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Write 3x2 + 12x - 1 = 0 in vertex form
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@ganeshie8
do you know what a "perfect square trinomial" is?
yes @jdoe0001
so let us group that first then \(\bf 3x^2 + 12x - 1 = 0\implies (3x^2 + 12x) - 1 = 0\implies 3(x^2+4x)-1=0 \\ \quad \\ 3(x^2+4x+{\color{red}{ \square }}^2)-1=0\) so... what do you think we need there to get a "perfect square trinomial" in the parentheses?
Vertex form is: f(x) = a(x - h)^2 + k, (3x^2 + 12x) - 1 = 0 3(x^2 + 4x) - 1 = 0 3(x^2 + 4x + 4) - 1 - 12 = 0 ... you added 3(4) to the equation, to keep it balanced you subtract 12, 3 * 4 = 12. -1 - 12 = -13 f(x) = 3(x + 2)^2 - 13
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