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what is the vertex of f(x)=x^2+2x-8?
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(-1, -9) = vertex!
Wuld you like to know how?
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)
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