Question

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

Answer 1

vertex: `(0,-4)`
x-intercepts: `+-2`
y-intercept: `(-4)`

Explanation:

General vertex form
`color(white)("XXX")f(x)=color(green)(m)(x-a)^2+b` with vertex at `(a,b)`

Writing given equation: `f(x)=x^2-4` in vertex form:
`color(white)("XXX")f(x)=1(x-0)^2+(-4)` with vertex at `(0,-4)`

x-intercepts are the possible values of `x` when `f(x)=0`
`color(white)("XXX")0=x^2-4`
`color(white)("XXX")rarr x=+-2`

y-intercept is the value of `y(=f(x))` when `x=0`
`color(white)("XXX")y=f(x=0) = 0^2-4=-4`

graph{x^2-4 [-5.717, 6.773, -4.885, 1.355]}