Quadratic Functions and Graphs Practice Questions — SAT

Question 1Question →

A quadratic function $g$ is defined by $g(x) = 3(x - 5)^2 - 4$ . If the graph of $y = g(x)$ in the $xy$ -plane is translated $6$ units up to produce the graph of $y = f(x)$ , what is the $y$ -coordinate of the vertex of the graph of $y = f(x)$ ?

Show answer & explanation

Answer: 2

Answer

The correct answer is 2.

The original function

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

is in vertex form

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

, where the vertex is

(h, k)

. Therefore, the vertex of the graph of

g

is

(5, -4)

. Translating a graph upward by

6

units is represented by adding

6

to the function, so

f(x) = g(x) + 6

. This transformation shifts the vertex from

(5, -4)

to

(5, -4 + 6)

, which is

(5, 2)

. The

y

-coordinate of this new vertex is

2

.

Step-by-Step Solution

1

Identify the vertex of the original quadratic function.

The vertex of the graph of

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

is

(5, -4)

.

A quadratic function written in vertex form

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

has its vertex at

(h, k)

.

2

Determine the vertex of the translated function.

The vertex of the graph of

y = f(x)

is

(5, 2)

.

Translating a graph

6

units up increases the

y

-coordinate of all points, including the vertex, by

6

.

Key Concept

Vertex form of a quadratic function and vertical translation of graphs.

Estimated Time:45s

Question 2Question →

In the $xy$ -plane, the graph of the quadratic function $f(x) = ax^2 + bx + c$ , where $a > 0$ , has vertex $V$ . The graph intersects the $x$ -axis at points $A$ and $B$ . If triangle $VAB$ is an equilateral triangle with an area of $12\sqrt{3}$ , what is the value of $a$ ?

\frac{1}{8}

\frac{1}{2}

1

2

Show answer & explanation

Answer: $\frac{1}{2}$

Answer

The correct answer is the value 1/2.

The value 1/2 is correct because the area of the equilateral triangle determines its side length to be

4\sqrt{3}

and its height to be

6

. Since the parabola opens upwards and its base lies on the

x

-axis, the vertex is

V(h, -6)

. The two

x

-intercepts are symmetric about the axis of symmetry

x = h

, meaning they are located at

h \pm 2\sqrt{3}

. Substituting one of these points into the vertex form

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

yields

0 = 12a - 6

, which gives

a = 1/2

.

Step-by-Step Solution

1

Determine the side length and height of the equilateral triangle from the given area.

The side length

s

is

4\sqrt{3}

, and the height of the triangle is

6

.

The area of an equilateral triangle with side length

s

is given by the formula

\text{Area} = \frac{s^2\sqrt{3}}{4}

. Setting this equal to

12\sqrt{3}

yields

s^2 = 48

, which means

s = \sqrt{48} = 4\sqrt{3}

. The height of the triangle is given by

h_{\text{tri}} = \frac{s\sqrt{3}}{2} = \frac{4\sqrt{3} \cdot \sqrt{3}}{2} = 6

.

2

Relate the height and side length of the triangle to the vertex and

x

-intercepts of the parabola.

The vertex of the parabola is

V(h, -6)

and the

x

-intercepts are at

x = h \pm 2\sqrt{3}

.

Because

a > 0

, the parabola opens upwards, meaning the vertex

V(h, k)

lies below the

x

-axis. Since the base of the triangle lies on the

x

-axis, the vertical distance from the vertex to the

x

-axis is the height of the triangle, so

k = -6

. The axis of symmetry

x = h

bisects the base

AB

, so the

x

-intercepts are located at a distance of half the side length,

\frac{s}{2} = 2\sqrt{3}

, to the left and right of

h

.

3

Write the quadratic function in vertex form and substitute one of the

x

-intercepts to solve for

a

.

a = \frac{1}{2}

The vertex form of the quadratic function is

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

. Substituting the

x

-intercept

(h + 2\sqrt{3}, 0)

into this equation gives

0 = a(h + 2\sqrt{3} - h)^2 - 6

, which simplifies to

0 = a(2\sqrt{3})^2 - 6 \implies 12a = 6 \implies a = \frac{1}{2}

.

Key Concept

Using the symmetry properties and vertex form of a quadratic function to relate the geometric features of its graph to its algebraic coefficients.

Alternative Method

Alternatively, you can use the relationship between the roots and coefficients. For any quadratic equation

ax^2 + bx + c = 0

with real roots, the distance between the roots is

\frac{\sqrt{b^2 - 4ac}}{|a|}

. Since the triangle is equilateral, the height is

\frac{\sqrt{3}}{2}

times this distance. The vertex is at

(-\frac{b}{2a}, -\frac{b^2-4ac}{4a})

. Setting the absolute value of the vertex

y

-coordinate equal to the height allows you to solve for

a

directly when combined with the area formula.

Estimated Time:3m 0s

Question 3Question →

The quadratic function $f$ is defined by $f(x) = -2x^2 + bx + c$ , where $b$ and $c$ are constants. In the $xy$ -plane, the vertex of the graph of $f$ has an $x$ -coordinate of $3$ . If $f(1) = 2$ , what is the $y$ -coordinate of the vertex of the graph of $f$ ?

Show answer & explanation

Answer: 10

Answer

10

The quadratic function in standard form is

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

, which has a leading coefficient of

a = -2

. The vertex form of a quadratic function is

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

, where

(h, k)

is the vertex of the parabola. Given that the

x

-coordinate of the vertex is

3

(so

h = 3

), we can write the function as

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

. Since the graph passes through the point

(1, 2)

, we substitute

x = 1

and

f(1) = 2

into the equation:

2 = -2(1-3)^2 + k

. Simplifying the expression gives

2 = -2(-2)^2 + k

, which becomes

2 = -2(4) + k

, or

2 = -8 + k

. Adding

8

to both sides yields

k = 10

. Thus, the

y

-coordinate of the vertex is

10

.

Step-by-Step Solution

1

Write the quadratic function in vertex form.

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

The standard form equation

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

shows that the leading coefficient

a = -2

. The vertex form is

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

, where

(h, k)

is the vertex. Since the

x

-coordinate of the vertex is

3

, we substitute

a = -2

and

h = 3

into the vertex form.

2

Substitute the point

(1, 2)

into the vertex form equation.

2 = -2(1-3)^2 + k

Since

f(1) = 2

, the graph passes through the point

(1, 2)

. Substituting these values allows us to solve for the unknown vertex

y

-coordinate,

k

.

3

Simplify the equation and solve for

k

.

k = 10

Simplifying the term

-2(1-3)^2

yields

-2(-2)^2 = -2(4) = -8

. This simplifies the equation to

2 = -8 + k

. Adding

8

to both sides gives

k = 10

.

Key Concept

Vertex form of a quadratic function

Question 4Question →

The quadratic function $f$ is defined by $f(x) = (x - 4)^2 + 7$ . What are the coordinates of the vertex of the graph of $y = f(x)$ in the $xy$ -plane?

(-4, 7)

(4, -7)

(4, 7)

(7, 4)

Show answer & explanation

Answer: $(4, 7)$

Answer

The coordinates of the vertex are

(4, 7)

.

The quadratic function is given in vertex form,

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

, where the vertex of the graph of the function in the

xy

-plane is

(h, k)

. Comparing

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

to this form, we get

h = 4

and

k = 7

. Therefore, the coordinates of the vertex are

(4, 7)

.

Step-by-Step Solution

1

Identify the standard vertex form of a quadratic function.

The vertex form of a quadratic function is

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

, where

(h, k)

is the vertex of the parabola.

This standard form allows direct extraction of the vertex coordinates by matching the variables.

2

Compare the given function

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

to the standard vertex form.

Comparing the two equations, we identify that

a = 1

,

h = 4

, and

k = 7

.

Matching coefficients and constants determines the values of the vertex parameters.

3

Write the coordinates of the vertex

(h, k)

using the identified values.

The vertex is at

(4, 7)

.

Substituting

h = 4

and

k = 7

into

(h, k)

yields the final vertex coordinates.

Key Concept

Identifying the vertex of a quadratic function from its vertex form.

Question 5Question →

The quadratic function $f$ is defined by $f(x) = (x - 5)(x - 11)$ . In the $xy$ -plane, the graph of $y = f(x)$ is a parabola with vertex $(h, k)$ . What is the value of $h$ ?

Show answer & explanation

Answer: 8

Answer

The value of

h

is

8

.

The given quadratic function is in factored form,

f(x) = (x - 5)(x - 11)

, which shows that the

x

-intercepts of the graph are at

x = 5

and

x = 11

. Because a parabola is symmetric, the

x

-coordinate of its vertex,

h

, is located exactly halfway between the two

x

-intercepts. Calculating the average of the

x

-intercepts gives

h = \frac{5 + 11}{2} = 8

.

Step-by-Step Solution

1

Find the

x

-intercepts of the parabola.

The

x

-intercepts are

x = 5

and

x = 11

.

The

x

-intercepts of a quadratic function in the form

f(x) = (x - r_1)(x - r_2)

occur at

x = r_1

and

x = r_2

because setting

f(x) = 0

yields these roots.

2

Calculate the

x

-coordinate of the vertex,

h

.

h = 8

Since a parabola is symmetric, its vertex lies on the axis of symmetry, which is the vertical line halfway between the

x

-intercepts. The average of

5

and

11

is

\frac{5 + 11}{2} = 8

.

Key Concept

Finding the vertex of a parabola from its factored form

Question 6Question →

In the $xy$ -plane, the vertex of the parabola $y = x^2 + bx + c$ is $V_1$ , and the vertex of the parabola $y = -x^2 + dx + e$ is $V_2$ , where $b$ , $c$ , $d$ , and $e$ are constants. The parabola $y = x^2 + bx + c$ passes through $V_2$ , and the parabola $y = -x^2 + dx + e$ passes through $V_1$ . If the distance between $V_1$ and $V_2$ is $2\sqrt{5}$ , and the line passing through $V_1$ and $V_2$ has a positive slope, what is the value of $b + d$ ?

2

4

8

16

Show answer & explanation

Answer: 4

Answer

4

The correct answer is 4. By writing the equations of the parabolas in vertex form,

y = (x - h_1)^2 + k_1

and

y = -(x - h_2)^2 + k_2

, and using the fact that each parabola passes through the other's vertex, we establish that the vertical difference between the vertices is equal to the square of their horizontal difference:

k_2 - k_1 = (h_2 - h_1)^2

. Substituting this into the distance formula

(h_2 - h_1)^2 + (k_2 - k_1)^2 = 20

yields a quadratic equation in terms of

(h_2 - h_1)^2

, which solves to

(h_2 - h_1)^2 = 4

. Since the slope of the line segment connecting the vertices is positive, the horizontal difference

h_2 - h_1

must be positive, which gives

h_2 - h_1 = 2

. Finally, expressing the linear coefficients in terms of the vertex coordinates using the vertex formula gives

b = -2h_1

and

d = 2h_2

, so

b + d = 2(h_2 - h_1) = 2(2) = 4

.

Step-by-Step Solution

1

Write each parabola in vertex form using the vertices

V_1(h_1, k_1)

and

V_2(h_2, k_2)

.

f(x) = (x - h_1)^2 + k_1

and

g(x) = -(x - h_2)^2 + k_2

Expressing the functions in vertex form allows us to easily use the condition that each parabola passes through the other's vertex.

2

Substitute the vertex

V_2(h_2, k_2)

into

f(x)

and

V_1(h_1, k_1)

into

g(x)

.

k_2 = (h_2 - h_1)^2 + k_1

and

k_1 = -(h_1 - h_2)^2 + k_2

, both of which simplify to

k_2 - k_1 = (h_2 - h_1)^2

.

Since the points lie on the respective curves, their coordinates must satisfy the equations, establishing a relation between horizontal difference

\Delta h = h_2 - h_1

and vertical difference

\Delta k = k_2 - k_1

.

3

Use the distance formula for

V_1

and

V_2

with the given distance

2\sqrt{5}

and substitute

\Delta k = (\Delta h)^2

.

(\Delta h)^2 + ((\Delta h)^2)^2 = (2\sqrt{5})^2 \implies (\Delta h)^4 + (\Delta h)^2 - 20 = 0

. Solving this quadratic in terms of

(\Delta h)^2

yields

(\Delta h)^2 = 4

.

The distance formula connects the horizontal and vertical differences, which we can solve using our relation from the previous step.

4

Determine the horizontal difference

\Delta h

using the slope condition.

Since the slope

m = \frac{\Delta k}{\Delta h}

is positive and we know

\Delta k = 4 > 0

, we have

\Delta h = h_2 - h_1 = 2

.

The positive slope restriction determines that the horizontal translation is to the right from

V_1

to

V_2

.

5

Relate the standard form coefficients

b

and

d

to the vertex coordinates to find

b + d

.

For

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

, the vertex x-coordinate is

h_1 = -\frac{b}{2} \implies b = -2h_1

. For

g(x) = -x^2 + dx + e

, the vertex x-coordinate is

h_2 = \frac{d}{2} \implies d = 2h_2

. Thus,

b + d = -2h_1 + 2h_2 = 2(h_2 - h_1) = 2(2) = 4

.

The vertex x-coordinate formula

h = -\frac{\text{coefficient of } x}{2 \times \text{coefficient of } x^2}

connects the standard form coefficients back to

h_1

and

h_2

.

Key Concept

Identifying vertex coordinates of quadratic functions and applying translations and the distance formula in the coordinate plane.

Question 7Question →

A parabola in the $xy$ -plane has its vertex at $(3, 8)$ and passes through the point $(5, 16)$ . If the equation of the parabola is written in the form $y = a(x - 3)^2 + 8$ , what is the value of the constant $a$ ?

Show answer & explanation

Answer: 2

Answer

2

Substituting the coordinates

(5, 16)

into

y = a(x - 3)^2 + 8

yields

16 = a(5 - 3)^2 + 8

. Simplifying the term

(5 - 3)^2

gives

4

, leading to the equation

16 = 4a + 8

. Subtracting 8 from both sides gives

8 = 4a

, and dividing by 4 gives the final answer of 2.

Step-by-Step Solution

1

Substitute the point

(5, 16)

into the equation.

16 = a(5 - 3)^2 + 8

Since the parabola passes through the point

(5, 16)

, these coordinates must satisfy the equation.

2

Simplify the squared term.

16 = 4a + 8

Subtract 3 from 5 to get 2, and square 2 to get 4.

3

Solve for the constant

a

.

a = 2

Subtract 8 from both sides to get

8 = 4a

, then divide by 4 to find

a

.

Key Concept

Determining a constant in a quadratic function given its vertex form and a point on the graph.

Question 8Question →

A projectile is launched from a platform. The height $h(t)$ , in meters, of the projectile $t$ seconds after it is launched is modeled by the function $h(t) = -5(t - 2)^2 + 24$ . What is the maximum height, in meters, reached by the projectile?

2

24

-2

4

Show answer & explanation

Answer: 24

Answer

The maximum height reached by the projectile is 24 meters.

The given quadratic function is in vertex form,

h(t) = a(t - d)^2 + c

, where the vertex of the parabola is

(d, c)

. For the function

h(t) = -5(t - 2)^2 + 24

, the vertex is at

(2, 24)

. Because the leading coefficient

-5

is negative, the parabola opens downward, which means the vertical coordinate of the vertex represents the maximum value of the function. Therefore, the maximum height reached by the projectile is 24 meters.

Step-by-Step Solution

1

Identify the form of the quadratic function.

The function

h(t) = -5(t - 2)^2 + 24

is given in vertex form, which is

h(t) = a(t - d)^2 + c

.

Recognizing the vertex form allows direct identification of the vertex of the parabola.

2

Determine the vertex of the parabola.

For

h(t) = -5(t - 2)^2 + 24

, the vertex is

(2, 24)

.

The vertex coordinates

(d, c)

represent the input and output values at the turning point of the parabola.

3

Determine whether the vertex represents a maximum or a minimum.

Since the coefficient

a = -5

is negative, the parabola opens downward, meaning the vertex is the highest point on the graph.

The sign of the leading coefficient determines the concavity of the parabola.

4

Extract the maximum value of the function.

The maximum value is the vertical coordinate of the vertex, which is 24.

The height of the projectile is represented by the output of the function, which is maximized at the vertex.

Key Concept

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

Question 9Question →

The graph of the quadratic function $g$ in the $xy$ -plane is a parabola with vertex $(2, -5)$ . Which of the following equations could define the function $g$ ?

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

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

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

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

Show answer & explanation

Answer: $g(x) = (x - 2)^2 - 5$

Answer

The equation

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

could define the function

g

.

The vertex form of a quadratic function is given by

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

, where the point

(h, k)

is the vertex of the parabola. We are given that the vertex is

(2, -5)

, which means

h = 2

and

k = -5

. Substituting these values into the vertex form equation gives

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

. Letting

a = 1

yields the equation

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

.

Step-by-Step Solution

1

Identify the vertex form of a quadratic function.

The vertex form of a quadratic function is written as

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

, where

(h, k)

is the vertex of the parabola and

a

is a non-zero constant.

Recognizing the vertex form allows us to directly substitute the coordinates of the given vertex.

2

Substitute the given vertex coordinates into the vertex form equation.

Given the vertex is

(2, -5)

, we substitute

h = 2

and

k = -5

into the vertex form:

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

. If we assume

a = 1

, this simplifies to

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

.

This matches one of the given choices to find the equation that could define the function.

Key Concept

Vertex form of a quadratic function

Estimated Time:45s

Question 10Question →

The function $f$ is defined by $f(x) = x^2 - 10x + 29$ . For what value of $x$ does $f(x)$ reach its minimum value?

Show answer & explanation

Answer: 5

Answer

The function reaches its minimum value at

x = 5

.

A quadratic function defined by

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

where

a > 0

reaches its minimum value at its vertex. The

x

-coordinate of the vertex is calculated using the formula

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

. For

f(x) = x^2 - 10x + 29

, the coefficients are

a = 1

and

b = -10

. Substituting these values into the formula gives

x = -\frac{-10}{2(1)} = 5

.

Step-by-Step Solution

1

Identify the coefficients of the quadratic function in standard form

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

.

For

f(x) = x^2 - 10x + 29

, the coefficients are

a = 1

and

b = -10

.

To use the vertex formula

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

.

2

Apply the vertex formula to calculate the

x

-coordinate where the function reaches its minimum.

x = 5

.

Since the coefficient of

x^2

is positive (

a = 1

), the parabola opens upward, and its minimum value occurs at the vertex.

Key Concept

Finding the x-coordinate of the vertex of a quadratic function

Question 11Question →

In the $xy$ -plane, the graph of $y = x^2$ is shifted $4$ units to the right and $9$ units up to produce the graph of $y = h(x)$ . Which of the following equations defines the function $h$ ?

h(x) = (x + 4)^2 + 9

h(x) = (x - 4)^2 - 9

h(x) = (x - 4)^2 + 9

h(x) = (x + 4)^2 - 9

Show answer & explanation

Answer: $h(x) = (x - 4)^2 + 9$

Answer

h(x) = (x - 4)^2 + 9

The parent function

y = x^2

has its vertex at

(0, 0)

. Shifting this graph

4

units to the right moves the x-coordinate of the vertex to

4

, and shifting it

9

units up moves the y-coordinate of the vertex to

9

. The new vertex is at

(4, 9)

. A parabola with a vertex of

(h, k)

and a leading coefficient of

1

is written in vertex form as

y = (x - h)^2 + k

. Substituting

h = 4

and

k = 9

gives the equation

h(x) = (x - 4)^2 + 9

.

Step-by-Step Solution

1

Identify the base function and the rules of graph translation.

The base function is

y = x^2

. Shifting a graph horizontally by

h

units changes the input

x

to

(x - h)

, and shifting vertically by

k

units adds

k

to the function value.

Applying transformations to the parent function changes its position in the coordinate plane while maintaining its shape.

2

Apply the horizontal translation of

4

units to the right.

Replacing

x

with

(x - 4)

in the parent function gives the intermediate equation

y = (x - 4)^2

.

A horizontal shift of

h

units to the right corresponds to replacing the input variable

x

with

(x - h)

.

3

Apply the vertical translation of

9

units up.

Adding

9

to the expression yields the final equation

h(x) = (x - 4)^2 + 9

.

A vertical shift of

k

units upward corresponds to adding

k

to the output of the function.

Key Concept

Quadratic graph transformations and translation rules

Question 12Question →

In the $xy$ -plane, the graph of the quadratic function $f(x) = -x^2 + bx + c$ has a vertex at $(h, k)$ . The graph of $f$ is translated $4$ units to the right and $5$ units up to form the graph of a quadratic function $g$ . If the graph of $g$ passes through the point $(1, 16)$ and has a $y$ -intercept of $13$ , what is the value of $k$ ?

Show answer & explanation

Answer: 12

Answer

The value of

k

is

12

.

By writing the function in vertex form as

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

, we can apply the transformations directly. Translating the graph

4

units to the right and

5

units up gives the function

g(x) = f(x-4) + 5 = -(x-h-4)^2 + k+5

. Using the given points

g(0) = 13

and

g(1) = 16

, we set up a system of equations:

-(h+4)^2 + k+5 = 13

and

-(h+3)^2 + k+5 = 16

. Solving these simultaneously gives

h = -2

and

k = 12

.

Step-by-Step Solution

1

Write the function

f(x)

in vertex form.

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

Since the leading coefficient of

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

is

-1

, the vertex form must have a coefficient of

-1

.

2

Represent the translated function

g(x)

.

g(x) = -(x-h-4)^2 + k+5

Translating a function

f(x)

by

4

units to the right and

5

units up yields

g(x) = f(x-4) + 5

.

3

Use the

y

-intercept of

g(x)

to write the first equation.

k - (h+4)^2 = 8

The

y

-intercept is the value of the function at

x = 0

, so

g(0) = 13

.

4

Use the point

(1, 16)

on

g(x)

to write the second equation.

k - (h+3)^2 = 11

The graph passing through

(1, 16)

means

g(1) = 16

.

5

Solve the system of equations for

h

and

k

.

h = -2

and

k = 12

Equating the expressions for

k

gives

(h+4)^2 + 8 = (h+3)^2 + 11

, which simplifies to

2h = -4

, so

h = -2

. Substituting

h = -2

into

k = (h+4)^2 + 8

yields

k = 12

.

Key Concept

Translating quadratic functions and solving systems of quadratic equations.

Question 13Question →

The quadratic function $f$ is defined by $f(x) = 3(x - 2)^2 + 5$ . In the $xy$ -plane, the graph of $y = f(x)$ intersects the $y$ -axis at the point $(0, c)$ . What is the value of $c$ ?

Show answer & explanation

Answer: 17

Answer

The correct answer is 17.

To find the

y

-intercept of the graph of

y = f(x)

, we substitute

x = 0

into the equation:

f(0) = 3(0 - 2)^2 + 5

. Simplifying the expression inside the parentheses gives

3(-2)^2 + 5

. Squaring

-2

gives

4

, making the expression

3(4) + 5 = 12 + 5 = 17

. Therefore, the graph intersects the

y

-axis at the point

(0, 17)

, which means the value of

c

is 17.

Step-by-Step Solution

1

Set

x = 0

to find the point where the graph intersects the

y

-axis.

f(0) = 3(0 - 2)^2 + 5

The

y

-intercept of any graph in the

xy

-plane occurs when the

x

-coordinate is equal to

0

.

2

Perform the arithmetic evaluation to find the value of

f(0)

.

f(0) = 17

First, evaluate the term inside the parentheses:

0 - 2 = -2

. Squaring

-2

gives

4

. Then, multiply

3

by

4

to get

12

. Finally, add

5

to obtain

17

.

Key Concept

Evaluating a quadratic function at

x = 0

to determine its

y

-intercept.

Question 14Question →

A quadratic function $f$ is defined by $f(x) = 2(x + 3)^2 - 8$ . What are the coordinates of the vertex of the graph of $y = f(x)$ in the $xy$ -plane?

(-3, -8)

(3, -8)

(-3, 8)

(3, 8)

Show answer & explanation

Answer: $(-3, -8)$

Answer

The correct answer is the coordinate pair

(-3, -8)

.

The vertex form of a quadratic function is

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

, where the vertex of the graph of the function in the

xy

-plane is

(h, k)

. For the function

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

, the expression inside the parentheses can be rewritten as

(x - (-3))

, showing that

h = -3

. The constant term outside the parentheses is

-8

, showing that

k = -8

. Therefore, the vertex of the graph of the function is

(-3, -8)

.

Step-by-Step Solution

1

Identify the standard vertex form of a quadratic function.

The vertex form is

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

, where

(h, k)

is the vertex of the parabola.

Comparing the given equation to this template allows us to read the coordinates directly.

2

Rewrite the given equation to align with the negative sign inside the parentheses of the vertex form.

The term

(x + 3)^2

can be rewritten as

(x - (-3))^2

, and the subtraction of

8

can be written as adding

-8

. This yields

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

.

Matching the template exactly prevents sign errors when identifying the values of

h

and

k

.

3

Extract the values of

h

and

k

.

h = -3

and

k = -8

, which means the vertex is located at

(-3, -8)

.

The vertex

(h, k)

coordinates represent the minimum point of this upward-opening parabola.

Key Concept

Identifying the vertex coordinates directly from the vertex form of a quadratic function.

Alternative Method

An alternative method is to expand the function into the standard form

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

. Expanding

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

gives

f(x) = 2(x^2 + 6x + 9) - 8

, which simplifies to

f(x) = 2x^2 + 12x + 10

. The

x

-coordinate of the vertex can be found using the formula

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

. Substituting

x = -3

back into the function gives

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

. This confirms the vertex is at

(-3, -8)

.

Estimated Time:45s

Question 15Question →

In the $xy$ -plane, the graph of the quadratic function $f(x) = a(x-3)(x-11)$ , where $a$ is a constant, has a vertex at $(h, k)$ . The graph of the quadratic function $g(x) = -(x-h)^2 + q$ has its vertex at $(h, k+8)$ . If the graphs of $f$ and $g$ intersect at a point with an $x$ -coordinate of $5$ , what is the value of $a$ ?

Show answer & explanation

Answer: 1

Answer

1

The value of

a

is 1. By finding the vertex of

f(x)

at

(7, -16a)

and using the vertex of

g(x)

at

(7, -16a+8)

, we evaluate both functions at

x=5

. Setting

f(5) = g(5)

yields the linear equation

-12a = -16a + 4

, which simplifies to

a = 1

.

Step-by-Step Solution

1

Find the x-coordinate of the vertex of

f(x)

h = 7

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

f(x) = a(x-r_1)(x-r_2)

is the midpoint of its x-intercepts,

r_1 = 3

and

r_2 = 11

.

2

Find the y-coordinate of the vertex of

f(x)

in terms of

a

k = -16a

Evaluate the function at the vertex x-coordinate:

f(7) = a(7-3)(7-11) = -16a

.

3

Write the equation of

g(x)

using the given vertex information

g(x) = -(x-7)^2 + k+8

The vertex of

g(x)

is given as

(h, k+8)

. Since

h = 7

, the vertex form of

g(x)

is

g(x) = -(x-7)^2 + k+8

.

4

Set up the equation for the intersection point at

x = 5

f(5) = g(5)

Since the graphs of

f

and

g

intersect at

x=5

, their y-values must be equal at this point.

5

Substitute expressions and solve for

a

a = 1

Evaluate both sides:

f(5) = a(5-3)(5-11) = -12a

, and

g(5) = -(5-7)^2 + k+8 = -4 + k+8 = k+4

. Substituting

k = -16a

gives

-12a = -16a + 4 \implies 4a = 4 \implies a = 1

.

Key Concept

Quadratic Functions and Graphs

Question 16Question →

The function $f$ is defined by $f(x) = -(x - 5)^2 + 4$ . What is the maximum value of the function $f$ ?

-4

4

5

-5

Show answer & explanation

Answer: $4$

Answer

The maximum value of the function is 4.

The quadratic function

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

is in vertex form

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

, where the vertex of the parabola is

(h, k) = (5, 4)

. Because the leading coefficient

a = -1

is negative, the parabola opens downward, which means the vertex is the highest point on the graph. The maximum value of the function is the y-coordinate of this vertex, which is 4.

Step-by-Step Solution

1

Identify the form of the quadratic function.

The function

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

is written in vertex form,

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

.

Recognizing the vertex form allows us to directly identify the vertex coordinates of the parabola.

2

Identify the vertex

(h, k)

and the direction the parabola opens.

The vertex is

(5, 4)

and the leading coefficient is

a = -1

.

Since the leading coefficient

a

is negative, the parabola opens downward, meaning the vertex is the highest point on the graph.

3

Determine the maximum value of the function.

The maximum value of the function is the y-coordinate of the vertex, which is 4.

The maximum value of a function refers to its largest output (y-value), which corresponds to the y-coordinate of its highest point.

Key Concept

Quadratic Functions and Graphs

Question 17Question →

The graph of the quadratic function $f$ in the $xy$ -plane has a vertex at $(4, 7)$ . If the function $g$ is defined by $g(x) = f(x) + 5$ , what is the $y$ -coordinate of the vertex of the graph of $y = g(x)$ ?

Show answer & explanation

Answer: 12

Answer

The correct answer is 12.

The vertex of the graph of the quadratic function

f

is

(4, 7)

. The function

g(x) = f(x) + 5

adds

5

to every output value of

f(x)

, which shifts the entire graph vertically upward by

5

units. Consequently, the vertex of the graph of

g

is translated from

(4, 7)

to

(4, 7 + 5)

, which simplifies to

(4, 12)

. Therefore, the

y

-coordinate of the vertex of the graph of

y = g(x)

is

12

.

Step-by-Step Solution

1

Identify the y-coordinate of the vertex of the graph of the function f.

The y-coordinate of the vertex of the graph of f is 7.

The vertex of the graph of f is given as (4, 7).

2

Determine the vertical translation from f(x) to g(x).

The graph of g(x) is shifted upward by 5 units relative to the graph of f(x).

The term + 5 in the definition g(x) = f(x) + 5 increases all y-values of the function by 5.

3

Calculate the y-coordinate of the vertex of the graph of g.

The new y-coordinate is 7 + 5 = 12.

Applying the vertical shift of 5 units to the original y-coordinate of 7 gives the new vertex's y-coordinate.

Key Concept

Vertical translations of quadratic functions and their graphs

Question 18Question →

The quadratic function $h$ is defined by $h(x) = (x + 2)^2 - 5$ . What are the coordinates of the vertex of the graph of $y = h(x)$ in the $xy$ -plane?

(-2, -5)

(2, -5)

(-2, 5)

(2, 5)

Show answer & explanation

Answer: $(-2, -5)$

Answer

The coordinates of the vertex of the graph are

(-2, -5)

.

The correct answer is the coordinate pair

(-2, -5)

. The vertex form of a quadratic function is given by

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

, where

(h, k)

is the vertex of the parabola. For the given equation

h(x) = (x + 2)^2 - 5

, we rewrite the squared expression as

(x - (-2))^2

to match the standard form. This shows that

h = -2

. The value of

k

is the constant term

-5

. Therefore, the vertex of the graph of

h

is

(-2, -5)

.

Step-by-Step Solution

1

Identify the general vertex form of a quadratic function.

The vertex form is

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

, where the vertex of the parabola is at the point

(h, k)

.

This general equation allows us to directly read the coordinates of the vertex by comparing the coefficients.

2

Compare the given function

h(x) = (x + 2)^2 - 5

to the vertex form.

By matching the terms, we get

x - h = x + 2

, which implies

h = -2

. The constant term is

k = -5

.

This identifies the values of

h

and

k

that make up the coordinates of the vertex

(h, k)

.

3

Write the vertex coordinates.

The vertex coordinates are

(h, k) = (-2, -5)

.

Combining the identified values gives the final coordinates of the vertex.

Key Concept

Identifying the vertex of a quadratic function directly from its vertex form equation,

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

.

Question 19Question →

A parabola in the $xy$ -plane has a vertex at $(3, -4)$ and passes through the point $(1, 0)$ . A second parabola is created by reflecting the original parabola across the $x$ -axis, then translating the resulting graph $4$ units to the left and $2$ units down. If this second parabola represents the graph of the function $g$ , which of the following equations defines $g(x)$ ?

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

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

g(x) = -(x + 1)^2 - 6

g(x) = (x + 1)^2 - 6

Show answer & explanation

Answer: $g(x) = -(x + 1)^2 + 2$

Answer

The equation

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

The correct equation is

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

. Since the vertex of the original parabola

f(x)

is

(3, -4)

and it passes through

(1, 0)

, its vertex form equation is

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

. Reflecting

f(x)

across the

x

-axis negates the entire function, resulting in

-(x - 3)^2 + 4

. Shifting this function

4

units to the left replaces

x

with

x + 4

, yielding

-(x + 1)^2 + 4

. Finally, shifting it

2

units down subtracts

2

from the entire expression, giving

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

.

Step-by-Step Solution

1

Determine the equation of the original parabola

f(x)

using its vertex form

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

.

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

Since the vertex is

(3, -4)

, the vertex form is

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

. Substituting the point

(1, 0)

gives

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

.

2

Reflect the graph of

f(x)

across the

x

-axis.

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

A reflection across the

x

-axis negates the entire function, so

y = -f(x)

.

3

Translate the reflected graph

4

units to the left.

-(x + 1)^2 + 4

Translating a function

h(x)

to the left by

c

units is represented by

h(x + c)

. Replacing

x

with

x + 4

in

-(x - 3)^2 + 4

gives

-((x + 4) - 3)^2 + 4 = -(x + 1)^2 + 4

.

4

Translate the graph

2

units down to find the final function

g(x)

.

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

Translating a function down by

d

units is represented by subtracting

d

from the function. Subtracting

2

from

-(x + 1)^2 + 4

gives

-(x + 1)^2 + 2

.

Key Concept

Applying transformations (reflections and horizontal/vertical translations) to quadratic functions in vertex form.

Question 20Question →

The quadratic function $f$ is defined by $f(x) = ax^2 + bx + c$ , where $a$ , $b$ , and $c$ are constants, and has its vertex at $(2, 4)$ . The function $g$ is defined by $g(x) = f(x - 3) + k$ , where $k$ is a constant. The vertex of the graph of $y = g(x)$ in the $xy$ -plane lies on the line $y = 2x - 3$ . If the product of the $x$ -intercepts of the graph of $y = g(x)$ is $11$ , what is the $y$ -intercept of the graph of $y = f(x)$ ?

2

6

-4

-6

Show answer & explanation

Answer: 2

Answer

2

The correct answer is 2. By writing the quadratic function

f(x)

in vertex form as

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

, we can apply the transformation rules to express

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

as having a vertex at

(5, 4 + k)

. Since this vertex lies on the line

y = 2x - 3

, substituting

x = 5

gives

4 + k = 7

, meaning

g(x) = a(x - 5)^2 + 7

. The roots of

g(x) = 0

are

5 \pm \sqrt{-7/a}

, and their product is

25 + 7/a

. Setting this product equal to 11 yields

a = -1/2

. Substituting

a = -1/2

back into the expression for

f(0) = 4a + 4

gives 2.

Step-by-Step Solution

1

Write the function

f(x)

in vertex form using its vertex

(2, 4)

.

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

Any quadratic function with a vertex at

(h, k)

can be written in the form

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

.

2

Determine the vertex of the transformed function

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

.

The vertex of

g(x)

is

(5, 4 + k)

.

The transformation

f(x - 3)

shifts the graph 3 units to the right, changing the

x

-coordinate of the vertex from 2 to 5. The addition of

k

shifts the graph vertically by

k

units, changing the

y

-coordinate of the vertex from 4 to

4 + k

.

3

Use the condition that the vertex of

y = g(x)

lies on the line

y = 2x - 3

to find the value of

k

and the vertex coordinates.

4 + k = 7 \implies k = 3

. The vertex of

g(x)

is

(5, 7)

.

Since the vertex

(5, 4 + k)

lies on the line

y = 2x - 3

, substituting

x = 5

into the line equation gives

y = 2(5) - 3 = 7

.

4

Write

g(x)

in vertex form and express its roots (the

x

-intercepts) in terms of

a

.

g(x) = a(x - 5)^2 + 7

. The roots are

x = 5 \pm \sqrt{-\frac{7}{a}}

.

Setting

g(x) = 0

gives

a(x - 5)^2 + 7 = 0

, which simplifies to

(x - 5)^2 = -\frac{7}{a}

, yielding

x = 5 \pm \sqrt{-\frac{7}{a}}

.

5

Calculate the product of the roots and set it equal to 11 to solve for

a

.

25 + \frac{7}{a} = 11 \implies a = -\frac{1}{2}

.

The product of the roots is

(5 + \sqrt{-\frac{7}{a}})(5 - \sqrt{-\frac{7}{a}}) = 25 - (-\frac{7}{a}) = 25 + \frac{7}{a}

. Setting this equal to 11 gives

\frac{7}{a} = -14

, so

a = -\frac{1}{2}

.

6

Evaluate

f(0)

to find the

y

-intercept of the graph of

f(x)

.

f(0) = 4(-\frac{1}{2}) + 4 = 2

.

The

y

-intercept is the value of the function at

x = 0

. Substituting

x = 0

and

a = -\frac{1}{2}

into

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

yields

4a + 4 = 2

.

Key Concept

Quadratic transformations and properties of roots in vertex form

Quadratic Functions and Graphs

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer