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

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Answer 1 · 2015-06-05T20:36:15

#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]}