How to graph a parabola #y = (x + 2)^2 + 2#?

1 Answer
Jun 5, 2015

#y = (x+2)^2+2#

The vertex of this parabola will be where #(x+2) = 0#, that is where #x = -2# and #y = 0+2 = 2#, that is at #(-2, 2)#.

The axis is vertical, given by the equation #x = -2#.

The intercept with the #y# axis will be where #x=0#, so

#y = (0+2)^2+2 = 4+2 = 6# - that is at #(0, 6)#

#y->oo# as #x->+-oo#

If you want any more points, just substitute them into the equation of the parabola.

graph{(x+2)^2+2 [-11.87, 8.13, -1.68, 8.32]}