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
48Cevap
C
12
D
24

Cevap

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

Adım Adım Çözüm

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.

Anahtar Kavram

Quadratic Functions and Graphs

Soru

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Konular

Bu konudan daha fazla soru