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

1 Answer
May 29, 2017

#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" :)