How do you find the intercept and vertex of #x^2+6y=0#?

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How do you find the intercept and vertex of #x^2+6y=0#?

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Answer 1 · 2016-03-15T10:18:31

Vertex is at#(0,0)# and Intercept is at origin#(0,0)#

Explanation:

#6*y= - x^2 or y= -1/6*x^2# or #y=-1/6*(x-0)^2+0#: Vertex is at #0,0#
putting x=0; we get y=0 So the intercept is at origin(0,0) graph{-1/6x^2 [-10, 10, -5, 5]}[Ans]

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How do you find the intercept and vertex of #x^2+6y=0#?

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