Question

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

Answer 1

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 `(y+1)^2`:
`x-2=1/8(y^2+2y+1)`
`x-2=1/8y^2+1/4y+1/8`
Then isolate x in the left member:
`x=1/8y^2+1/4y+1/8+2`
`x=1/8y^2+1/4y+17/8`
`x=1/8(y^2+2y+17)`
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 -> x=17/8`
`x=0 -> notexists y in RR`