Question

What is the y-value of the vertex of the function f(x)=-(x-3)(x+11)?
0
-8

Answer 1

49

Explanation:

The given function is
`f(x)=-(x-3)(x+11)`.

We expand to get:

`f(x)=-x^2-8x+33`

We complete the square to obtain the vertex form as follows:

`f(x)=-(x^2+8x)+33`

`f(x)=-(x^2+8x+16)--16+33`

`f(x)=-(x^2+8x+16)+16+33`

`f(x)=-(x+4)^2+49`

This function is now of the form:

`f(x)=a(x-h)+k`, where (h,k)=(-4,49) is the vertex.

The y-value of the vertex is therefore 49