How do you write the vertex form equation of the parabola #y= x^2 +6x +7#?

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Answer 1 · 2016-04-12T02:56:59

#y=(x+3)^2-2#

Given -

#y=x^2+6x+7#
#y=ax^2+bx+c#

Vertex form of the equation is -

#y=a(x-h)^2+k#
#a=1#
#h=(-b)/(2a)=(-6)/(2 xx 1)=(-6)/2=-3 #

#k# is the value of the function

#k=(-3)^2+6(-3)+7#
#k=9-18+7#
#k=16-18=-2#

#a=1#
#h=-3#
#k=-2#

Substitute these values in #y=a(x-h)^2+k#

#y=1(x-(-3))^2+(-2)#

#y=(x+3)^2-2#