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

1 Answer
Dec 6, 2016

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#