Question

How do you find the x and y intercepts for `y=(x-8)^2-4`?

Answer 1

For the `y` axis just calculate `y` for `x=0`.
And the same goes for the `x` axis.

Explanation:

If you find yourself in the `y` axis, you'll notice that there's no horizontal movement, which means that your `x` is equal to `0`.
Thus, if you want to find the `y` axis interception you will have to impose the condition of `x=0`, this way you will have:

`x=0 ->y=(0-8)^2-4=(-8)^2-4=64-4=60 ->` Interception point `(0, 60)`

And the same reasoning is applied for the `x` axis. No vertical movement means `y=0`, in this case you will have to solve the complete second degree equation:

`(x-8)^2-4=0 -> x^2-2*8*x+64-4=0 -> x^2-16x+60=0 -> x=(16+-sqrt((-16)^2-4*1*60))/2 -> x=(16+-sqrt(16))/2=(16+-4)/2`

This gives us two interception points for the `x` axis:

`x=(16+4)/2=10` and `x=(16-4)/2=6`