How do you rewrite y = x^2 + 14x + 29y=x2+14x+29 in vertex form?

2 Answers
Nallasivam V
Nov 26, 2017

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

Explanation:

Given -

y=x^2+14x+29y=x2+14x+29

Vertex form of the equation is -

y=a(x-h)^2-ky=a(x−h)2−k

Where -

a -a− is the coefficient of x^2x2
h-h− is the x-coordinate of the vertex
k-k− is the y-coordinate of the vertex

First, find the vertex of the given equation

x=(-b)/(2a)=(-14)/2=-7x=−b2a=−142=−7

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

Vertex (-7, -20)(−7,−20)

a=1a=1

Substitute these values in the formula

y=(x-(-7))^2-(-20)y=(x−(−7))2−(−20)

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

EZ as pi
Nov 26, 2017

y = (x+7)^2 -20y=(x+7)2−20

Explanation:

Vertex form is y=a(x+b)^2+cy=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=x2+14x+72−72+29 ←(±142)2

y = (x^2 +14x+49) + (-49+29)y=(x2+14x+49)+(−49+29)

y = (x+7)^2 -20y=(x+7)2−20

This is vertex form.

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

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