In the xy-plane, the graph of the quadratic function f(x) = a(x-3)(x-11), where a is a con… — SAT

In the $xy$ -plane, the graph of the quadratic function $f(x) = a(x-3)(x-11)$ , where $a$ is a constant, has a vertex at $(h, k)$ . The graph of the quadratic function $g(x) = -(x-h)^2 + q$ has its vertex at $(h, k+8)$ . If the graphs of $f$ and $g$ intersect at a point with an $x$ -coordinate of $5$ , what is the value of $a$ ?

Answer: 1

Answer

The value of

a

is 1. By finding the vertex of

f(x)

(7, -16a)

and using the vertex of

g(x)

(7, -16a+8)

, we evaluate both functions at

x=5

. Setting

f(5) = g(5)

yields the linear equation

-12a = -16a + 4

, which simplifies to

a = 1

Step-by-Step Solution

Find the x-coordinate of the vertex of

f(x)

h = 7

The x-coordinate of the vertex of a quadratic function of the form

f(x) = a(x-r_1)(x-r_2)

is the midpoint of its x-intercepts,

r_1 = 3

and

r_2 = 11

Find the y-coordinate of the vertex of

f(x)

in terms of

a

k = -16a

Evaluate the function at the vertex x-coordinate:

f(7) = a(7-3)(7-11) = -16a

Write the equation of

g(x)

using the given vertex information

g(x) = -(x-7)^2 + k+8

The vertex of

g(x)

is given as

(h, k+8)

. Since

h = 7

, the vertex form of

g(x)

g(x) = -(x-7)^2 + k+8

Set up the equation for the intersection point at

x = 5

f(5) = g(5)

Since the graphs of

f

and

g

intersect at

x=5

, their y-values must be equal at this point.

Substitute expressions and solve for

a

a = 1

Evaluate both sides:

f(5) = a(5-3)(5-11) = -12a

, and

g(5) = -(5-7)^2 + k+8 = -4 + k+8 = k+4

. Substituting

k = -16a

gives

-12a = -16a + 4 \implies 4a = 4 \implies a = 1

Key Concept

Quadratic Functions and Graphs

Question

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Step-by-Step Solution

Key Concept

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