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

May 29, 2017

$f \left(x\right) = \frac{1}{8} {x}^{2} - \frac{1}{4} x + \frac{33}{8}$

#### Explanation:

First, break it down:
Using the quadratic rule ${\left(a - b\right)}^{2} = {a}^{2} + {b}^{2} - 2 a b$ , we get:

${\left(x - 1\right)}^{2} + 32 = 8 y \iff {x}^{2} + 1 - 2 x + 32 = 8 y$

$\iff \frac{{x}^{2} + 1 - 2 x + 32}{8} = \frac{1}{8} {x}^{2} - \frac{1}{4} x + \frac{33}{8} = y$

To graph it, you can turn it into a function:
$f \left(x\right) = \frac{1}{8} {x}^{2} - \frac{1}{4} x + \frac{33}{8}$

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" :)