Question

What is the vertex of `y= x^2 + 12x + 9`?

Answer 1

`y=x^2+12x+9`
`=>y=x^2+12x+6^2-36+9`
`=>y=(x+6)^2-27`
`=>y+27=(x+6)^2`
putting y+27 =Y and x+6=X
we have
`Y=X^2`
`=>4xx1/4Y=X^2`
Vertex of this Equation
is `(0,0)`

So the actual vertex
putting X=9 and Y=0
`x=-6 and y=-27`

graph{x^2+12*x+9 [-58.53, 58.57, -29.24, 29.27]}