Draw the graph of #x=y^2+2y+2#?

Question

1599 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-20T09:25:49

Please see below.

As #x=y^2+2y+2#

or #x=y^2+2y+1+1#

or #x=(y+1)^2+1#

Hence, this is a parabola whose vertex is #(1,-1)# and axis of symmetry is #y+1=0#

Let us select some points around #y=-1# say #y=-4,-3-2,-1,0,1,2#

for these corresponding values of #x# are #x=10,5,2,1,2,5,10#

and draw points #(10,-4)#, #(5,-3)#, #(2,-2)#, #(1,-1)#, #(2,0)#, #(5,1)# and #(10,2)#. Joining them forms a curve as follows.

graph{y^2+2y+1+1-x=0 [-2.71, 17.29, -5.72, 4.28]}