In the xy-plane, the graph of the quadratic function f(x) = -x^2 + 6x + 7 has a vertex at … — SAT

In the $xy$ -plane, the graph of the quadratic function $f(x) = -x^2 + 6x + 7$ has a vertex at $(h, k)$ . If the graph of $f$ is translated $4$ units to the right and $3$ units up to produce the graph of the function $g$ , what is the maximum value of $g$ ?

Answer: 19

Answer

The maximum value of the function

g

19

The vertex of the original quadratic function

f(x) = -x^2 + 6x + 7

(3, 16)

. Since the coefficient of

x^2

is negative, the graph opens downward, making

16

the maximum value of the function. Translating the graph

3

units up shifts all

y

-values up by

3

, which increases the maximum value to

16 + 3 = 19

. The horizontal translation of

4

units to the right shifts the graph horizontally but does not affect the maximum output value.

Step-by-Step Solution

Find the vertex of the original quadratic function

f(x) = -x^2 + 6x + 7

The vertex of

f(x)

(3, 16)

Rewrite the function in vertex form,

f(x) = -(x-3)^2 + 16

, to identify the vertex

(h, k)

(3, 16)

Determine the maximum value of the original function

f(x)

The maximum value of

f(x)

16

Since the leading coefficient is negative, the parabola opens downward, and the maximum value occurs at the vertex's

y

-coordinate.

Apply the vertical translation to find the maximum value of the new function

g(x)

The maximum value of

g

19

Translating the graph

3

units up increases all

y

-values, including the maximum value, by

3

, so

16 + 3 = 19

. The horizontal translation of

4

units to the right does not affect the maximum value.

Key Concept

Quadratic Functions and Graphs

Question

Answer

Step-by-Step Solution

Key Concept

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