Question

what is the vertex of f(x)=x^2+2x-8?

Answer 1

(-1, -9) = vertex!

Answer 2

Wuld you like to know how?

Answer 3

step 1; isolate the ax^2+bx terms by a common factor; here, that's 1:

f(x) = 1(x^2 +2x) -8

step 2; look at your new bx term; take b; half it; square it; add and subtract this number:

f(x) = (x^2+2x +1^2 -1^2) -8

f(x) = (x^2+2x+1-1) -8

NOTE that x^2+2x+1 is factored as (x+1)(x+1) = (x+1)^2. BUT what do we do with the -1?

step 3; get rid of the -1 by multiplying it by the brackets' coefficient (1):

f(x) = (x+1)^2 (1)(-1) -8

f(x) = (x+1)^2 -1 -8

f(x) = (x+1)^2 -9

NOTE our equation is now in vertex form, y= a(x-h)^2 +k, where (-h, k) is the vertex;

(-(+1), -9)

Answer 4

Dont forget to give a medal for best answer!