What is the vertex of $y= 2|x|^2 – 1$ ?

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Answer 1 · 2016-01-30T09:51:25

Vertex : $(0,-1)$

$y=2abs(x)^2-1$

This should give us a parabola and this equation is same as

$y=2x^2-1$ as $abs(x)^2$ and $x^2$ would give the same value as on squaring we would get only the positive value.

The vertex of $y=2x^2-1$ can be found by comparing it with the vertex form
$y=a(x-h)^2+k$ where $(h,k)$ is the vertex

$y=2(x-0)^2-1$
$y=a(x-h)^2 +k$

We can see $h=0$ and $k=-1$

Vertex is $(0,-1)$

What is the vertex of y= 2|x|^2 – 1 y= 2|x|^2 – 1 ?