What is the vertex of $y = 3 x^{2} + 12 x - 16$ $y=3x^2 +12x - 16$ ?

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Answer 1 · 2018-03-06T14:15:56

$(- 2, - 28)$ $(-2, -28)$

To find the x-coordinate of the vertex, you do $- \frac{b}{2 a}$ $-b/(2a)$

Where $a = 3, b = 12, c = - 16$ $a = 3 , b=12 , c=-16$

You then take that answer. Here that is $- \frac{12}{6} = - 2$ $-12/6=-2$ , and then input that value as the $x$ $x$ value.

$3 {(- 2)}^{2} + 12 (- 2) - 6 = 12 - 24 - 16 = - 28$ $3(-2)^2+12(-2)-6=12-24-16=-28$

So the coordinates are $(- 2, - 28)$ $(-2,-28)$

What is the vertex of y=3x^2 +12x - 16 y=3x2+12x−16y=3x^2 +12x - 16 ?