Question

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

Answer 1

Explanation:

hello :

f(x)=-(x-3)(x+11) = -(x²+11x-3x-33)

f(x) = -x²-8x+33 = -(x²+8x-33)

f(x) = - (x²+8x+16-16-33)=-(x²+8x+16) + 49

f(x) = - (x+4)²+49 .... vertex form

the y-value of the vertex of the function is 49 when x= - 4

Answer 2

y = - 49

Explanation:

The vertex lies on the axis of symmetry which is at the midpoint of the zeros.

To find the zeros, let f(x) = 0, that is

(x - 3)(x + 11) = 0

Equate each factor to zero and solve for x

x - 3 = 0 ⇒ x = 3

x + 11 = 0 ⇒ x = - 11

`x_(vertex)` =
`(-11+3)/(2)` =
`(-8)/(2)` = - 4

To find the y- coordinate of the vertex evaluate f(- 4)

f(- 4) = (- 4 - 3)(- 4+ 11) = (- 7)(7) = - 49