A quadratic function f has its vertex at (4, 12) and a y-intercept at (0, -4) in the xy-pl… — SAT

A quadratic function $f$ has its vertex at $(4, 12)$ and a $y$ -intercept at $(0, -4)$ in the $xy$ -plane. The function $g$ is defined by $g(x) = f(x + 2) + k$ , where $k$ is a constant. If the $y$ -intercept of the graph of $g$ is $(0, 15)$ , what is the value of $k$ ?

Answer: 7

Answer

The quadratic function

f(x)

is determined to be

f(x) = -(x - 4)^2 + 12

by substituting the vertex

(4, 12)

and the

y

-intercept

(0, -4)

into the vertex form. The transformation

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

translates the function horizontally left by 2 units and vertically by

k

units, resulting in

g(x) = -(x - 2)^2 + 12 + k

. Using the

y

-intercept of

g

, which is

(0, 15)

, we substitute

x = 0

to get

15 = -(0 - 2)^2 + 12 + k

, simplifying to

15 = 8 + k

, which yields

k = 7

Step-by-Step Solution

Write the vertex form of

f(x)

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

The vertex form of a quadratic function with vertex

(h, k_{vertex})

is given by

f(x) = a(x - h)^2 + k_{vertex}

Determine the value of the coefficient

a

a = -1

, so

f(x) = -(x - 4)^2 + 12

Substitute the coordinates of the

y

-intercept

(0, -4)

into the vertex form equation to solve for

a

Express the transformed function

g(x)

in terms of

x

and

k

g(x) = -(x - 2)^2 + 12 + k

Apply the translation rules: substituting

x + 2

for

x

shifts the graph left by 2 units, and adding

k

shifts the graph vertically by

k

units.

Solve for the constant

k

k = 7

Use the

y

-intercept of

g(x)

, which is

(0, 15)

, so

g(0) = 15

. Setting

15 = -(0 - 2)^2 + 12 + k

simplifies to

15 = 8 + k

, which yields

k = 7

Key Concept

Quadratic function vertex form and transformations

Question

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Step-by-Step Solution

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