Question

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

Answer 1

vertex at `(x,y)=(0,-2)`
y-intercept: `y=0`
x-intercepts: `x=-sqrt(2)` and `x=+sqrt(2)`

Explanation:

One way to find the vertex is to convert the given equation into an explicit vertex form
`color(white)("XXX")y=(x-color(red)a)^2+color(blue)(b)` with vertex at `color(red)a,color(blue)b)`

`y=x^2-2`
is equivalent to
`color(white)("XXX")y=(x-color(red)0)^2+color(blue)((-2))`
the vertex form with vertex at `(color(red)0,color(blue)((-2)))`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The y-intercept is the value of `y` when `x=0`

In this case `y=color(green)x^2-2` with `color(green)x=color(green)0`
becomes `y=color(green)0^2-2=-2`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The x-intercepts are the values of `x` which are possible when `y=0`

In this case `color(magenta)y=x^2-2` with `color(magenta)y=color(magenta)0`
becomes `color(magenta)0=x^2-2`
`color(white)("XXX")rarr x^2=2`
`color(white)("XXX")rarr x=+-sqrt(2)`