How do you find the VERTEX of a parabola #y= 3/4x^2#?

1 Answer
VNVDVI
Apr 1, 2018

#(0,0)#

Explanation:

In general, the vertex of a parabola in the form #y=ax^2# where #a# is any number is #(0,0),# as the parabola has not been shifted up, down, left, or right at all.

We can prove this by comparing #y=3/4x^2# to the standard form of a quadratic, #y=ax^2+bx+c#

In this case, #a=3/4, b=0, c=0#

The #x-#coordinate of the vertex is given by #-b/(2a)#. In this case, it would be #(-0/(2*3/4))=0#.

The #y-#coordinate of the vertex is given by plugging in the result of #-b/(2a)# into the equation. In this case, #-b/(2a)=0, y=3/4(0^2)=0#

So, the vertex is #(0,0)#

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