Question

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

Answer 1

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

Explanation:

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

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

Explanation:

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)`