Quadratic Functions and Graphs Practice Questions — SAT — Page 2

Question 21Question →

A quadratic function $f$ is defined by $f(x) = a(x - h)^2 + k$ , where $a$ , $h$ , and $k$ are constants. The graph of $y = f(x)$ in the $xy$ -plane passes through the points $(0, 5)$ and $(4, 5)$ . If the minimum value of $f(x)$ for $0 \le x \le 3$ is $1$ , what is the value of $f(6)$ ?

Show answer & explanation

Answer: 17

Answer

The value of

f(6)

is

17

.

The correct answer is

17

. Since the quadratic function passes through

(0, 5)

and

(4, 5)

, its axis of symmetry is the line

x = 2

, which means the vertex

x

-coordinate is

h = 2

. For the interval

0 \le x \le 3

, this vertex is within the bounds. An upward-opening parabola has its minimum value at its vertex, so the minimum value of

1

must be the

y

-coordinate of the vertex, giving

k = 1

. The function can then be written as

f(x) = a(x - 2)^2 + 1

. Substituting

(0, 5)

into this equation gives

5 = a(0 - 2)^2 + 1

, which simplifies to

4a = 4

, meaning

a = 1

. The fully determined function is

f(x) = (x - 2)^2 + 1

. Evaluating this at

x = 6

yields

f(6) = (6 - 2)^2 + 1 = 17

.

Step-by-Step Solution

1

Determine the axis of symmetry of the parabola.

h = 2

Since the points

(0, 5)

and

(4, 5)

have the same

y

-value, they are symmetric across the vertical line of symmetry of the parabola. The

x

-coordinate of the vertex

h

is the average of the

x

-coordinates of these two points:

h = \frac{0 + 4}{2} = 2

.

2

Determine the

y

-coordinate of the vertex

k

using the minimum value over the interval

[0, 3]

.

k = 1

The vertex

x

-coordinate

x = 2

lies within the interval

0 \le x \le 3

. For a quadratic function that opens upward, the minimum value occurs at the vertex. Thus, the minimum value on this interval is the vertex

y

-coordinate, so

k = 1

.

3

Substitute the point

(0, 5)

into the vertex form of the equation to solve for

a

.

a = 1

Using the vertex form

f(x) = a(x - 2)^2 + 1

, substitute

x = 0

and

f(x) = 5

:

5 = a(0 - 2)^2 + 1 \implies 5 = 4a + 1 \implies 4a = 4 \implies a = 1

.

4

Evaluate

f(6)

using the complete quadratic function formula.

f(6) = 17

Substitute

x = 6

into the equation

f(x) = (x - 2)^2 + 1

:

f(6) = (6 - 2)^2 + 1 = 4^2 + 1 = 16 + 1 = 17

.

Key Concept

Using symmetry properties and interval extrema to determine the equation of a quadratic function in vertex form.

Question 22Question →

The graph of the quadratic function $f$ in the $xy$ -plane is a parabola with vertex $(3, 12)$ . If the graph passes through the point $(5, 8)$ , what is the $y$ -value of the point on the graph where $x = 1$ ?

Show answer & explanation

Answer: 8

Answer

The correct answer is

8

.

The vertex of the parabola is given as

(3, 12)

, which means the axis of symmetry is the line

x = 3

. Since a parabola is symmetric with respect to its axis of symmetry, any two points on the parabola that are equidistant from this line must share the same

y

-coordinate. The given point has an

x

-coordinate of

5

, which is

5 - 3 = 2

units to the right of the axis of symmetry. The target point has an

x

-coordinate of

1

, which is

3 - 1 = 2

units to the left of the axis of symmetry. Because both points are exactly

2

units away from the axis of symmetry, their

y

-coordinates are equal. Therefore, the

y

-value of the point where

x = 1

is

8

.

Step-by-Step Solution

1

Determine the axis of symmetry of the parabola.

The axis of symmetry is the vertical line

x = 3

.

The vertex of a parabola

(h, k)

always lies on its axis of symmetry, which is

x = h

.

2

Find the horizontal distance from the axis of symmetry to the given point.

The distance from

x = 3

to

x = 5

is

|5 - 3| = 2

units.

This measures how far the point is horizontally from the line of symmetry.

3

Find the horizontal distance from the axis of symmetry to the target point.

The distance from

x = 3

to

x = 1

is

|1 - 3| = 2

units.

This determines if the target point is symmetric to the given point.

4

Equate the y-values using symmetry.

The

y

-value at

x = 1

is

8

.

Because both

x = 5

and

x = 1

are

2

units away from the axis of symmetry, their corresponding

y

-values must be identical.

Key Concept

Symmetry of Quadratic Graphs

Question 23Question →

The function $f$ is defined by $f(x) = (x - 5)^2 + 3$ . If the graph of $y = f(x)$ is translated 4 units down in the $xy$ -plane to create the graph of the function $g$ , what is the vertex of the graph of $g$ ?

(1, 3)

(-5, -1)

(5, -1)

(5, 7)

Show answer & explanation

Answer: $(5, -1)$

Answer

The vertex of the graph of

g

is

(5, -1)

.

The vertex of a quadratic function of the form

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

is

(h, k)

. For the function

f(x) = (x - 5)^2 + 3

, the vertex is

(5, 3)

. Translating the graph of a function vertically downward by 4 units decreases the

y

-coordinate of every point on the graph by 4, while the

x

-coordinate remains unchanged. Applying this transformation to the vertex coordinates yields

(5, 3 - 4)

, which simplifies to

(5, -1)

.

Step-by-Step Solution

1

Identify the vertex of the original function

f(x) = (x - 5)^2 + 3

.

The vertex of the parabola

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

is

(h, k)

. For

f(x) = (x - 5)^2 + 3

, the vertex is

(5, 3)

.

Finding the original vertex is the starting point for applying the translation.

2

Apply the vertical translation of 4 units downward to the vertex coordinates.

A downward translation of 4 units subtracts 4 from the

y

-coordinate of the vertex:

(5, 3 - 4) = (5, -1)

.

Translating a graph vertically downward decreases the

y

-value of all its points by the translation distance.

Key Concept

Identifying the vertex of a quadratic function in vertex form and applying vertical translations.

Estimated Time:45s

Question 24Question →

In the $xy$ -plane, the graph of the quadratic function $f(x) = ax^2 + bx + c$ , where $a$ , $b$ , and $c$ are constants, has vertex $(3, 12)$ and passes through the point $(1, 0)$ . The function $g$ is defined by $g(x) = f(x + d) - 4$ , where $d$ is a constant. If the $y$ -intercept of the graph of $g$ is $(0, 5)$ and the vertex of the graph of $g$ lies in the second quadrant, what is the value of $d$ ?

Show answer & explanation

Answer: 4

Answer

The value of

d

is 4.

First, the equation of the function

f(x)

is determined in vertex form. Since the vertex of

f

is

(3, 12)

, we write

f(x) = a(x - 3)^2 + 12

. Substituting the point

(1, 0)

into this equation gives

0 = a(1 - 3)^2 + 12

, which yields

a = -3

. Thus,

f(x) = -3(x - 3)^2 + 12

. The transformation

g(x) = f(x + d) - 4

shifts the graph of

f

left by

d

units and down by 4 units, so the vertex of the graph of

g

is

(3 - d, 8)

. For this vertex to lie in the second quadrant, the

x

-coordinate must be negative, meaning

3 - d < 0

, or

d > 3

. The

y

-intercept of

g

is

(0, 5)

, so

g(0) = 5

. Since

g(0) = f(d) - 4

, we have

f(d) = 9

. Substituting

d

into

f(x)

gives

-3(d - 3)^2 + 12 = 9

, which simplifies to

(d - 3)^2 = 1

. Solving for

d

gives

d = 4

or

d = 2

. Since

d > 3

, the value of

d

must be 4.

Step-by-Step Solution

1

Write the function

f(x)

in vertex form and substitute the point

(1, 0)

to solve for

a

.

f(x) = -3(x - 3)^2 + 12

The vertex

(h, k)

is given as

(3, 12)

, and the point

(1, 0)

lies on the graph.

2

Determine the vertex of

g(x)

based on the horizontal and vertical translations of

f(x)

.

The vertex of

g

is

(3 - d, 8)

.

The transformation

g(x) = f(x + d) - 4

shifts the vertex of

f(x)

left by

d

units and down by 4 units.

3

Establish the constraint on

d

using the quadrant of the vertex of

g

.

d > 3

For the vertex

(3 - d, 8)

to lie in the second quadrant, the

x

-coordinate must be negative.

4

Set up an equation for

d

using the

y

-intercept of

g(x)

.

f(d) = 9

The

y

-intercept is

(0, 5)

, so

g(0) = 5

. Substituting this into

g(x) = f(x + d) - 4

gives

5 = f(d) - 4

.

5

Solve

f(d) = 9

for

d

.

d = 2

or

d = 4

Substituting

d

into

f(x)

gives

-3(d - 3)^2 + 12 = 9

, which simplifies to

(d - 3)^2 = 1

.

6

Select the correct value of

d

that satisfies the quadrant constraint.

d = 4

Since

d > 3

, the value

d = 2

is discarded, leaving

d = 4

as the only valid solution.

Key Concept

Vertex form and transformations of quadratic functions

Question 25Question →

A parabola passes through the point $(0, 12)$ on the $y$ -axis and intersects the $x$ -axis at two distinct points, $P$ and $Q$ . The line connecting $P$ to the $y$ -intercept has a slope of $2$ , while the line connecting $Q$ to the $y$ -intercept has a slope of $-6$ . What is the maximum $y$ -value achieved by this parabola?

Show answer & explanation

Answer: 16

Answer

The maximum

y

-value achieved by the parabola is 16.

The slopes of the lines connecting the

x

-intercepts

P(x_1, 0)

and

Q(x_2, 0)

to the

y

-intercept

C(0, 12)

determine their coordinates. The slope of the line through

P

and

C

is

\frac{12 - 0}{0 - x_1} = 2

, which gives

x_1 = -6

. The slope of the line through

Q

and

C

is

\frac{12 - 0}{0 - x_2} = -6

, which gives

x_2 = 2

. The factored form of the parabola is

y = a(x + 6)(x - 2)

. Using the

y

-intercept

C(0, 12)

, we find

12 = a(6)(-2)

, meaning

a = -1

. The

x

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

\frac{-6 + 2}{2} = -2

. Substituting

x = -2

into the equation gives

y = -(-2 + 6)(-2 - 2) = 16

. Since the leading coefficient is negative, this represents the maximum

y

-value.

Step-by-Step Solution

1

Find the coordinates of

P(x_1, 0)

and

Q(x_2, 0)

using the slopes of the lines from these points to the

y

-intercept

C(0, 12)

.

P = (-6, 0)

and

Q = (2, 0)

The slope formula between

(x_1, 0)

and

(0, 12)

is

\frac{12 - 0}{0 - x_1} = 2

, which gives

x_1 = -6

. Similarly, for the second line,

\frac{12 - 0}{0 - x_2} = -6

, which gives

x_2 = 2

.

2

Write the quadratic equation in factored form using the

x

-intercepts.

y = a(x + 6)(x - 2)

Since the parabola intersects the

x

-axis at

(-6, 0)

and

(2, 0)

, these points are the roots of the quadratic function.

3

Solve for the leading coefficient

a

by substituting the

y

-intercept

(0, 12)

into the equation.

a = -1

Substituting

x = 0

and

y = 12

gives

12 = a(0 + 6)(0 - 2)

, which simplifies to

12 = -12a

, so

a = -1

.

4

Determine the

x

-coordinate of the vertex.

x = -2

The axis of symmetry of a parabola lies exactly midway between its

x

-intercepts:

x = \frac{-6 + 2}{2} = -2

.

5

Calculate the maximum

y

-value by evaluating the function at the

x

-coordinate of the vertex.

y = 16

Substituting

x = -2

into

y = -(x + 6)(x - 2)

gives

y = -(-2 + 6)(-2 - 2) = -(4)(-4) = 16

.

Key Concept

Using

x

-intercepts and coordinate geometry to find the vertex of a parabola.

Question 26Question →

In the $xy$ -plane, the graph of the quadratic function $f(x) = -(x - d)^2 + d^2$ , where $d$ is a positive constant, has vertex $V$ and intersects the $x$ -axis at points $P$ and $Q$ . If the area of triangle $PVQ$ is $64$ , what is the value of $d$ ?

2

4

8

16

Show answer & explanation

Answer: 4

Answer

The value of

d

is

4

.

The function

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

is in vertex form,

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

, so its vertex is

V(d, d^2)

. Since

d > 0

, the vertex is in the first quadrant, and the height of the triangle is

d^2

. Setting

f(x) = 0

gives the

x

-intercepts

P(0,0)

and

Q(2d,0)

, meaning the base of the triangle has a length of

2d

. Using the area of a triangle formula, the area is

\frac{1}{2} \times 2d \times d^2 = d^3

. Since the area is given as

64

, we set

d^3 = 64

, which yields

d = 4

.

Step-by-Step Solution

1

Identify the vertex

V

of the quadratic function

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

.

The vertex is

V(d, d^2)

.

The function is written in vertex form,

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

, where the vertex is

(h, k)

.

2

Find the

x

-intercepts

P

and

Q

of the function by setting

f(x) = 0

.

x = 0

and

x = 2d

.

Setting

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

gives

(x - d)^2 = d^2

, which simplifies to

x - d = \pm d

.

3

Calculate the area of triangle

PVQ

in terms of

d

.

The area is

d^3

.

The base of the triangle along the

x

-axis is the distance between the intercepts,

2d - 0 = 2d

. The height is the

y

-coordinate of the vertex,

d^2

. The area is

\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} (2d)(d^2) = d^3

.

4

Solve for

d

using the given area of

64

.

d = 4

.

Setting the area expression

d^3

equal to

64

and taking the cube root of both sides gives

d = 4

.

Key Concept

Finding the vertex and intercepts of a quadratic function in vertex form and applying geometric formulas to analyze the graph.

Question 27Question →

A quadratic function $f$ is defined by $f(x) = a(x - h)^2 + k$ , where $a$ , $h$ , and $k$ are constants, and its graph in the $xy$ -plane has vertex $(h, k)$ in the first quadrant. The $y$ -intercept of the graph of $f$ is $(0, 4)$ . The function $g$ is defined by $g(x) = f(x - 2) + 12$ . If the $y$ -intercept of the graph of $g$ is $(0, 8)$ , and the vertex of the graph of $f$ lies on the line $y = 5x$ , what is the value of $k$ ?

4

5

\frac{20}{3}

20

Show answer & explanation

Answer: 5

Answer

The value of

k

is 5.

By using the given vertex relation

k = 5h

, we can substitute this into the

y

-intercept equations for

f(x)

and

g(x)

. The

y

-intercept of

f

is at

(0,4)

, which translates to the equation

ah^2 + k = 4

. The

y

-intercept of

g

is at

(0,8)

, which leads to

a(h+2)^2 + k = -4

. Eliminating the parameter

a

by equating the two representations of

a

in terms of

h

yields the quadratic equation

3h^2 + h - 4 = 0

. Factoring this equation gives

(3h+4)(h-1)=0

. Since the vertex is in the first quadrant, both coordinates must be positive, which restricts

h

to the positive value

h = 1

. Substituting

h = 1

back into the vertex relation

k = 5h

yields

k = 5

.

Step-by-Step Solution

1

Use the vertex line relation to express

k

in terms of

h

.

k = 5h

Since the vertex

(h, k)

lies on the line

y = 5x

, substituting the coordinates into the line's equation gives

k = 5h

.

2

Set up an equation using the

y

-intercept of

f(x)

.

ah^2 + 5h = 4 \implies a = \frac{4 - 5h}{h^2}

The

y

-intercept of

f(x)

is

(0, 4)

, so

f(0) = a(0-h)^2 + k = 4

. Substituting

k = 5h

gives

ah^2 + 5h = 4

.

3

Set up an equation using the

y

-intercept of

g(x)

.

a(h + 2)^2 + 5h = -4 \implies a = \frac{-5h - 4}{(h+2)^2}

The function

g(x) = f(x - 2) + 12 = a(x - 2 - h)^2 + k + 12

. The

y

-intercept is

(0, 8)

, so

g(0) = a(-2 - h)^2 + k + 12 = 8

. Simplifying and substituting

k = 5h

gives

a(h+2)^2 + 5h = -4

.

4

Equate the expressions for

a

and solve the resulting quadratic equation for

h

.

3h^2 + h - 4 = 0 \implies (3h + 4)(h - 1) = 0

Equating the two expressions for

a

gives

\frac{4 - 5h}{h^2} = \frac{-(5h + 4)}{(h + 2)^2}

. Cross-multiplying yields

(4 - 5h)(h^2 + 4h + 4) = -h^2(5h + 4) \implies 16 - 4h - 16h^2 - 5h^3 = -5h^3 - 4h^2 \implies 12h^2 + 4h - 16 = 0 \implies 3h^2 + h - 4 = 0

.

5

Determine

h

and compute

k

based on the quadrant constraints.

h = 1 \implies k = 5

Since the vertex

(h, k)

is in the first quadrant, we must have

h > 0

. Thus,

h = 1

is the only valid solution. Using

k = 5h

, we find

k = 5(1) = 5

.

Key Concept

Analyzing quadratic vertex form and transformations using algebraic systems.

Estimated Time:3m 0s

Question 28Question →

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 $(h, k)$ that lies on the line $y = 4x + 8$ . If the graph of $y = f(x)$ has $x$ -intercepts at $x = -2$ and $x = 6$ , what is the value of $c$ ?

Show answer & explanation

Answer: 12

Answer

The value of c is 12.

The correct answer is 12. The axis of symmetry of the quadratic function lies midway between its x-intercepts at

x = -2

and

x = 6

, giving an x-coordinate of

h = 2

for the vertex. Since the vertex lies on the line

y = 4x + 8

, its y-coordinate is

k = 4(2) + 8 = 16

. Substituting the vertex

(2, 16)

into the vertex form of a quadratic function gives

f(x) = a(x - 2)^2 + 16

. Using the x-intercept at

(6, 0)

to solve for

a

gives

0 = a(6 - 2)^2 + 16

, which yields

a = -1

. Evaluating the function at

x = 0

to find the constant term

c

gives

c = f(0) = -(0 - 2)^2 + 16 = 12

.

Step-by-Step Solution

1

Determine the x-coordinate of the vertex of the quadratic function.

h = 2

The axis of symmetry of a parabola lies midway between its x-intercepts. The midpoint of the intercepts

x = -2

and

x = 6

is calculated as

\frac{-2 + 6}{2} = 2

.

2

Determine the y-coordinate of the vertex using the given linear equation.

k = 16

Since the vertex

(h, k)

lies on the line

y = 4x + 8

, substituting

x = h = 2

into the line equation gives

k = 4(2) + 8 = 16

.

3

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

a

.

a = -1

The vertex form is

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

. Substituting the vertex

(2, 16)

yields

f(x) = a(x - 2)^2 + 16

. Using the x-intercept

(6, 0)

, substitute

x = 6

and

f(x) = 0

into the equation:

0 = a(6 - 2)^2 + 16

, which simplifies to

16a = -16

, resulting in

a = -1

.

4

Expand the quadratic function or evaluate it at

x = 0

to find the value of

c

.

c = 12

In standard form

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

, the constant term

c

is equal to

f(0)

. Substituting

x = 0

into

f(x) = -(x - 2)^2 + 16

yields

f(0) = -(0 - 2)^2 + 16 = -4 + 16 = 12

.

Key Concept

Using symmetry and the vertex form of a quadratic function to determine its standard form coefficients.

Alternative Method

Alternatively, since the x-intercepts are

-2

and

6

, the quadratic function can be written in factored form as

f(x) = a(x + 2)(x - 6)

. Expanding this gives

f(x) = a(x^2 - 4x - 12) = ax^2 - 4ax - 12a

. Comparing this to the standard form

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

, we see that the x-coordinate of the vertex is

h = -\frac{b}{2a} = -\frac{-4a}{2a} = 2

. Using the line equation

y = 4x + 8

at

x = 2

, we find the vertex y-coordinate is

k = 16

. Since the vertex is

(2, 16)

, we evaluate the factored form at

x = 2

:

16 = a(2 + 2)(2 - 6) \Rightarrow 16 = -16a \Rightarrow a = -1

. The constant term

c

is

-12a

, so

c = -12(-1) = 12

.

Estimated Time:2m 30s

Question 29Question →

A parabola defined by the equation $y = ax^2 + bx + c$ , where $a$ , $b$ , and $c$ are constants, has its vertex in the second quadrant of the coordinate plane. If the parabola passes through the point $(0,0)$ , which of the following inequalities must be true?

ab > 0

ab < 0

a > 0

b > 0

Show answer & explanation

Answer: $ab > 0$

Answer

The inequality

ab > 0

must be true.

Since the parabola passes through the point

(0,0)

and its vertex

(h, k)

lies in the second quadrant where

k > 0

, the vertex must represent a maximum value. Therefore, the parabola opens downward, so the leading coefficient

a

is negative. The x-coordinate of the vertex,

h

, is also negative because it lies in the second quadrant. The vertex x-coordinate is defined by

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

, which means

b = -2ah

. Because both

a

and

h

are negative, their product

ah

is positive, which makes

b

negative when multiplied by

-2

. Finally, since both

a

and

b

are negative, their product

ab

must be positive.

Step-by-Step Solution

1

Determine the signs of the vertex coordinates from its quadrant.

The vertex

(h, k)

lies in the second quadrant, which means

h < 0

and

k > 0

.

Points in the second quadrant have negative x-coordinates and positive y-coordinates.

2

Determine the value of the constant

c

.

c = 0

Since the parabola passes through

(0,0)

, substituting

x = 0

must result in

y = 0

.

3

Determine the sign of the leading coefficient

a

.

a < 0

The vertex is at

(h, k)

with

k > 0

, and the graph passes through

(0, 0)

where the y-value is

0

. Since the maximum value must be at least the value at any other point, the vertex

(h,k)

is a maximum. A parabola with a maximum opens downward, so

a < 0

.

4

Determine the sign of the coefficient

b

using the vertex formula.

b < 0

The x-coordinate of the vertex is

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

, which can be rewritten as

b = -2ah

. Since

a < 0

and

h < 0

, their product

ah

is positive. Multiplying this positive product by

-2

yields a negative value for

b

.

5

Find the sign of the product

ab

.

ab > 0

Since both

a

and

b

are negative, their product

ab

must be positive.

Key Concept

Analyzing quadratic coefficients and vertex properties in the coordinate plane.

Question 30Question →

The function $f$ is defined by $f(x) = x^2 - 6x + c$ , where $c$ is a constant. In the $xy$ -plane, the graph of $f$ has vertex $A$ . The function $g$ is defined by $g(x) = -f(x - 6)$ , and its graph has vertex $B$ . If the distance between points $A$ and $B$ is $10$ , and $c > 10$ , what is the value of $c$ ?

Show answer & explanation

Answer: 13

Answer

The value of

c

is

13

.

Completing the square for

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

gives

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

, which shows that vertex

A

is located at

(3, c - 9)

. The transformation

g(x) = -f(x - 6)

translates the graph

6

units to the right and reflects it vertically, giving vertex

B

the coordinates

(3 + 6, -(c - 9)) = (9, 9 - c)

. Using the distance formula, the distance between

A

and

B

is

\sqrt{(9 - 3)^2 + ((9 - c) - (c - 9))^2} = \sqrt{36 + (18 - 2c)^2}

. Setting this distance equal to

10

and squaring both sides gives

36 + (18 - 2c)^2 = 100

, which simplifies to

(18 - 2c)^2 = 64

. Taking the square root of both sides gives

18 - 2c = 8

or

18 - 2c = -8

, yielding solutions of

c = 5

or

c = 13

. Since the question specifies that

c > 10

, the correct value must be

13

.

Step-by-Step Solution

1

Rewrite the function

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

in vertex form by completing the square.

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

, which gives the coordinates of vertex

A

as

(3, c - 9)

.

Completing the square reveals the vertex

(h, k)

of a quadratic function in the form

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

.

2

Determine the vertex

B

of the graph of

g(x) = -f(x - 6)

by applying transformations to vertex

A(3, c - 9)

.

The horizontal shift of

f(x - 6)

moves the vertex to

(3 + 6, c - 9) = (9, c - 9)

. The reflection of

-f(x - 6)

negates the

y

-coordinate of the vertex, resulting in

B(9, 9 - c)

.

The transformation

f(x - h)

shifts a graph right by

h

units, and the transformation

-f(x)

reflects it across the

x

-axis.

3

Use the distance formula to set up an equation for the distance between

A(3, c - 9)

and

B(9, 9 - c)

.

\sqrt{(9 - 3)^2 + ((9 - c) - (c - 9))^2} = 10 \implies \sqrt{6^2 + (18 - 2c)^2} = 10

.

The distance

d

between points

(x_1, y_1)

and

(x_2, y_2)

is given by

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

.

4

Solve the radical equation

\sqrt{36 + (18 - 2c)^2} = 10

for

c

.

36 + (18 - 2c)^2 = 100 \implies (18 - 2c)^2 = 64 \implies 18 - 2c = 8

or

18 - 2c = -8

. This yields

c = 5

or

c = 13

.

Squaring both sides eliminates the square root, allowing us to solve the resulting quadratic equation.

5

Apply the constraint

c > 10

to choose the correct value for

c

.

c = 13

.

The question specifies that

c

must be greater than

10

, which excludes

c = 5

.

Key Concept

Using vertex form of quadratic equations to determine vertex coordinates and applying transformations (horizontal shifts and vertical reflections) to find key graphical points.

Question 31Question →

In the $xy$ -plane, the graph of the quadratic function $f(x) = -2x^2 + 12x - 10$ is translated $4$ units to the left and $k$ units up, where $k$ is a constant, to produce the graph of a new quadratic function $g$ . If the graph of $g$ passes through the origin $(0,0)$ , what is the value of $k$ ?

90

10

-6

3

Show answer & explanation

Answer: -6

Answer

The correct answer is

-6

.

The correct answer is

-6

. By completing the square on the original quadratic function, we rewrite

f(x) = -2(x^2 - 6x) - 10

as

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

. A translation of

4

units to the left is represented by replacing

x

with

x+4

, and a translation of

k

units up is represented by adding

k

, giving

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

. Since the graph of

g

passes through the origin, we substitute

(0,0)

into the equation:

0 = -2(0+1)^2 + 8 + k

, which simplifies to

0 = 6 + k

, yielding

k = -6

.

Step-by-Step Solution

1

Rewrite the function

f(x)

in vertex form by completing the square.

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

Converting the standard form

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

to vertex form

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

helps easily identify the vertex of the parabola. First, factor out

-2

from the variable terms:

f(x) = -2(x^2 - 6x) - 10

. To complete the square inside the parentheses, add and subtract

9

(since

(\frac{-6}{2})^2 = 9

):

f(x) = -2(x^2 - 6x + 9 - 9) - 10

. This simplifies to

f(x) = -2((x-3)^2 - 9) - 10 = -2(x-3)^2 + 18 - 10 = -2(x-3)^2 + 8

.

2

Determine the equation of the translated function

g(x)

.

g(x) = -2(x+1)^2 + 8 + k

Translating a function

4

units to the left is represented by replacing

x

with

x + 4

. Translating a function

k

units up is represented by adding

k

to the entire function. Therefore,

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

.

3

Substitute the point

(0,0)

into

g(x)

to solve for

k

.

k = -6

Since the graph of

g

passes through the origin

(0,0)

, we have

g(0) = 0

. Substituting

x=0

yields

0 = -2(0+1)^2 + 8 + k \Rightarrow 0 = -2(1) + 8 + k \Rightarrow 0 = 6 + k \Rightarrow k = -6

.

Key Concept

Quadratic transformations and translations in the coordinate plane using vertex form.

Question 32Question →

A parabola in the $xy$ -plane has vertex $(2, 11)$ and passes through the point $(5, -7)$ . The equation of the parabola is $y = ax^2 + bx + c$ , where $a$ , $b$ , and $c$ are constants. What is the value of $a + b + c$ ?

Show answer & explanation

Answer: 9

Answer

The value of

a + b + c

is

9

.

The correct value of

a + b + c

is

9

. This is found by writing the parabola's equation in vertex form as

y = -2(x - 2)^2 + 11

and expanding it to standard form

y = -2x^2 + 8x + 3

, which gives the coefficients

a = -2

,

b = 8

, and

c = 3

. Alternatively, substituting

x = 1

directly into the vertex form gives

f(1) = a(1)^2 + b(1) + c = -2(1 - 2)^2 + 11 = 9

.

Step-by-Step Solution

1

Write the equation of the parabola in vertex form using the given vertex

(2, 11)

.

y = a(x - 2)^2 + 11

The vertex form of a quadratic function is

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

, where

(h, k)

is the vertex of the parabola.

2

Substitute the coordinates of the point

(5, -7)

into the vertex form equation to solve for the constant

a

.

a = -2

Since the point

(5, -7)

lies on the parabola, substituting

x = 5

and

y = -7

allows us to solve the linear equation

-7 = a(5 - 2)^2 + 11

for

a

.

3

Expand the vertex form equation

y = -2(x - 2)^2 + 11

into standard form

y = ax^2 + bx + c

to identify the coefficients

a

,

b

, and

c

.

y = -2x^2 + 8x + 3

, which gives

a = -2

,

b = 8

, and

c = 3

.

Expanding the squared term and distributing the coefficient

-2

converts the equation to standard form, making it easy to read off the coefficients.

4

Calculate the sum of the coefficients

a + b + c

.

a + b + c = 9

Adding the identified coefficients:

-2 + 8 + 3 = 9

.

Key Concept

Writing and converting quadratic functions between vertex form

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

and standard form

y = ax^2 + bx + c

.

Alternative Method

Instead of expanding the vertex form equation to find the individual coefficients

a

,

b

, and

c

, recognize that the expression

a + b + c

is equal to

f(1)

for the quadratic function

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

. After finding

a = -2

using the vertex form

f(x) = a(x - 2)^2 + 11

, directly evaluate

f(1) = -2(1 - 2)^2 + 11 = 9

.

Estimated Time:2m 30s

Question 33Question →

The quadratic function $f$ is defined by $f(x) = ax^2 + bx + c$ , where $a$ , $b$ , and $c$ are constants. In the $xy$ -plane, the graph of $f$ is a parabola with vertex $(3, -5)$ that passes through the point $(0, 4)$ . What is the value of $a + b + c$ ?

11

-9

-1

-7

Show answer & explanation

Answer: -1

Answer

-1

The vertex form of a quadratic function is

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

, where

(h, k)

is the vertex. Substituting the given vertex

(3, -5)

gives

f(x) = a(x - 3)^2 - 5

. Since the graph passes through the point

(0, 4)

, substituting

x = 0

and

f(x) = 4

yields the equation

4 = a(0 - 3)^2 - 5

, which simplifies to

9a = 9

, or

a = 1

. The function is therefore defined by

f(x) = (x - 3)^2 - 5

. The expression

a + b + c

represents the sum of the coefficients of the quadratic function in standard form

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

. Evaluating the function at

x = 1

gives

f(1) = a(1)^2 + b(1) + c = a + b + c

. Substituting

x = 1

into our vertex form equation yields

f(1) = (1 - 3)^2 - 5 = (-2)^2 - 5 = 4 - 5 = -1

. Therefore, the value of

a + b + c

is

-1

.

Step-by-Step Solution

1

Write the quadratic function in vertex form using the given vertex

(3, -5)

.

f(x) = a(x - 3)^2 - 5

, where

a

is a constant.

The vertex form of a quadratic function is

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

, where

(h, k)

is the vertex.

2

Substitute the coordinates of the given point

(0, 4)

into the vertex form equation to solve for

a

.

4 = a(0 - 3)^2 - 5

, which simplifies to

4 = 9a - 5

, and solving for

a

gives

a = 1

.

Since the graph passes through

(0, 4)

, substituting these coordinates into the function's equation must yield a true statement.

3

Find the value of

a + b + c

by evaluating

f(1)

.

f(1) = 1(1 - 3)^2 - 5 = 1(-2)^2 - 5 = 4 - 5 = -1

.

For any quadratic function

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

, evaluating the function at

x = 1

gives

f(1) = a(1)^2 + b(1) + c = a + b + c

.

Key Concept

Quadratic Functions and Graphs

Question 34Question →

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 $(h, k)$ . The function $g$ is defined by $g(x) = f(x - 3) + 4$ . The graph of $g$ passes through the origin $(0,0)$ , and its vertex lies on the line $y = x$ in the first quadrant. What is the value of $f(0)$ ?

-1

-7

-3

-21

Show answer & explanation

Answer: -7

Answer

-7

The correct answer is

-7

. Representing

f(x)

in vertex form as

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

tells us that translating the function 3 units right and 4 units up shifts the vertex from

(h, k)

to

(h+3, k+4)

. Since this vertex lies on the line

y=x

, we have

h+3 = k+4

, or

k = h-1

. Since the vertex of

g

lies in the first quadrant, its coordinates must be positive, meaning

h > -3

and

k > -4

. Using the fact that the graph of

g

passes through the origin, we have

g(0) = 0 \implies f(-3) + 4 = 0 \implies f(-3) = -4

. Substituting

x = -3

into the vertex form of

f(x)

gives

-(-3-h)^2 + k = -4 \implies (h+3)^2 - k = 4

. Substituting

k = h-1

yields

(h+3)^2 - (h-1) = 4 \implies h^2 + 5h + 6 = 0

, which factors into

(h+2)(h+3) = 0

. The solution

h = -3

is discarded because it places the vertex of

g

at

(0,0)

, which is not in the first quadrant. Therefore,

h = -2

and

k = -3

. Calculating

f(0)

gives

-h^2 + k = -(-2)^2 + (-3) = -7

.

Step-by-Step Solution

1

Determine the vertex form of

f(x)

and the vertex of

g(x)

.

Since

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

has its vertex at

(h, k)

, its vertex form is

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

. The function

g(x) = f(x - 3) + 4

represents a horizontal shift of

f

by 3 units to the right and a vertical shift by 4 units up. Thus, the vertex of

g

is

(h + 3, k + 4)

.

Understanding translations allows us to write the coordinates of the new vertex in terms of the original vertex variables.

2

Use the line

y = x

and quadrant constraints to find a relation between

h

and

k

.

Since the vertex of

g

lies on the line

y = x

, we set its coordinates equal:

h + 3 = k + 4

, which simplifies to

k = h - 1

. Additionally, because the vertex is in the first quadrant, we must have

h + 3 > 0

and

k + 4 > 0

.

The geometric placement of the vertex on the line

y = x

constrains its coordinate values.

3

Apply the condition that the graph of

g

passes through the origin.

g(0) = 0 \implies f(-3) + 4 = 0 \implies f(-3) = -4

. Substituting

x = -3

into the vertex form of

f(x)

gives

-(-3 - h)^2 + k = -4

, which simplifies to

(h + 3)^2 - k = 4

.

The point

(0,0)

lying on the graph of

g

provides an equation to solve for the vertex parameters.

4

Substitute

k = h - 1

into the quadratic equation and solve for

h

.

(h + 3)^2 - (h - 1) = 4 \implies h^2 + 6h + 9 - h + 1 = 4 \implies h^2 + 5h + 6 = 0

. Factoring gives

(h + 2)(h + 3) = 0

, so

h = -2

or

h = -3

.

Substituting the linear relation into the quadratic equation isolates the variable

h

.

5

Filter the solutions using the first quadrant constraint and calculate

f(0)

.

If

h = -3

, then

k = -4

, giving the vertex of

g

at

(0, 0)

, which is not in the first quadrant. If

h = -2

, then

k = -3

, giving the vertex of

g

at

(1, 1)

, which is in the first quadrant. Thus,

h = -2

and

k = -3

. The value of

f(0)

is

f(0) = -(0 - h)^2 + k = -h^2 + k = -(-2)^2 + (-3) = -4 - 3 = -7

.

The quadrant condition uniquely determines the correct vertex coordinates, allowing us to find the y-intercept of the original function.

Key Concept

Quadratic Functions and Graphs

Question 35Question →

In the $xy$ -plane, the graph of the quadratic function $f(x) = x^2 - 4x - 5$ intersects the $x$ -axis at the points $(p, 0)$ and $(q, 0)$ and has vertex $(h, k)$ . What is the area of the triangle with vertices at $(p, 0)$ , $(q, 0)$ , and $(h, k)$ ?

Show answer & explanation

Answer: 27

Answer

The area of the triangle is 27.

To find the area of the triangle, we first determine the coordinates of its vertices. The base of the triangle lies on the

x

-axis, with endpoints at the

x

-intercepts of the function

f(x) = x^2 - 4x - 5

. Solving

x^2 - 4x - 5 = 0

by factoring gives

(x - 5)(x + 1) = 0

, so the intercepts are at

x = -1

and

x = 5

. The distance between these two points is

5 - (-1) = 6

, which is the base of the triangle. The third vertex is the vertex of the parabola. The

x

-coordinate of the vertex is

h = -\frac{b}{2a} = -\frac{-4}{2(1)} = 2

. Substituting

x = 2

into the function gives the

y

-coordinate:

k = f(2) = 2^2 - 4(2) - 5 = -9

. The height of the triangle is the vertical distance from the

x

-axis to the vertex, which is

|-9| = 9

. The area of the triangle is

\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 9 = 27

.

Step-by-Step Solution

1

Find the

x

-intercepts of the parabola.

The intercepts are

(-1, 0)

and

(5, 0)

.

Setting

f(x) = 0

gives

x^2 - 4x - 5 = 0

. Factoring the quadratic equation yields

(x - 5)(x + 1) = 0

, which gives

x = 5

and

x = -1

.

2

Calculate the base of the triangle.

The base length is

6

.

The base of the triangle is the segment along the

x

-axis between the two intercepts. The distance between

(-1, 0)

and

(5, 0)

is

5 - (-1) = 6

.

3

Find the vertex

(h, k)

of the parabola.

The vertex is at

(2, -9)

.

The

x

-coordinate of the vertex is the midpoint of the intercepts:

h = \frac{-1 + 5}{2} = 2

. The

y

-coordinate is

k = f(2) = 2^2 - 4(2) - 5 = 4 - 8 - 5 = -9

.

4

Calculate the area of the triangle.

The area is

27

.

The height of the triangle is the distance from the

x

-axis to the vertex, which is

|k| = |-9| = 9

. Using the formula for the area of a triangle,

\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 9 = 27

.

Key Concept

Finding the

x

-intercepts and vertex of a quadratic function to solve geometric problems in the coordinate plane.

Question 36Question →

For the quadratic function $f$ , the table below shows three points that lie on its graph in the $xy$ -plane, where $k$ is a constant.

$x$	$f(x)$
$2$	$0$
$6$	$0$
$1$	$-15$

If the vertex of the graph of $y = f(x)$ is $(4, k)$ , what is the value of $k$ ?

Show answer & explanation

Answer: 12

Answer

The value of

k

is

12

.

The table indicates that the function has

x

-intercepts at

x = 2

and

x = 6

because

f(2) = 0

and

f(6) = 0

. Consequently, the quadratic function can be written in factored form as

f(x) = a(x - 2)(x - 6)

for some constant

a

. Using the point

(1, -15)

to find

a

, we substitute

x = 1

and

f(1) = -15

, which yields

-15 = a(1 - 2)(1 - 6)

, or

-15 = 5a

, so

a = -3

. Since the vertex of the graph is

(4, k)

, the value of

k

is the function value at the vertex's

x

-coordinate, which is

f(4)

. Evaluating the function gives

k = -3(4 - 2)(4 - 6) = -3(2)(-2) = 12

.

Step-by-Step Solution

1

Write the quadratic function in factored form using the given

x

-intercepts.

f(x) = a(x - 2)(x - 6)

The table shows that

f(2) = 0

and

f(6) = 0

, meaning the graph has

x

-intercepts at

x = 2

and

x = 6

.

2

Substitute the point

(1, -15)

into the factored equation to find the value of the constant

a

.

a = -3

Substituting

x = 1

and

f(x) = -15

gives

-15 = a(1 - 2)(1 - 6) \implies -15 = 5a \implies a = -3

.

3

Calculate the value of

k

by finding the function value at the vertex

x = 4

.

k = 12

Since the vertex is

(4, k)

, the value of

k

is

f(4)

. Substituting

x = 4

into

f(x) = -3(x - 2)(x - 6)

yields

k = -3(4 - 2)(4 - 6) = -3(2)(-2) = 12

.

Key Concept

Using intercepts and an additional point to determine the equation of a quadratic function, and evaluating it at the vertex.

Question 37Question →

A manufacturer models the daily profit, $P(x)$ , in dollars, from producing and selling $x$ units of a product using the function $P(x) = -2x^2 + kx - 800$ , where $k$ is a constant. If the maximum daily profit is $1,000$ dollars, what is the number of units that must be sold to achieve this maximum profit?

10

30

60

120

Show answer & explanation

Answer: 30

Answer

The manufacturer must sell 30 units to achieve the maximum daily profit.

The correct answer is 30. The maximum value of a downward-opening quadratic function occurs at its vertex

(h, q)

. Given that the maximum value is 1000, we write the function in vertex form:

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

. Expanding this gives

P(x) = -2x^2 + 4hx - 2h^2 + 1000

. Comparing this to the given function

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

, the constant term must satisfy

-2h^2 + 1000 = -800

. Solving for

h

yields

2h^2 = 1800

, so

h^2 = 900

. Taking the positive square root because the number of units must be positive gives

h = 30

.

Step-by-Step Solution

1

Write the quadratic function in vertex form and identify the given parameters.

The vertex form of a quadratic function is

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

, where

(h, q)

is the vertex. Here,

a = -2

and the maximum profit

q = 1000

, so the function is

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

.

Since the coefficient of the

x^2

term is negative (

a = -2

), the parabola opens downward, meaning the vertex represents the maximum value of the function.

2

Expand the vertex form equation to compare its coefficients with the standard form

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

.

P(x) = -2(x^2 - 2hx + h^2) + 1000 = -2x^2 + 4hx - (2h^2 - 1000)

Expanding the vertex form allows us to equate corresponding coefficients and constant terms between the two forms of the quadratic function.

3

Equate the constant terms from both expressions to solve for the vertex

x

-coordinate

h

.

-2h^2 + 1000 = -800 \Rightarrow -2h^2 = -1800 \Rightarrow h^2 = 900 \Rightarrow h = 30

(since the number of units must be positive).

By setting the constant terms equal, we can isolate and solve for

h

, which represents the number of units that maximizes the daily profit.

Key Concept

Quadratic functions in vertex form and their standard form equivalents.

Question 38Question →

In the $xy$ -plane, the vertex of the parabola with equation $y = -2x^2 + bx + c$ , where $b$ and $c$ are constants, lies on the line $y = 3x + 5$ . If the parabola has its maximum value at $x = -2$ , what is the value of $c$ ?

3

-9

-7

-1

Show answer & explanation

Answer: -9

Answer

-9

The maximum value of the quadratic function

y = -2x^2 + bx + c

occurs at its vertex. Since the maximum occurs at

x = -2

, the

x

-coordinate of the vertex is

-2

. The vertex lies on the line

y = 3x + 5

, so substituting

x = -2

into this equation gives the

y

-coordinate of the vertex:

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

. Thus, the vertex of the parabola is

(-2, -1)

. The vertex form of a quadratic function is

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

, where

(h, k)

is the vertex. Substituting

a = -2

,

h = -2

, and

k = -1

gives

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

. Expanding this equation yields

y = -2(x^2 + 4x + 4) - 1 = -2x^2 - 8x - 9

. Comparing this to the standard form

y = -2x^2 + bx + c

shows that

c = -9

.

Step-by-Step Solution

1

Identify the x-coordinate of the vertex of the parabola

The x-coordinate of the vertex is

h = -2

.

For a quadratic function in standard form with a negative leading coefficient, the maximum value occurs at the vertex. Thus, the x-coordinate of the vertex is

-2

.

2

Determine the y-coordinate of the vertex using the equation of the line

The y-coordinate of the vertex is

k = -1

.

Since the vertex

(h, k)

lies on the line

y = 3x + 5

and

h = -2

, substituting

x = -2

yields

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

.

3

Write the quadratic equation in vertex form and expand it to find c

y = -2x^2 - 8x - 9

, which gives

c = -9

.

Using the vertex form

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

with

a = -2

,

h = -2

, and

k = -1

, the equation is

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

. Expanding this yields

y = -2(x^2 + 4x + 4) - 1 = -2x^2 - 8x - 9

. Comparing this with

y = -2x^2 + bx + c

shows that

c = -9

.

Key Concept

Vertex form of a quadratic function and coordinate geometry.

Estimated Time:2m 0s

Question 39Question →

For the quadratic function $f(x) = -x^2 + bx + c$ , where $b$ and $c$ are positive constants, the maximum value of $f(x)$ is $k$ . The distance between the two $x$ -intercepts of the graph of $y = f(x)$ in the $xy$ -plane is equal to $\frac{2}{3}k$ . If the graph of $y = f(x)$ passes through the point $(1, 8)$ , what is the value of $b$ ?

Show answer & explanation

Answer: 4

Answer

The correct answer is 4.

The correct answer is 4. By expressing the maximum value

k

and the distance between the

x

-intercepts

d

in terms of the constants

b

and

c

, we set up the equation

d = \frac{2}{3}k

. This simplifies to

b^2 + 4c = 36

. Using the point

(1, 8)

, we establish

c = 9 - b

. Substituting this into the first equation yields

b^2 - 4b = 0

, giving

b = 4

as the only positive solution.

Step-by-Step Solution

1

Find the maximum value

k

of the function

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

in terms of

b

and

c

.

k = \frac{b^2}{4} + c

The maximum value of a quadratic function with a negative leading coefficient occurs at its vertex, where

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

.

2

Find the distance

d

between the

x

-intercepts of the graph in terms of

b

and

c

.

d = \sqrt{b^2 + 4c}

The

x

-intercepts are the roots of

-x^2 + bx + c = 0

, which are

x = \frac{b \pm \sqrt{b^2 + 4c}}{2}

. The distance between them is the difference of these roots.

3

Use the relation

d = \frac{2}{3}k

to find the value of the expression

b^2 + 4c

.

b^2 + 4c = 36

Substituting the expressions for

d

and

k

gives

\sqrt{b^2 + 4c} = \frac{1}{6}(b^2 + 4c)

. Solving this radical equation yields

b^2 + 4c = 36

.

4

Use the point

(1, 8)

to express

c

in terms of

b

.

c = 9 - b

Since the graph passes through

(1, 8)

, substituting

x = 1

and

y = 8

into

y = -x^2 + bx + c

yields

8 = -1 + b + c

, which simplifies to

c = 9 - b

.

5

Substitute

c = 9 - b

into

b^2 + 4c = 36

and solve for

b

.

b = 4

Substituting yields

b^2 + 4(9 - b) = 36 \implies b^2 - 4b = 0

. Solving for

b

gives

b = 0

or

b = 4

. Since

b

is a positive constant, we have

b = 4

.

Key Concept

Quadratic Functions and Graphs

Question 40Question →

In the $xy$ -plane, the graph of the quadratic function $f(x) = -x^2 + 6x + 7$ has a vertex at $(h, k)$ . If the graph of $f$ is translated $4$ units to the right and $3$ units up to produce the graph of the function $g$ , what is the maximum value of $g$ ?

Show answer & explanation

Answer: 19

Answer

The maximum value of the function

g

is

19

.

The vertex of the original quadratic function

f(x) = -x^2 + 6x + 7

is

(3, 16)

. Since the coefficient of

x^2

is negative, the graph opens downward, making

16

the maximum value of the function. Translating the graph

3

units up shifts all

y

-values up by

3

, which increases the maximum value to

16 + 3 = 19

. The horizontal translation of

4

units to the right shifts the graph horizontally but does not affect the maximum output value.

Step-by-Step Solution

1

Find the vertex of the original quadratic function

f(x) = -x^2 + 6x + 7

.

The vertex of

f(x)

is

(3, 16)

.

Rewrite the function in vertex form,

f(x) = -(x-3)^2 + 16

, to identify the vertex

(h, k)

as

(3, 16)

.

2

Determine the maximum value of the original function

f(x)

.

The maximum value of

f(x)

is

16

.

Since the leading coefficient is negative, the parabola opens downward, and the maximum value occurs at the vertex's

y

-coordinate.

3

Apply the vertical translation to find the maximum value of the new function

g(x)

.

The maximum value of

g

is

19

.

Translating the graph

3

units up increases all

y

-values, including the maximum value, by

3

, so

16 + 3 = 19

. The horizontal translation of

4

units to the right does not affect the maximum value.

Key Concept

Quadratic Functions and Graphs

Quadratic Functions and Graphs

Answer

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