The graph of the quadratic function f in the xy-plane has its vertex at (3, -4) and passes… — SAT

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 $g(4)$ ?

A
16
48Answer
C
12
D
24

Answer

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)

(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

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

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.

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

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

Answer

Step-by-Step Solution

Key Concept

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