Question 1179013
sketch

{{{f(x)=x^2-2x}}}-> this is a parabola, make a table with several points and draw a parabola through

{{{x}}}|{{{y}}}
{{{-2}}}|{{{8}}}->{{{f(x)=(-2)^2-2(-2)=4+4=8}}}
{{{-1}}}|{{{3}}}->{{{f(x)=(-1)^2-2(-1)=1+2=3}}}
{{{0}}}|{{{0}}}->{{{f(x)=0^2-2(0)=0}}}
{{{1}}}|{{{-1}}}->{{{f(x)=(1)^2-2(1)=1-2=-1}}}
{{{2}}}|{{{0}}}->{{{f(x)=(2)^2-2(2)=4-4=0}}}

draw inverse

recall: The inverse of a function has all the same points as the original function, except that the {{{x}}}'s and {{{y}}}'s have been reversed.

{{{8}}}|{{{-2}}}
{{{3}}}|{{{-1}}}
{{{0}}}|{{{0}}}
{{{-1}}}|{{{1}}}
{{{0}}}|{{{2}}}


sketch both on a graph:

{{{drawing( 600, 600, -10, 10, -10, 10,

circle(-2,8,.12),locate(-2,8,p(-2,8)),
circle(-1,3,.12),locate(-1,3,p(-1,3)),
circle(1,-1,.12),locate(1,-1,p(1,-1)),
circle(0,0,.12),locate(0.3,0.3,p(0,0)),
circle(2,0,.12),locate(2,0.3,p(2,0)),

circle(8,-2,.12),locate(8,-2,p(8,-2)),
circle(3,-1,.12),locate(3,-1,p(3,-1)),
circle(-1,1,.12),locate(-1,1,p(-1,1)),
circle(0,2,.12),locate(0,2,p(0,2)),
graph( 600, 600, -10, 10, -10, 10,x^2-2x,1-sqrt(x + 1), 1 + sqrt(x + 1))) }}}