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() = 1/2 x2 +6 - 32
B.f() = -x2 + 12x -32
C.f() = 1/2 x2 + 6x -16
D f() = -x2 + 12x -6​

Answer 1

Option B
`f(x)=-x^(2)+12x-32`

Explanation:

we know that

For the given values, the quadratic function is a vertical parabola open downward (vertex is a maximum)

The equation in vertex form is equal to

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

where

(h,k) is the vertex

a is a coefficient

we have

(h,k)=(6,4)

so

`f(x)=a(x-6)^(2) +4`

Find the value of a

For x=8, y=0 -----> the y-intercept

substitute

`0=a(8-6)^(2) +4`

`0=a(2)^(2) +4`

`0=4a +4`

`a=-1`

substitute

`f(x)=-(x-6)^(2) +4`

Convert to standard form

`f(x)=-(x^(2)-12x+36) +4`

`f(x)=-x^(2)+12x-36+4`

`f(x)=-x^(2)+12x-32`

The graph in the attached figure