What is the the vertex of $y = 6(x - 2)^2 - 8$ ?

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Answer 1 · 2016-02-25T11:36:32

$"Vertex"->(x,y)->(2,-8)$

Tony B

The equation in this vertex form gives you the value of x for the vertex.

Consider the -2 from $(x-2)$

Apply $(-1)xx(-2)=+2$

$color(blue)(x_("vertex")=+2)$

Substitute $x=2$ into the equation to find $y_("vertex")$

$y_("vertex") = 6( 2-2)^2-8$

$y_("vertex") = 6(0)^2-8$

$color(blue)(y_("vertex") = -8$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$color(green)("Vertex"->(x,y)->(2,-8)$

What is the the vertex of y = 6(x - 2)^2 ­- 8y = 6(x - 2)^2 ­- 8?