Question

How do you find the vertex and intercepts for `y=x^2 -4`?

Answer 1

I found:
1) y-intercept `(0, -4)`;
2) x-intercepts `(-2,0) and (2,0)`;
3) vertex at `(0,-4)`

Explanation:

Your function is a Quadratic and can be represented by a Parabola:

We first find the intercepts:

1) y-axis:
set `x=0`
to get: `y=-4`

2) x-axis (if exists):
set `y=0`
solve: `x^2-4=0`
that gives you `x=+-2`
so you have 2 intercepts on the `x` axis.

We now find the vertex:
The `x` coordinate of the vertex can be found as:
`x_v=-b/(2a)`

where `b and a` refer to the numerical coefficients of the general Quadratic Function:
`y=ax^2+bx+c`
in your case we have:
`a=1`
`b=0`
`c=-4`
so that `x_v=-0/2=0`
the `y` coordinate of the vertex can be found substituting `x_v=0` into the original function that gives us:
`y_v=-4`

Graphically we can "see" these points:
graph{x^2-4 [-10, 10, -5, 5]}