What is the vertex form of #y=x^2+2x+15#?

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Answer 1 · 2016-01-10T05:00:07

#y=(x+1)^2+14#

Given _

#y=x^2+2x+15#

The vertex form of the equation is -

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

If we know the values of #a,h and k# we can change the given equation into a vertex form.

Find the vertex #(h, k)#

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

#a=1#
#h= (-b)/(2a)=(-2)/(2 xx 1)=-1#

#k=(-1)^2+2(-1)+15=1-2+15=14#

Now substitute the values of #a,h and k# in the vertex form of the equation.

#y=(1)(x-(-1))^2+14#
#y=(x+1)^2+14#