Question

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

Answer 1

Answer:

49

Step-by-step 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