practice Questions — SAT — Page 3

Question 41Question →

The graph of the quadratic function $f(x) = -2(x - 3)^2 + a$ in the $xy$ -plane has a $y$ -intercept at $(0, -10)$ , where $a$ is a constant. What is the maximum value of $f(x)$ ?

-28

-10

3

8

Show answer & explanation

Answer: $8$

Answer

The maximum value of the function is

8

.

The correct answer is

8

. Since the

y

-intercept of the graph is

(0, -10)

, we substitute

x = 0

into the function:

f(0) = -2(0 - 3)^2 + a = -10

. Simplifying this yields

-2(9) + a = -10

, which becomes

-18 + a = -10

. Adding

18

to both sides gives

a = 8

. The equation of the function is therefore

f(x) = -2(x - 3)^2 + 8

. Since this is in vertex form, the vertex is

(3, 8)

. Because the coefficient of the squared term is negative, the parabola opens downward, and the maximum value of the function is the

y

-coordinate of the vertex, which is

8

.

Step-by-Step Solution

1

Use the given

y

-intercept to set up an equation for the constant

a

.

f(0) = -10 \implies -2(0 - 3)^2 + a = -10

The

y

-intercept occurs where the input

x

is equal to

0

.

2

Simplify the expression and solve for

a

.

-2(-3)^2 + a = -10 \implies -2(9) + a = -10 \implies -18 + a = -10 \implies a = 8

Squaring

-3

yields

9

, and multiplying by

-2

gives

-18

. Adding

18

to both sides isolates

a

.

3

Identify the vertex of the quadratic function and determine the maximum value.

The function is

f(x) = -2(x - 3)^2 + 8

. The vertex of this parabola is

(3, 8)

. Since the leading coefficient

-2

is negative, the parabola opens downward, meaning the

y

-coordinate of the vertex,

8

, is the maximum value of

f(x)

.

A quadratic function in vertex form

f(x) = a(x - h)^2 + k

has its vertex at

(h, k)

. If

a < 0

, the maximum value is

k

.

Key Concept

Identifying the vertex and maximum value of a quadratic function from its vertex form and

y

-intercept.

Question 42Question →

A quadratic function $f$ has a vertex at $(h, k)$ , where $h$ and $k$ are constants. In the $xy$ -plane, the graph of $y = f(x)$ contains the points $(-1, 3)$ and $(7, 3)$ . The function $g$ is defined by $g(x) = f(x + 2) - 4$ , and its graph has a vertex at $(p, q)$ . If $p + q = 5$ , what is the value of $k$ ?

Show answer & explanation

Answer: 8

Answer

The value of k is 8.

The axis of symmetry of the quadratic function f is located at the midpoint of the x-values of any two points on the graph that have the same y-value. Since the graph of f contains (-1, 3) and (7, 3), the x-coordinate of its vertex, h, is equal to the midpoint (-1 + 7) / 2 = 3. Thus, the vertex of f is (3, k). The definition of g(x) = f(x + 2) - 4 indicates a horizontal translation of 2 units to the left and a vertical translation of 4 units down. Applying these translations to the vertex (3, k) of f gives the vertex of g as (3 - 2, k - 4) = (1, k - 4). Since the vertex of g is defined as (p, q), we have p = 1 and q = k - 4. Substituting these values into the equation p + q = 5 yields 1 + (k - 4) = 5, which simplifies to k - 3 = 5, or k = 8.

Step-by-Step Solution

1

Find the x-coordinate of the vertex of the function f

h = 3

Because the points (-1, 3) and (7, 3) have the same y-coordinate, the axis of symmetry (and thus the x-coordinate of the vertex) must lie exactly halfway between x = -1 and x = 7.

2

Determine the vertex of the function g(x)

(p, q) = (1, k - 4)

The function g(x) = f(x + 2) - 4 represents a translation of the graph of f by 2 units to the left and 4 units down. Since the vertex of f is (3, k), the vertex of g is (3 - 2, k - 4).

3

Solve for k using the given sum p + q = 5

k = 8

Substitute p = 1 and q = k - 4 into the equation p + q = 5 to get 1 + k - 4 = 5, which simplifies to k = 8.

Key Concept

Quadratic functions and graphs, including symmetry and translations.

Question 43Question →

In the $xy$ -plane, the graph of the quadratic function $f(x) = -2(x - d)^2 + 8$ , where $d$ is a positive constant, intersects the $x$ -axis at the point $(2, 0)$ . The function $g$ is defined by $g(x) = f(x + 3) - 5$ . What is the $y$ -coordinate of the $y$ -intercept of the graph of $g$ in the $xy$ -plane?

-15

-95

1

11

Show answer & explanation

Answer: 1

Answer

The

y

-coordinate of the

y

-intercept of the graph of

g

is

1

.

To find the correct answer, we first substitute the point

(2,0)

into

f(x) = -2(x - d)^2 + 8

to obtain

0 = -2(2 - d)^2 + 8

. Simplifying this equation gives

(2 - d)^2 = 4

, which has solutions

d = 0

and

d = 4

. Since the problem states that

d

is a positive constant, we have

d = 4

, which means

f(x) = -2(x - 4)^2 + 8

. We then find the equation for the transformed function

g(x)

by replacing

x

with

x + 3

and subtracting

5

, which yields

g(x) = f(x + 3) - 5 = -2(x + 3 - 4)^2 + 8 - 5 = -2(x - 1)^2 + 3

. To find the

y

-coordinate of the

y

-intercept, we evaluate

g(x)

at

x = 0

, giving

g(0) = -2(0 - 1)^2 + 3 = -2(1) + 3 = 1

.

Step-by-Step Solution

1

Use the given

x

-intercept

(2,0)

of

f(x)

to solve for the constant

d

.

d = 4

Substitute

x = 2

and

f(x) = 0

into the equation

f(x) = -2(x - d)^2 + 8

to get

0 = -2(2 - d)^2 + 8

. Dividing by

-2

and isolating the squared term gives

(2 - d)^2 = 4

. Taking the square root of both sides gives

2 - d = 2

or

2 - d = -2

, which yields

d = 0

or

d = 4

. Since

d

must be a positive constant, we must choose

d = 4

.

2

Determine the equation of the transformed function

g(x)

.

g(x) = -2(x - 1)^2 + 3

Substitute

d = 4

into

f(x)

to get

f(x) = -2(x - 4)^2 + 8

. Since

g(x) = f(x + 3) - 5

, substitute

x + 3

for

x

in

f(x)

and subtract

5

:

g(x) = -2((x + 3) - 4)^2 + 8 - 5

, which simplifies to

g(x) = -2(x - 1)^2 + 3

.

3

Find the

y

-coordinate of the

y

-intercept of the graph of

g

.

1

The

y

-intercept of a graph is the point where

x = 0

. Substitute

x = 0

into the equation for

g(x)

to find

g(0) = -2(0 - 1)^2 + 3 = -2(1) + 3 = 1

.

Key Concept

Applying horizontal and vertical translations to quadratic equations and identifying graph features.

Question 44Question →

A quadratic function $f$ is defined by $f(x) = a(x - 4)(x - 10)$ , where $a$ is a positive constant. In the $xy$ -plane, the graph of $f$ has a vertex with a $y$ -coordinate of $-18$ . What is the value of $a$ ?

Show answer & explanation

Answer: 2

Answer

The value of

a

is 2.

The vertex of a parabola lies on the axis of symmetry, which is located midway between its x-intercepts. Since the function is

f(x) = a(x - 4)(x - 10)

, the x-intercepts are at

x = 4

and

x = 10

. The midpoint of these values is

\frac{4 + 10}{2} = 7

. Thus, the x-coordinate of the vertex is 7. Since the y-coordinate of the vertex is given as

-18

, the vertex is at

(7, -18)

. Substituting these coordinates into the function equation yields

-18 = a(7 - 4)(7 - 10)

, which simplifies to

-18 = a(3)(-3)

, or

-18 = -9a

. Solving for

a

gives

a = 2

.

Step-by-Step Solution

1

Find the x-coordinate of the vertex using the symmetry of the parabola.

The x-coordinate of the vertex is 7.

The x-coordinate of the vertex lies on the axis of symmetry, which is the midpoint of the x-intercepts

x = 4

and

x = 10

.

2

Identify the coordinates of the vertex.

The vertex is at

(7, -18)

.

The y-coordinate of the vertex is given as

-18

, and its x-coordinate was calculated as 7.

3

Substitute the vertex coordinates into the function definition to solve for

a

.

a = 2

Substituting

(7, -18)

into

f(x) = a(x - 4)(x - 10)

gives

-18 = a(7 - 4)(7 - 10)

, which simplifies to

-18 = -9a

.

Key Concept

Finding the vertex of a quadratic function from its factored form and solving for a leading coefficient.

Alternative Method

Alternatively, the function

f(x) = a(x - 4)(x - 10)

can be expanded to standard form:

f(x) = a(x^2 - 14x + 40)

. The x-coordinate of the vertex can be found using the formula

x = -\frac{b}{2a}

, which gives

x = -\frac{-14a}{2a} = 7

. Then, substitute

x = 7

and

f(7) = -18

to solve for

a

.

Estimated Time:1m 30s

Question 45Question →

A quadratic function $g$ is defined by $g(x) = 2x^2 - 12x + k$ , where $k$ is a constant. In the $xy$ -plane, the graph of $g$ has its vertex on the line $y = -5$ . What is the value of $k$ ?

-59

-23

-5

13

Show answer & explanation

Answer: 13

Answer

13

The

x

-coordinate of the vertex for the quadratic function

g(x) = ax^2 + bx + c

is found using the formula

x = -\frac{b}{2a}

. Substituting

a = 2

and

b = -12

yields

x = -\frac{-12}{2(2)} = 3

. Evaluating

g(3)

gives the

y

-coordinate of the vertex:

g(3) = 2(3)^2 - 12(3) + k = -18 + k

. Since the vertex lies on the line

y = -5

, setting

-18 + k = -5

and solving for

k

gives

k = 13

.

Step-by-Step Solution

1

Find the

x

-coordinate of the vertex of the parabola.

The

x

-coordinate is

3

.

For a quadratic function

g(x) = ax^2 + bx + c

, the

x

-coordinate of the vertex is given by

x = -\frac{b}{2a}

. Here,

a = 2

and

b = -12

, so

x = -\frac{-12}{2(2)} = 3

.

2

Evaluate the function at

x = 3

to find the

y

-coordinate of the vertex in terms of

k

.

The

y

-coordinate is

-18 + k

.

Substituting

x = 3

into the function gives

g(3) = 2(3)^2 - 12(3) + k = 18 - 36 + k = -18 + k

.

3

Set the

y

-coordinate of the vertex equal to

-5

and solve for

k

.

k = 13

.

Since the vertex lies on the line

y = -5

, the

y

-coordinate of the vertex must be

-5

. Thus,

-18 + k = -5 \implies k = 13

.

Key Concept

Determining the vertex of a quadratic function and using its coordinates to solve for unknown parameters.

Question 46Question →

The graph of the quadratic function $f$ in the $xy$ -plane has its vertex at $(3, -4)$ and passes through the point $(1, 8)$ . The function $g$ is defined by $g(x) = f(x - h) + k$ , where $h$ and $k$ are constants. If the graph of $g$ has its vertex at $(0, 0)$ , what is the value of $g(4)$ ?

16

48

12

24

Show answer & explanation

Answer: 48

Answer

48

The quadratic function

f(x)

can be written in vertex form as

f(x) = a(x - 3)^2 - 4

. Substituting the point

(1, 8)

gives

8 = a(1 - 3)^2 - 4

, which simplifies to

12 = 4a

, so

a = 3

. This means

f(x) = 3(x - 3)^2 - 4

. The function

g(x) = f(x - h) + k

has its vertex at

(0, 0)

, which is a translation of the vertex of

f(x)

from

(3, -4)

to

(0, 0)

. This shifts the function to

g(x) = 3x^2

. Evaluating

g(4)

yields

3(4)^2 = 3(16) = 48

.

Step-by-Step Solution

1

Write the vertex form of the quadratic function

f(x)

.

f(x) = a(x - 3)^2 - 4

The vertex of the parabola is given as

(3, -4)

, so we can substitute these coordinates into the vertex form

f(x) = a(x - h)^2 + k

.

2

Substitute the point

(1, 8)

into the equation to solve for the constant

a

.

8 = a(1 - 3)^2 - 4 \implies 12 = 4a \implies a = 3

Since the graph passes through

(1, 8)

, these coordinates must satisfy the equation of the function.

3

Determine the equation of the transformed function

g(x)

.

g(x) = 3x^2

The vertex of

f(x)

is at

(3, -4)

. Shifting this vertex to

(0, 0)

removes the horizontal and vertical offsets, simplifying the function to

g(x) = ax^2

with the same leading coefficient

a = 3

.

4

Evaluate

g(4)

.

g(4) = 3(4)^2 = 3(16) = 48

Substitute

x = 4

into the equation for

g(x)

to find the required value.

Key Concept

Quadratic Functions and Graphs

Question 47Question →

A parabola in the $xy$ -plane has vertex $(3, 18)$ and passes through the origin. If the equation of the parabola is written in the form $y = ax^2 + bx + c$ , where $a$ , $b$ , and $c$ are constants, what is the value of $a + b$ ?

Show answer & explanation

Answer: 10

Answer

10

The vertex form of a quadratic function with vertex

(h, k)

is

y = a(x - h)^2 + k

. Substituting the given vertex

(3, 18)

yields the equation

y = a(x - 3)^2 + 18

. Since the parabola passes through the origin, we can substitute the point

(0, 0)

into the equation to find the value of

a

:

0 = a(0 - 3)^2 + 18 \implies 9a = -18 \implies a = -2

. Substituting

a = -2

back into the vertex form and expanding gives

y = -2(x - 3)^2 + 18 = -2(x^2 - 6x + 9) + 18 = -2x^2 + 12x

. Comparing this to the standard form

y = ax^2 + bx + c

, we identify

a = -2

and

b = 12

. The sum of these constants is

a + b = -2 + 12 = 10

.

Step-by-Step Solution

1

Write the equation of the parabola in vertex form.

y = a(x - 3)^2 + 18

The vertex form of a quadratic function is

y = a(x - h)^2 + k

, where

(h, k)

is the vertex.

2

Substitute the coordinates of the origin

(0, 0)

to solve for

a

.

a = -2

The parabola passes through the origin

(0, 0)

, so these coordinates must satisfy the equation.

3

Expand the vertex form equation into standard form

y = ax^2 + bx + c

.

y = -2x^2 + 12x

Expanding the equation allows us to identify the coefficients

a

and

b

directly.

4

Identify the values of

a

and

b

, and calculate

a + b

.

a + b = 10

Comparing

y = -2x^2 + 12x

to

y = ax^2 + bx + c

gives

a = -2

and

b = 12

, so

a + b = -2 + 12 = 10

.

Key Concept

Vertex form of quadratic functions and conversion to standard form

Question 48Question →

In the $xy$ -plane, the graph of the quadratic function $f$ has its vertex at the point $(4, -3)$ . The function $g$ is defined by $g(x) = f(x + 2) + 5$ . Which of the following ordered pairs represents the vertex of the graph of $g$ ?

(6, 2)

(2, -8)

(2, 2)

(6, -8)

Show answer & explanation

Answer: $(2, 2)$

Answer

(2, 2)

The vertex of

f(x)

is given as

(4, -3)

. The translation

g(x) = f(x + 2) + 5

shifts the graph

2

units to the left and

5

units upward. Applying these transformations to the vertex of

f(x)

yields the new vertex

(4 - 2, -3 + 5) = (2, 2)

.

Step-by-Step Solution

1

Identify the vertex of the function

f(x)

The vertex of

f(x)

is

(4, -3)

This is given in the problem and serves as the starting point for the transformations.

2

Determine the effect of the horizontal translation

f(x + 2)

The vertex shifts

2

units to the left, changing the

x

-coordinate from

4

to

4 - 2 = 2

For any function

f(x)

, the graph of

f(x + c)

is shifted

c

units to the left when

c > 0

.

3

Determine the effect of the vertical translation

+ 5

The vertex shifts

5

units upward, changing the

y

-coordinate from

-3

to

-3 + 5 = 2

Adding a constant to a function shifts its graph vertically upward by that constant's value.

4

Combine the translated coordinates to find the vertex of

g(x)

The vertex of the graph of

g(x)

is

(2, 2)

The horizontal shift results in an

x

-coordinate of

2

, and the vertical shift results in a

y

-coordinate of

2

.

Key Concept

Quadratic Transformations and Vertex Shifts

Question 49Question →

The graph of the quadratic function $f$ in the $xy$ -plane has $x$ -intercepts at $(-2, 0)$ and $(8, 0)$ . If the maximum value of $f(x)$ is $25$ , what is the value of $f(0)$ ?

Show answer & explanation

Answer: 16

Answer

16

The axis of symmetry of the quadratic function

f

lies halfway between the

x

-intercepts

x = -2

and

x = 8

, which is at

x = \frac{-2 + 8}{2} = 3

. Since the function has a maximum value of

25

, this maximum must occur at the vertex, giving the vertex coordinates

(3, 25)

. In vertex form, the function is

f(x) = a(x - 3)^2 + 25

. Substituting the

x

-intercept

(8, 0)

into the function yields

0 = a(8 - 3)^2 + 25

, which simplifies to

25a = -25

, or

a = -1

. Therefore, the equation of the function is

f(x) = -(x - 3)^2 + 25

. Evaluating this at

x = 0

gives

f(0) = -(0 - 3)^2 + 25 = -9 + 25 = 16

.

Step-by-Step Solution

1

Find the

x

-coordinate of the vertex (axis of symmetry)

x = 3

The axis of symmetry of a parabola is located exactly halfway between its

x

-intercepts:

x = \frac{-2 + 8}{2} = 3

.

2

Determine the vertex coordinates

(3, 25)

The maximum value of the quadratic function occurs at its vertex, so the

y

-coordinate of the vertex is the maximum value

25

.

3

Write the vertex form of the quadratic function

f(x) = a(x - 3)^2 + 25

The vertex form of a quadratic function is

f(x) = a(x - h)^2 + k

, where

(h, k)

is the vertex.

4

Solve for the leading coefficient

a

a = -1

Substitute the

x

-intercept

(8, 0)

into the vertex form:

0 = a(8 - 3)^2 + 25 \implies 25a = -25 \implies a = -1

.

5

Find the value of

f(0)

f(0) = 16

Substitute

x = 0

into the function:

f(0) = -(0 - 3)^2 + 25 = -9 + 25 = 16

.

Key Concept

Using

x

-intercepts and the maximum value to determine the vertex and equation of a quadratic function.

Question 50Question →

An object is launched from a platform. The function $h(t) = -5t^2 + 30t + 12$ models the height of the object, in meters, $t$ seconds after it was launched. What is the maximum height, in meters, reached by the object?

3

12

57

147

Show answer & explanation

Answer: 57

Answer

The maximum height reached by the object is 57 meters.

The given function

h(t) = -5t^2 + 30t + 12

represents a parabola opening downward because the coefficient of

t^2

is negative. The maximum value of this function occurs at its vertex. The time

t

at the vertex is determined using the formula

t = -\frac{b}{2a} = -\frac{30}{2(-5)} = 3

seconds. Substituting

t = 3

back into the function yields the maximum height:

h(3) = -5(3)^2 + 30(3) + 12 = -45 + 90 + 12 = 57

meters.

Step-by-Step Solution

1

Identify the coefficients of the quadratic function in standard form

h(t) = at^2 + bt + c

.

a = -5

,

b = 30

, and

c = 12

.

These coefficients are needed to calculate the vertex of the parabola.

2

Calculate the time

t

at which the maximum height occurs using the vertex formula

t = -\frac{b}{2a}

.

t = -\frac{30}{2(-5)} = 3

seconds.

Since the leading coefficient

a = -5

is negative, the parabola opens downward, meaning its vertex represents the maximum value.

3

Substitute

t = 3

back into the height function to find the maximum height.

h(3) = -5(3)^2 + 30(3) + 12 = -45 + 90 + 12 = 57

meters.

Evaluating the function at the time of the vertex gives the corresponding maximum height.

Key Concept

The maximum value of a quadratic function

f(x) = ax^2 + bx + c

(where

a < 0

) is the vertical coordinate of its vertex, which occurs at

x = -\frac{b}{2a}

.

Question 51Question →

In the $xy$ -plane, the graph of the quadratic function $f(x) = -x^2 + bx + c$ , where $b$ and $c$ are constants, has its vertex at $(4, 25)$ . If the positive $x$ -intercept of the graph of $f$ is $(d, 0)$ , what is the value of $d$ ?

Show answer & explanation

Answer: 9

Answer

The value of

d

is

9

.

The vertex form of a quadratic function is

f(x) = a(x-h)^2 + k

, where

(h, k)

is the vertex. Since the vertex is

(4, 25)

and the coefficient of

x^2

is

-1

, the function is

f(x) = -(x-4)^2 + 25

. Setting

f(x) = 0

to find the

x

-intercepts yields

-(x-4)^2 + 25 = 0

, which simplifies to

(x-4)^2 = 25

. Taking the square root of both sides gives

x - 4 = 5

or

x - 4 = -5

. Solving these equations gives

x = 9

or

x = -1

. The positive

x

-intercept is

(9, 0)

, so the value of

d

is

9

.

Step-by-Step Solution

1

Write the function in vertex form.

f(x) = -(x-4)^2 + 25

The vertex form of a quadratic function is

f(x) = a(x-h)^2 + k

, where

(h, k)

is the vertex. Here, the vertex is

(4, 25)

and the coefficient of

x^2

is

-1

, so

a = -1

,

h = 4

, and

k = 25

.

2

Set

f(x) = 0

to find the

x

-intercepts.

-(x-4)^2 + 25 = 0

The

x

-intercepts of a graph are the points where the function value is equal to 0.

3

Solve the equation for

x

.

x = 9

or

x = -1

Rearranging the equation gives

(x-4)^2 = 25

. Taking the square root of both sides gives

x-4 = 5

or

x-4 = -5

, which solves to

x = 9

or

x = -1

.

4

Identify the positive

x

-intercept coordinate

d

.

d = 9

The question asks for the positive

x

-intercept

(d, 0)

, which corresponds to

x = 9

.

Key Concept

Vertex form of a quadratic function and finding

x

-intercepts

Question 52Question →

The graph of the quadratic function $f$ in the $xy$ -plane has its vertex at $(2, -5)$ . If the graph passes through the point $(5, 13)$ , what is the value of $f(-1)$ ?

-23

-3

13

-11

Show answer & explanation

Answer: 13

Answer

13

The vertex of the parabola is

(2, -5)

, which indicates that the axis of symmetry is the vertical line

x = 2

. The given point has an

x

-coordinate of

5

, which is a distance of

3

units from the axis of symmetry (

5 - 2 = 3

). The target point has an

x

-coordinate of

-1

, which is also a distance of

3

units from the axis of symmetry (

2 - (-1) = 3

). Because a parabola is perfectly symmetric about its axis of symmetry, any two points that are the same horizontal distance from this line must have the same

y

-coordinate. Thus,

f(-1)

must be equal to

f(5)

, which is

13

. Alternatively, one can find the specific equation of the quadratic function by substituting the vertex and the point

(5, 13)

into the vertex form

f(x) = a(x - 2)^2 - 5

, yielding

a = 2

. Evaluating

f(-1) = 2(-1 - 2)^2 - 5

gives

13

.

Step-by-Step Solution

1

Identify the axis of symmetry from the given vertex.

The axis of symmetry is

x = 2

.

For any quadratic function with a vertex at

(h, k)

, the vertical line

x = h

is the axis of symmetry of its parabolic graph.

2

Determine the horizontal distance from the axis of symmetry to the given point

x = 5

and the target point

x = -1

.

The distance for

x = 5

is

5 - 2 = 3

units. The distance for

x = -1

is

2 - (-1) = 3

units.

Checking if the two

x

-coordinates are symmetric with respect to the line

x = 2

allows us to use the symmetry property of parabolas.

3

Apply the symmetry property to find the function value.

Since both

x = 5

and

x = -1

are equidistant from the axis of symmetry, their function values are equal:

f(-1) = f(5) = 13

.

Points on a parabola that are equidistant from the axis of symmetry have the same

y

-coordinate.

Key Concept

Symmetry of quadratic functions about their vertex axis of symmetry

Question 53Question →

For the quadratic function $f$ , the table shows some values of $x$ and their corresponding values of $f(x)$ .

$x$	$f(x)$
$1$	$15$
$3$	$3$
$5$	$15$

What is the value of $f(0)$ ?

Show answer & explanation

Answer: 30

Answer

30

Since the function values at

x = 1

and

x = 5

are both

15

, the axis of symmetry of the quadratic function must be located at the midpoint of these values, which is

x = 3

. The vertex of the function must therefore have an

x

-coordinate of

3

. From the table,

f(3) = 3

, which means the vertex is

(3, 3)

. The vertex form of the quadratic function is

f(x) = a(x - 3)^2 + 3

. To find the value of

a

, substitute the point

(1, 15)

into the equation:

15 = a(1 - 3)^2 + 3

, which simplifies to

15 = 4a + 3

, giving

a = 3

. The equation of the function is

f(x) = 3(x - 3)^2 + 3

. Substituting

x = 0

yields

f(0) = 3(0 - 3)^2 + 3 = 3(9) + 3 = 30

.

Step-by-Step Solution

1

Find the axis of symmetry.

The axis of symmetry is

x = 3

.

Since

f(1) = 15

and

f(5) = 15

, the function values are equal at these points. Because a quadratic function is symmetric, the axis of symmetry is the midpoint of these two

x

-values:

\frac{1 + 5}{2} = 3

.

2

Identify the vertex.

The vertex is

(3, 3)

.

The vertex of a quadratic function lies on its axis of symmetry. Since the axis of symmetry is

x = 3

, the

x

-coordinate of the vertex is

3

. From the table,

f(3) = 3

, so the vertex is

(3, 3)

.

3

Write the function in vertex form and solve for the leading coefficient.

f(x) = 3(x - 3)^2 + 3

The vertex form of a quadratic function is

f(x) = a(x - h)^2 + k

, where

(h, k)

is the vertex. Substituting

(3, 3)

, we get

f(x) = a(x - 3)^2 + 3

. Using the point

(1, 15)

from the table, we substitute

x = 1

and

f(x) = 15

:

15 = a(1 - 3)^2 + 3 \Rightarrow 15 = 4a + 3 \Rightarrow 12 = 4a \Rightarrow a = 3

.

4

Evaluate the function at

x = 0

.

f(0) = 30

Substitute

x = 0

into the function:

f(0) = 3(0 - 3)^2 + 3 = 3(9) + 3 = 27 + 3 = 30

.

Key Concept

Quadratic Functions and Graphs

Question 54Question →

In the $xy$ -plane, the graph of the quadratic function $f(x) = ax^2 + bx + c$ , where $a$ , $b$ , and $c$ are constants, has a vertex at $(2, -3)$ and passes through the point $(4, 5)$ . If the graph of a second quadratic function, $g$ , is obtained by translating the graph of $f$ horizontally by $3$ units to the right and vertically by $5$ units up, what is the value of $g(5)$ ?

-8

2

8

74

Show answer & explanation

Answer: 2

Answer

The value of

g(5)

is

2

.

The vertex of the graph of

f

is at

(2, -3)

, which means

f(2) = -3

. The graph of

g

is obtained by translating the graph of

f

by

3

units to the right and

5

units up, so its equation is

g(x) = f(x - 3) + 5

. To find

g(5)

, we substitute

x = 5

into this relation, which gives

g(5) = f(5 - 3) + 5 = f(2) + 5

. Since

f(2) = -3

, we have

g(5) = -3 + 5 = 2

. Alternatively, translating the vertex of

f

at

(2, -3)

by

3

units to the right and

5

units up gives the vertex of

g

at

(2 + 3, -3 + 5) = (5, 2)

. Since the vertex of the parabola

g

occurs at

x = 5

, the value of

g(5)

is the

y

-coordinate of the vertex, which is

2

.

Step-by-Step Solution

1

Identify the vertex of the function

f

and write its vertex form equation.

The vertex of

f

is

(2, -3)

, so the vertex form of the function is

f(x) = a(x - 2)^2 - 3

.

This allows us to find the specific equation of

f(x)

if needed, and also tells us that the value of

f(2)

is

-3

.

2

Set up the equation for the translated function

g(x)

.

A horizontal translation of

3

units to the right and a vertical translation of

5

units up is represented by

g(x) = f(x - 3) + 5

.

This defines the function

g

in terms of the function

f

using standard translation rules.

3

Evaluate

g(5)

using the relation from step 2.

g(5) = f(5 - 3) + 5 = f(2) + 5

.

This simplifies the calculation by using the value of

f

at the vertex.

4

Substitute the value of

f(2)

to find

g(5)

.

g(5) = -3 + 5 = 2

.

Since the vertex of

f

is at

(2, -3)

, we know

f(2) = -3

, which yields the final result.

Key Concept

Quadratic functions can be analyzed and transformed using their vertex form and function translation rules.

Estimated Time:1m 30s

Question 55Question →

The graph of the quadratic function $f(x) = x^2 - 6x + c$ , where $c$ is a constant, has its vertex at $(h, k)$ in the $xy$ -plane. If the graph of $f$ is translated $3$ units to the right and $2$ units down, the vertex of the translated graph lies on the line $y = 2x$ . What is the value of $c$ ?

Show answer & explanation

Answer: 23

Answer

23

To find the constant

c

, we first determine the vertex of the function

f(x) = x^2 - 6x + c

. The x-coordinate of the vertex of a parabola in the form

y = ax^2 + bx + c

is given by

x = -\frac{b}{2a}

. For this function,

x = -\frac{-6}{2(1)} = 3

. Substituting this back into the function gives the y-coordinate of the vertex:

f(3) = (3)^2 - 6(3) + c = c - 9

. Thus, the original vertex is at

(3, c - 9)

. Translating the graph

3

units to the right increases the x-coordinate of the vertex by

3

, making it

3 + 3 = 6

. Translating the graph

2

units down decreases the y-coordinate of the vertex by

2

, making it

(c - 9) - 2 = c - 11

. The problem states that this new vertex

(6, c - 11)

lies on the line

y = 2x

. Substituting these coordinates into the linear equation gives

c - 11 = 2(6)

, which simplifies to

c - 11 = 12

. Adding

11

to both sides gives the value of

c

as

23

.

Step-by-Step Solution

1

Find the vertex

(h, k)

of the original quadratic function

f(x) = x^2 - 6x + c

.

The vertex is at

(3, c - 9)

.

The x-coordinate of the vertex of a quadratic function

y = ax^2 + bx + c

is given by

h = -\frac{b}{2a}

. For

f(x) = x^2 - 6x + c

, we have

h = -\frac{-6}{2(1)} = 3

. Substituting

x = 3

into the function gives the y-coordinate:

k = f(3) = 3^2 - 6(3) + c = c - 9

.

2

Determine the coordinates of the vertex after translating the graph

3

units to the right and

2

units down.

The new vertex is at

(6, c - 11)

.

A translation of

3

units to the right increases the x-coordinate of the vertex by

3

, so the new x-coordinate is

3 + 3 = 6

. A translation of

2

units down decreases the y-coordinate of the vertex by

2

, so the new y-coordinate is

(c - 9) - 2 = c - 11

.

3

Set up an equation using the line

y = 2x

and solve for

c

.

c = 23

Since the translated vertex

(6, c - 11)

lies on the line

y = 2x

, substituting

x = 6

and

y = c - 11

into the line's equation must satisfy it:

c - 11 = 2(6) \Rightarrow c - 11 = 12 \Rightarrow c = 23

.

Key Concept

Determining the vertex of a quadratic function and applying translations to its graph.

Estimated Time:1m 30s

Question 56Question →

The graph of the quadratic function $f$ in the $xy$ -plane has its vertex at $(4, 12)$ . The function is defined by $f(x) = -(x - c)(x - d)$ , where $c$ and $d$ are constants. What is the value of the product $cd$ ?

Show answer & explanation

Answer: 4

Answer

The value of the product

cd

is

4

.

The

x

-coordinate of the vertex of a quadratic function of the form

f(x) = -(x-c)(x-d)

is the average of the

x

-intercepts

c

and

d

. Since the vertex is

(4, 12)

, we have

\frac{c+d}{2} = 4

, which means

c+d = 8

. Substituting the vertex

(4, 12)

into the function gives

12 = -(4-c)(4-d)

, which expands to

12 = -(16 - 4(c+d) + cd)

. Substituting

c+d = 8

gives

12 = -(16 - 32 + cd) = -(-16 + cd) = 16 - cd

. Solving for

cd

yields

cd = 4

.

Step-by-Step Solution

1

Find the sum of the constants

c

and

d

using the

x

-coordinate of the vertex.

c + d = 8

The graph of a quadratic function in the form

f(x) = -(x-c)(x-d)

has a vertical line of symmetry at the

x

-coordinate of its vertex, which is the midpoint of its

x

-intercepts

c

and

d

. Therefore,

\frac{c+d}{2} = 4

, which simplifies to

c + d = 8

.

2

Substitute the vertex coordinates

(4, 12)

into the function definition.

(4-c)(4-d) = -12

Since

(4, 12)

is the vertex, the point lies on the graph of

f

, meaning

f(4) = 12

. Substituting

x = 4

into the function gives

12 = -(4-c)(4-d)

, which simplifies to

(4-c)(4-d) = -12

.

3

Expand the expression

(4-c)(4-d)

and substitute

c+d = 8

to solve for

cd

.

cd = 4

Expanding

(4-c)(4-d) = -12

gives

16 - 4(c+d) + cd = -12

. Substituting

c+d = 8

yields

16 - 32 + cd = -12

, which simplifies to

-16 + cd = -12

. Adding

16

to both sides gives

cd = 4

.

Key Concept

Quadratic Functions and Graphs

Question 57Question →

The graph of the quadratic function $f(x) = ax^2 + bx + c$ , where $a$ , $b$ , and $c$ are constants, is a parabola in the $xy$ -plane that passes through the points $(-3, 22)$ and $(9, 22)$ . If the minimum value of $f(x)$ is $4$ , what is the value of $f(1)$ ?

Show answer & explanation

Answer: 6

Answer

The correct answer is 6.

The correct answer is 6. The axis of symmetry of the parabola is halfway between the points with equivalent y-values: x = 3. Using the minimum value of 4, the vertex is identified as (3, 4). Writing the equation in vertex form as f(x) = a(x - 3)^2 + 4 and substituting (9, 22) yields a = 0.5. Evaluating the function f(x) = 0.5(x - 3)^2 + 4 at x = 1 yields 6.

Step-by-Step Solution

1

Determine the axis of symmetry of the parabola.

The axis of symmetry is

x = 3

.

Because the parabola passes through the points

(-3, 22)

and

(9, 22)

, which have the same

y

-coordinate, the axis of symmetry must lie halfway between their

x

-coordinates:

x = \frac{-3 + 9}{2} = 3

.

2

Write the quadratic function in vertex form.

f(x) = a(x - 3)^2 + 4

Since the function has a minimum value of

4

, the vertex of the upward-opening parabola is at

(3, 4)

.

3

Solve for the leading coefficient

a

.

a = 0.5

Substitute the point

(9, 22)

into the vertex form equation:

22 = a(9 - 3)^2 + 4

, which simplifies to

18 = 36a

, so

a = 0.5

.

4

Evaluate

f(1)

.

f(1) = 6

Substitute

x = 1

into the completed function

f(x) = 0.5(x - 3)^2 + 4

to get

f(1) = 0.5(1 - 3)^2 + 4 = 0.5(4) + 4 = 6

.

Key Concept

Using symmetry, vertex form, and given points to determine a quadratic function's equation and evaluate it.

Question 58Question →

The height of a diver, in meters, above the pool surface $t$ seconds after leaving the diving board is modeled by the function $f(t) = -4.9(t - 1)^2 + 10$ . If the diving board is moved so that the diver's trajectory is shifted $0.5$ seconds later in time and the maximum height is increased by $2$ meters, which of the following functions $g$ models the diver's new trajectory?

g(t) = -4.9(t - 0.5)^2 + 12

g(t) = -4.9(t - 1.5)^2 + 8

g(t) = -4.9(t - 1.5)^2 + 12

g(t) = -4.9(t + 0.5)^2 + 12

Show answer & explanation

Answer: $g(t) = -4.9(t - 1.5)^2 + 12$

Answer

The function

g(t) = -4.9(t - 1.5)^2 + 12

models the diver's new trajectory.

The correct answer represents the translated quadratic function. Since the original function

f(t) = -4.9(t - 1)^2 + 10

has its vertex at

(1, 10)

, shifting the trajectory

0.5

seconds later in time moves the vertex horizontally to the right to

t = 1.5

seconds, replacing

(t - 1)

with

(t - 1.5)

. Increasing the maximum height by

2

meters shifts the vertex vertically upward to

12

meters, replacing the constant term

10

with

12

. This results in the function

g(t) = -4.9(t - 1.5)^2 + 12

.

Step-by-Step Solution

1

Identify the vertex and its meaning in the original function

f(t) = -4.9(t - 1)^2 + 10

.

The vertex is

(1, 10)

, indicating that the maximum height of

10

meters occurs at

t = 1

second.

The vertex form of a quadratic function is

y = a(t - h)^2 + k

, where

(h, k)

is the vertex representing the extreme value.

2

Apply the horizontal translation of

0.5

seconds later in time.

The new

t

-coordinate of the vertex is

1 + 0.5 = 1.5

seconds, which changes the term

(t - 1)

to

(t - 1.5)

.

A horizontal shift to the right by

c

units is represented by replacing the variable

t

with

t - c

.

3

Apply the vertical translation of

2

meters upward.

The new

y

-coordinate of the vertex is

10 + 2 = 12

meters, which changes the constant term of the function to

12

.

A vertical shift upward by

d

units is represented by adding

d

to the function value.

4

Combine the horizontal and vertical translations to write the new equation.

g(t) = -4.9(t - 1.5)^2 + 12

.

Substituting the shifted vertex coordinates

(1.5, 12)

into the vertex form while keeping the same vertical stretch/direction factor

a = -4.9

yields the final equation.

Key Concept

Vertex form and transformations of quadratic functions

Question 59Question →

The graph of the quadratic function $f(x) = x^2 - 4x + 7$ is translated $3$ units to the right and $2$ units down in the $xy$ -plane to form the graph of the function $g(x) = x^2 + px + q$ , where $p$ and $q$ are constants. What is the value of $q$ ?

Show answer & explanation

Answer: 26

Answer

The value of

q

is

26

.

To find the constant term

q

of the translated quadratic function

g(x) = x^2 + px + q

, we first determine the vertex of the original function

f(x) = x^2 - 4x + 7

. Completing the square gives

f(x) = (x - 2)^2 + 3

, which identifies the vertex of the parabola as

(2, 3)

. Translating the graph

3

units to the right and

2

units down shifts the vertex to

(2 + 3, 3 - 2) = (5, 1)

. Because the translation does not affect the shape of the parabola, the leading coefficient remains

1

. The vertex form of the new function is

g(x) = (x - 5)^2 + 1

. Expanding this expression yields

g(x) = x^2 - 10x + 25 + 1 = x^2 - 10x + 26

. Comparing this to

g(x) = x^2 + px + q

, we find that

q = 26

.

Step-by-Step Solution

1

Convert the original function

f(x) = x^2 - 4x + 7

into vertex form,

f(x) = a(x - h)^2 + k

, to identify its vertex

(h, k)

.

f(x) = (x - 2)^2 + 3

, which represents a parabola with vertex

(2, 3)

.

Finding the vertex of the original function allows us to apply the translation directly to the vertex coordinates.

2

Apply the translation of

3

units to the right and

2

units down to the coordinates of the vertex

(2, 3)

.

The new vertex is

(2 + 3, 3 - 2) = (5, 1)

.

Translating a graph shifts its vertex by the corresponding horizontal and vertical amounts.

3

Write the equation of the translated function

g(x)

in vertex form using the new vertex

(5, 1)

and the original leading coefficient

a = 1

.

g(x) = (x - 5)^2 + 1

A translation does not change the shape or vertical stretch of the parabola, so the coefficient of

x^2

remains

1

.

4

Expand the vertex form of

g(x)

into standard form,

g(x) = x^2 + px + q

, to determine the constant term

q

.

g(x) = x^2 - 10x + 26

, which means

q = 26

.

Expanding the equation allows us to compare it directly with the standard form of

g(x)

and identify the value of the constant term.

Key Concept

Vertex form and translations of quadratic functions

Alternative Method

Alternatively, the translation can be applied directly to the variable

x

in the function equation. Translating a function

f(x)

by

3

units to the right and

2

units down yields

g(x) = f(x - 3) - 2

. Substituting

x - 3

into the original function gives:

g(x) = (x - 3)^2 - 4(x - 3) + 7 - 2

. Simplifying this expression:

g(x) = (x^2 - 6x + 9) - (4x - 12) + 5 = x^2 - 10x + 26

. This directly shows that the constant term

q

is

26

.

Estimated Time:1m 30s

Question 60Question →

A company's daily profit, $P(x)$ , in dollars, from selling $x$ units of a product is modeled by the function $P(x) = -2x^2 + kx - 800$ , where $k$ is a constant. If the company achieves its maximum daily profit of $\$1000$ when it sells $30$ units, what is the value of $k$ ?

120

60

-120

180

Show answer & explanation

Answer: 120

Answer

120

The maximum value of a quadratic function occurs at its vertex. Given that the vertex is at

(30, 1000)

and the coefficient of the squared term is

-2

, the profit function can be written in vertex form as

P(x) = -2(x - 30)^2 + 1000

. Expanding this expression yields

P(x) = -2(x^2 - 60x + 900) + 1000 = -2x^2 + 120x - 800

. Comparing this to the given equation

P(x) = -2x^2 + kx - 800

, the coefficient of

x

must be 120. Alternatively, using the vertex formula

h = -b/(2a)

with

h = 30

and

a = -2

gives

30 = -k/(2 \times -2)

, which simplifies to

30 = k/4

and results in

k = 120

.

Step-by-Step Solution

1

Identify the vertex coordinates from the given context.

The vertex of the profit parabola is at

(h, d) = (30, 1000)

.

The maximum profit of

\$1000

occurs when

30

units are sold, representing the peak of the downward-opening parabola.

2

Substitute the vertex and the leading coefficient

a = -2

into the vertex form of a quadratic function,

P(x) = a(x - h)^2 + d

.

The equation becomes

P(x) = -2(x - 30)^2 + 1000

.

The vertex form allows direct substitution of the vertex coordinates to build the function's equation.

3

Expand the vertex form equation into standard form.

P(x) = -2(x^2 - 60x + 900) + 1000 = -2x^2 + 120x - 800

.

Expanding the equation allows direct comparison of terms with the given standard form

P(x) = -2x^2 + kx - 800

.

4

Compare the coefficient of the

x

term in the expanded equation to the coefficient of the

x

term in the given equation.

k = 120

.

Corresponding coefficients of identical functions must be equal, allowing us to determine the value of the constant

k

.

Key Concept

Quadratic Functions and Graphs

Quadratic Functions and Graphs

Answer

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Common Mistakes

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