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

1 Answer
Jan 26, 2016

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]}