Y=2x^2+4x-5 into vertex form?

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Answer 1 · 2018-05-23T04:58:50

Given -

#y=2x^2+4x-5#

Vertex form of the equation is -

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

Where -

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

Vertex

#h=(_b)/(2a)=(-4)/(2 xx2)=(-4)/4=-1#

#k=2(-1)^2+4(-1)-5#

#k=2-4-5=-7#

Then-

#y=2(x+1)^2-7#