Question

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

Answer 1

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

Explanation:

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