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

In the $xy$ -plane, the graph of the quadratic function $f(x) = -x^2 + bx + c$ has a vertex at $(h, k)$ . The graph of $f$ is translated $4$ units to the right and $5$ units up to form the graph of a quadratic function $g$ . If the graph of $g$ passes through the point $(1, 16)$ and has a $y$ -intercept of $13$ , what is the value of $k$ ?

Answer: 12

Answer

The value of

k

12

By writing the function in vertex form as

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

, we can apply the transformations directly. Translating the graph

4

units to the right and

5

units up gives the function

g(x) = f(x-4) + 5 = -(x-h-4)^2 + k+5

. Using the given points

g(0) = 13

and

g(1) = 16

, we set up a system of equations:

-(h+4)^2 + k+5 = 13

and

-(h+3)^2 + k+5 = 16

. Solving these simultaneously gives

h = -2

and

k = 12

Step-by-Step Solution

Write the function

f(x)

in vertex form.

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

Since the leading coefficient of

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

-1

, the vertex form must have a coefficient of

-1

Represent the translated function

g(x)

g(x) = -(x-h-4)^2 + k+5

Translating a function

f(x)

4

units to the right and

5

units up yields

g(x) = f(x-4) + 5

Use the

y

-intercept of

g(x)

to write the first equation.

k - (h+4)^2 = 8

The

y

-intercept is the value of the function at

x = 0

, so

g(0) = 13

Use the point

(1, 16)

g(x)

to write the second equation.

k - (h+3)^2 = 11

The graph passing through

(1, 16)

means

g(1) = 16

Solve the system of equations for

h

and

k

h = -2

and

k = 12

Equating the expressions for

k

gives

(h+4)^2 + 8 = (h+3)^2 + 11

, which simplifies to

2h = -4

, so

h = -2

. Substituting

h = -2

into

k = (h+4)^2 + 8

yields

k = 12

Key Concept

Translating quadratic functions and solving systems of quadratic equations.

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