Question

How do you graph the parabola `y=x^2 -4` using vertex, intercepts and additional points?

Answer 1

See below

Explanation:

`y=x^2-4`

Since `y` has no term in `x` we know that the axis of symmetry of it's parabolic graph is the line `x=0`

`y(0) = -4 -> y_"vertex" = (0,-4)`

We also know that `y_"vertex"` will be an absolute extremum of `y`

Since `x^2>=0 forall x in RR -> y_"vertex" = y_"min"`

Finally, we need to find the zeros of `y`

`x-4=0 -> x=+-2`

`:. y` has `x-` intercepts at `(-2,0) and (+2,0)`

Using this information we are able to graph `y` as below.

NB: In practice we would probably need a few extra points where `x<-2 and x>+2`

graph{x^2-4 [-14.24, 14.24, -7.11, 7.13]}