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

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:

$x = 0 \to y = {\left(0 - 8\right)}^{2} - 4 = {\left(- 8\right)}^{2} - 4 = 64 - 4 = 60 \to$ Interception point $\left(0 , 60\right)$

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:

$x = \frac{16 + 4}{2} = 10$ and $x = \frac{16 - 4}{2} = 6$