Question 1204828
{{{f(x) = x^2 - 2x }}}
{{{f(x) = (x^2 - 2x +b^2)-b^2}}}
{{{f(x) = (x^2 - 2x +1^2)-1^2}}}
{{{f(x) = (x - 1)^2-1}}} => it’s parabola with vertex at ({{{1}}},{{{-1}}})

axes of symmetry is {{{x=1}}}
the domain of {{{f(x)}}} is all real numbers

To find the {{{x}}}-intercepts, we can set {{{f(x)= 0}}}

{{{x^2 - 2x =0}}}

{{{x(x - 2) =0}}}

the {{{x}}}-intercepts are

{{{x=0}}}
{{{x=2}}}

make table

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

plot points and draw a graph

{{{drawing( 600, 600, -10, 10, -10, 10, 
circle(1,-1,.12),circle(3,3,.12),circle(-2,8,.12),circle(2,0,.12),circle(-1,3,.12),
 graph( 600, 600, -10, 10, -10, 10, x^2 - 2x ) )}}}




The reciprocal of this would be 

{{{f(x)  = 1/( a(x - h)^2 + k)}}}

This is a rational function.

{{{f(x) = 1/((x - 1)^2-1  )}}}

vertical asymptote will be at {{{x=0}}} and {{{x=2}}} (x-intercepts of given function)

make table


{{{x}}}|{{{y}}}
{{{0}}}|{{{0}}}
{{{2}}}|{{{0}}}
{{{1}}}|{{{-1}}}
{{{-1}}}|{{{1/3}}}
{{{3}}}|{{{1/3}}}
{{{-2}}}|{{{1/8}}}

plot the points and graph


{{{drawing( 600, 600, -10, 10, -10, 10, 
circle(1,-1,.12),circle(3,3,.12),circle(-2,8,.12),circle(2,0,.12),circle(-1,3,.12),
circle(3,1/3,.12),circle(-2,1/8,.12),circle(-1,1/33,.12),blue(line(0.1,10,0.1,-10)),blue(line(2,10,2,-10)),
 graph( 600, 600, -10, 10, -10, 10, x^2 - 2x ,1/((x - 1)^2-1  )) )}}}