How do you write the vertex form equation of the parabola #y= x^2 +6x +7#?
1 Answer
Apr 12, 2016
#y=(x+3)^2-2#
Explanation:
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=(-3)^2+6(-3)+7#
#k=9-18+7#
#k=16-18=-2#
#a=1#
#h=-3#
#k=-2#
Substitute these values in
#y=1(x-(-3))^2+(-2)#
#y=(x+3)^2-2#