How do you write #y = - 3x^2 + 5x - 2# in vertex form?

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Answer 1 · 2016-06-02T09:55:05

#y=-3(x-5/6)^2+1/12#

Equation in standard form

#y=-3x^2+5x-2#

Vertex form of the equation is -

#y=a(x-h)^2+k#

Where -

#a# is coefficient of #x^2#
#h# is the x-coordinate of the vertex
#k# is y-coordinate of the vertex

#h=(-b)/(2a)=(-5)/(2 xx -3)=(-5)/(-6)=5/6#

#k=-3(5/6)^2+5(5/6)-2#

#k=-3(25/36)+25/6-2=-25/12+25/6-2=1/12#

The equation is -

#y=-3(x-5/6)^2+1/12#