In the xy-plane, the graph of the quadratic function f(x) = -x^2 + bx + c, where b and c a… — SAT

In the $xy$ -plane, the graph of the quadratic function $f(x) = -x^2 + bx + c$ , where $b$ and $c$ are constants, has its vertex at $(h, k)$ . The function $g$ is defined by $g(x) = f(x - 3) + 4$ . The graph of $g$ passes through the origin $(0,0)$ , and its vertex lies on the line $y = x$ in the first quadrant. What is the value of $f(0)$ ?

A
-1
-7Answer
C
-3
D
-21

Answer

-7

The correct answer is

-7

. Representing

f(x)

in vertex form as

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

tells us that translating the function 3 units right and 4 units up shifts the vertex from

(h, k)

(h+3, k+4)

. Since this vertex lies on the line

y=x

, we have

h+3 = k+4

, or

k = h-1

. Since the vertex of

g

lies in the first quadrant, its coordinates must be positive, meaning

h > -3

and

k > -4

. Using the fact that the graph of

g

passes through the origin, we have

g(0) = 0 \implies f(-3) + 4 = 0 \implies f(-3) = -4

. Substituting

x = -3

into the vertex form of

f(x)

gives

-(-3-h)^2 + k = -4 \implies (h+3)^2 - k = 4

. Substituting

k = h-1

yields

(h+3)^2 - (h-1) = 4 \implies h^2 + 5h + 6 = 0

, which factors into

(h+2)(h+3) = 0

. The solution

h = -3

is discarded because it places the vertex of

g

(0,0)

, which is not in the first quadrant. Therefore,

h = -2

and

k = -3

. Calculating

f(0)

gives

-h^2 + k = -(-2)^2 + (-3) = -7

Step-by-Step Solution

Determine the vertex form of

f(x)

and the vertex of

g(x)

Since

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

has its vertex at

(h, k)

, its vertex form is

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

. The function

g(x) = f(x - 3) + 4

represents a horizontal shift of

f

by 3 units to the right and a vertical shift by 4 units up. Thus, the vertex of

g

(h + 3, k + 4)

Understanding translations allows us to write the coordinates of the new vertex in terms of the original vertex variables.

Use the line

y = x

and quadrant constraints to find a relation between

h

and

k

Since the vertex of

g

lies on the line

y = x

, we set its coordinates equal:

h + 3 = k + 4

, which simplifies to

k = h - 1

. Additionally, because the vertex is in the first quadrant, we must have

h + 3 > 0

and

k + 4 > 0

The geometric placement of the vertex on the line

y = x

constrains its coordinate values.

Apply the condition that the graph of

g

passes through the origin.

g(0) = 0 \implies f(-3) + 4 = 0 \implies f(-3) = -4

. Substituting

x = -3

into the vertex form of

f(x)

gives

-(-3 - h)^2 + k = -4

, which simplifies to

(h + 3)^2 - k = 4

The point

(0,0)

lying on the graph of

g

provides an equation to solve for the vertex parameters.

Substitute

k = h - 1

into the quadratic equation and solve for

h

(h + 3)^2 - (h - 1) = 4 \implies h^2 + 6h + 9 - h + 1 = 4 \implies h^2 + 5h + 6 = 0

. Factoring gives

(h + 2)(h + 3) = 0

, so

h = -2

h = -3

Substituting the linear relation into the quadratic equation isolates the variable

h

Filter the solutions using the first quadrant constraint and calculate

f(0)

h = -3

, then

k = -4

, giving the vertex of

g

(0, 0)

, which is not in the first quadrant. If

h = -2

, then

k = -3

, giving the vertex of

g

(1, 1)

, which is in the first quadrant. Thus,

h = -2

and

k = -3

. The value of

f(0)

f(0) = -(0 - h)^2 + k = -h^2 + k = -(-2)^2 + (-3) = -4 - 3 = -7

The quadrant condition uniquely determines the correct vertex coordinates, allowing us to find the y-intercept of the original function.

Key Concept

Quadratic Functions and Graphs

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