How do you find the vertex and intercepts for $y = x^2 - 2$ ?

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Answer 1 · 2018-04-14T12:38:09

$(-sqrt2,0), (sqrt2,0)$ and $(0,-2)$ are the intercepts
$(0,-2)$ is also the vertex of the parabola

For x-intercepts, $y=0$

$0=x^2-2$
$2=x^2$
$x=+-sqrt2$ ie $(-sqrt2,0), (sqrt2,0)$

For y-intercepts, $x=0$

$y=0-2$
$y=-2$ ie $(0,-2)$

Vertex: $x=-b/(2a)$ = $x=-(0)/1$ = $x=0$

Therefore, the vertex is the same as the y-intercept

How do you find the vertex and intercepts for y = x^2 - 2y = x^2 - 2?