Question

select the quadratic function with a graph that has the following features.

• x-intercept at (8,0)
• y-intercept at (0,-32)
• maximum value at (6,4)
• axis of symmetry at $x = 6$

a. $f(x) = -\frac{1}{2}x^{2} + 6x - 32$
b. $f(x) = -x^{2} + 12x - 32$
c. $f(x) = -x^{2} + 12x - 36$
d. $f(x) = -\frac{1}{2}x^{2} + 6x - 16$

Explanation:

Step1: Use vertex form of quadratic

The vertex (maximum point) is $(6,4)$, so the vertex form is $f(x)=a(x-6)^2 + 4$, where $a<0$ (since it's a maximum).

Step2: Use y-intercept to find $a$

Substitute $(0,-32)$ into the vertex form:
$-32 = a(0-6)^2 + 4$
$-32 = 36a + 4$
$36a = -32 - 4 = -36$
$a = \frac{-36}{36} = -1$

Step3: Expand to standard form

Substitute $a=-1$ into vertex form:
$f(x)=-1(x-6)^2 + 4$
$f(x)=-(x^2 -12x +36) +4$
$f(x)=-x^2 +12x -36 +4$
$f(x)=-x^2 +12x -32$

Step4: Verify x-intercept

Substitute $f(x)=0$:
$0=-x^2 +12x -32$
$x^2 -12x +32=0$
$(x-8)(x-4)=0$
Roots are $x=8$ and $x=4$, so $(8,0)$ is an x-intercept, which matches.

Step5: Verify axis of symmetry

For $f(x)=-x^2 +12x -32$, axis of symmetry is $x=-\frac{b}{2a}=-\frac{12}{2(-1)}=6$, which matches.

Answer:

B. $f(x) = -x^2 + 12x - 32$