# How to graph a parabola x-2=1/8(y+1)^2?

Jun 9, 2015

Isolate x, find vertex, intersections and so on as the two variables (x and y) were inverted. You will have a rotated parabola.

#### Explanation:

First you should expand the ${\left(y + 1\right)}^{2}$:
$x - 2 = \frac{1}{8} \left({y}^{2} + 2 y + 1\right)$
$x - 2 = \frac{1}{8} {y}^{2} + \frac{1}{4} y + \frac{1}{8}$
Then isolate x in the left member:
$x = \frac{1}{8} {y}^{2} + \frac{1}{4} y + \frac{1}{8} + 2$
$x = \frac{1}{8} {y}^{2} + \frac{1}{4} y + \frac{17}{8}$
$x = \frac{1}{8} \left({y}^{2} + 2 y + 17\right)$
Now you find the vertex and intersections:
V((4ac-b^2)/(4a);-b/(2a))=(1/8*(4*17-4)/4;1/8*(-2)/2)=(2;-1/8)
Intersections:
$y = 0 \to x = \frac{17}{8}$
$x = 0 \to \neg \exists y \in \mathbb{R}$