Question

How do you find the vertex of `y=x^2 + 4x + 2`?

Answer 1

Convert the given equation into vertex form to find:
vertex at `(-2,-2)`

Explanation:

The general vertex form for a parabola is
`color(white)("XXX")y=m(x-color(red)(a))^2+color(blue)(b)`
with its vertex at `(color(red)(a),color(blue)(b))`

Given
`color(white)("XXX")y=x^2+4x+2`
Completing the square
`color(white)("XXX")y=x^2+4xcolor(green)(+4)+2color(green)(-4)`

`color(white)("XXX")y=(x+2)^2-2`

`color(white)("XXX")y=(x-(color(red)(-2)))^2+(color(blue)(-2))`

which is in vertex form with the vertex at `(color(red)(-2),color(blue)(-2))`

Just to help verify this result, here is the graph of `y=x^2+4x+2`:
graph{x^2+4x+2 [-8.702, 2.397, -3.11, 2.437]}