How do you find the axis of symmetry and vertex point of the function: #y=3x^2+12x-2#?

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Answer 1 · 2015-10-12T01:09:59

Its vertex is #(-2, -14)#
its axis of symmetry is #(x=-2)#

#y=3x^2+12x-2#

Find the vertex point first.

#x=(-b)/(2a)=(-12)/(2 xx 3)=(-12)/6=-2#

At #x= -2#

#y=3(-2^2)+12(-2)-2#
#y=12-24-2#
#y=-14#

Its vertex is #(-2, -14)#
its axis of symmetry is #(x=-2)#

graph{3x^2+12x-2 [-32.47, 32.5, -16.24, 16.22]}