How do you rewrite #y = x^2 + 14x + 29# in vertex form?

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Answer 1 · 2017-11-26T03:09:41

#y=(x+7)^2+20#

Given -

#y=x^2+14x+29#

Vertex form of the equation is -

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

Where -

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

First, find the vertex of the given equation

#x=(-b)/(2a)=(-14)/2=-7#

#y=(-7)^2+14(-7)+29=49-98+29=-20#

Vertex #(-7, -20)#

#a=1#

Substitute these values in the formula

#y=(x-(-7))^2-(-20)#

#y=(x+7)^2+20#

Answer 2 · 2017-11-26T06:13:46

#y = (x+7)^2 -20#

Vertex form is #y=a(x+b)^2+c#

Use the process of completing the square

#y = x^2 +14x color(red)(+7^2 -7^2) +29" "larr color(red)((+-14/2)^2)#

#y = (x^2 +14x+49) + (-49+29)#

#y = (x+7)^2 -20#

This is vertex form.

The vertex will be at #(-7,-20)#