Question

How do you graph `(x – 1)^2 + 32 = 8y`?

Answer 1

`f(x)=1/8x^2-1/4x+33/8`

Explanation:

First, break it down:
Using the quadratic rule `(a-b)^2=a^2+b^2-2ab` , we get:

`(x-1)^2+32=8y iff x^2+1-2x+32=8y`

`iff (x^2+1-2x+32)/8=1/8x^2-1/4x+33/8=y`

To graph it, you can turn it into a function:
`f(x)=1/8x^2-1/4x+33/8`

graph{1/8x^2-1/4x+33/8 [-58.4, 81.8, -5.9, 64.2]}

This is a graph of "y". I think this is what you want as an answer. I would be surprised if you had to make a graph that showed "8y" :)