How to graph a parabola x-2=1/8(y+1)^2x−2=18(y+1)2?

1 Answer
Luca F.
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 (y+1)^2(y+1)2:
x-2=1/8(y^2+2y+1)x−2=18(y2+2y+1)
x-2=1/8y^2+1/4y+1/8x−2=18y2+14y+18
Then isolate x in the left member:
x=1/8y^2+1/4y+1/8+2x=18y2+14y+18+2
x=1/8y^2+1/4y+17/8x=18y2+14y+178
x=1/8(y^2+2y+17)x=18(y2+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)V(4ac−b24a;−b2a)=(18⋅4⋅17−44;18⋅−22)=(2;−18)
Intersections:
y=0 -> x=17/8y=0→x=178
x=0 -> notexists y in RR
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