In graphing $y = 4 {(x - 3)}^{2} - 2$ $y=4(x-3)^2 -2$ , the vertex of the parabola is at what point?

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Answer 1 · 2016-05-06T20:49:16

Vertex $\to (x, y) \to (3, - 2)$ $->(x,y)->(3,-2)$

Reading directly from the vertex form equation

$x_{vertex} = (- 1) \times (- 3) = + 3$ $x_("vertex")=(-1)xx(-3) = +3$
$y_{vertex} = - 2$ $y_("vertex")=-2$

So Vertex $\to (x, y) \to (3, - 2)$ $->(x,y)->(3,-2)$

In graphing y=4(x−3)2−2y=4(x-3)^2 -2, the vertex of the parabola is at what point?