What is the vertex of #y= -6x^2 + 4x + 3#?

1 Answer
Alan P.
Dec 21, 2015

Vertex: #(1/3, 3 2/3)#

Explanation:

Probably the easiest way to do this is to convert the equation into "vertex form": #y=m(x-a)^2+b# with vertex at #(a,b)#

Given:
#color(white)("XXX")y=-6x^2+4x+3#

Extract the #m# factor
#color(white)("XXX")y= (-6)(x^2-2/3x)+3#

Complete the square
#color(white)("XXX")y=(-6)(x^2-2/3x+(1/3)^2)+3-(-6)*(1/3)^2#

Rewrite with a squared binomial and simplified constant
#color(white)("XXX")y=(-6)(x-1/3)^2+ 3 2/3#

which is in vertex form with vertex at #(1/3, 3 2/3)#

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