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Find the vertex form of the following equation by completing the square. y=3x^2+6x+8 A. y=(x+4)^2 B. y=3(x+2)^2+4 C. y=3(x+1)^2+5 D. y=(x+3)^2+5 Please Help
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Put it in vertex form! f(x) = a(x -h)^2 + k
Take the "x" term and divide it by 2 and square it. 3(x^2+2x + 1) +8 -3 3(x^2 + 2x + 1) + 5 f(x) = 3(x+2)^2 + 5
Sorry typo, f(x) = 3(x + 1)^2 + 5
ok yeah I just got that thanks!!
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