A quadratic function f is defined by f(x) = a(x - h)^2 + k, where a, h, and k are constant… — SAT

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 $k$ ?

A
4
5Answer
C
$\frac{20}{3}$
D
20

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

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

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)

(0, 4)

, so

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

. Substituting

k = 5h

gives

ah^2 + 5h = 4

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

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

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

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