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

1 Answer
Jun 9, 2015

Answer:

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#
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