#f(x)=x^2/{(x-2)^2#

This function has a vertical asymptote at #x=2#, because the denominator is zero when #x=2#.

It approaches #1# from above as x goes to #+oo # (horizontal asymptote) and approaches #1# from below as x goes to #-oo #, because for large values #x^2~=(x-2)^2# with #x^2>(x-2)^2# for #x>0# and #x^2<(x-2)^2# for #x<0#.

To find max/min we need the first and second derivatives.

#{d f(x)}/dx =d/dx ( x^2/{(x-2)^2})# Use the quotient rule!

#{d f(x)}/dx = ( {(d/dx x^2) (x-2)^2 - x^2 ( d/dx (x-2)^2) }/{(x-2)^4})#.

Using rule for powers and the chain rule we get:

#{d f(x)}/dx = {(2x) (x-2)^2 - x^2 (2*(x-2)*1) }/(x-2)^4#.

We now neaten up a bit ...

#{d f(x)}/dx = {2x(x^2-4x+4) - x^2(2x-4) }/(x-2)^4#

#{d f(x)}/dx = {2x^3-8x^2+8x - 2x^3+4x^2 }/(x-2)^4#

#{d f(x)}/dx = {-4x^2 + 8x }/(x-2)^4#

Now the second derivative, done like the first.

#{d^2 f(x)}/dx^2 = {d/dx(-4x^2 + 8x)(x-2)^4 - (-4x^2 + 8x)(d/dx ((x-2)^4))}/(x-2)^8#

#{d^2 f(x)}/dx^2 = {(-8x + 8)(x-2)^4 - (-4x^2 + 8x)(4(x-2)^3 *1)}/(x-2)^8#

#{d^2 f(x)}/dx^2 = {(-8x + 8)(x-2)^4 - (-4x^2 + 8x)(4(x-2)^3 *1)}/(x-2)^8#

It's ugly but we only need to plug and and note where it's badly behaved.

#{d f(x)}/dx = {-4x^2 + 8x }/(x-2)^4# This function is undefined at #x=2#, that asymptote, but looks fine everywhere else.

We want to know were the max/min are ...

we set #{d f(x)}/dx=0#

#{-4x^2 + 8x }/(x-2)^4=0# this is zero when the numerator is zero and if the denominator is not.

#-4x^2 + 8x=0#

#4x( -x+2)=0# or #4x(2-x)=0# This is zero at #x=0# and #x=2#, but we cannot have a max/min were the derivative/function are undefined, so the only possibility is #x=0#.

"the second derivative test"

Now we look at the second derivative, ugly as it is ...

#{d^2 f(x)}/dx^2 = {(-8x + 8)(x-2)^4 - (-4x^2 + 8x)(4(x-2)^3)}/(x-2)^8#

Like the function and the first derivative this is undefined at #x=2#, but looks fine everywhere else.

We plug #x=0# into #{d^2 f(x)}/dx^2#

#{d^2 f(0)}/dx^2=#

# {(-8*0 + 8)(0-2)^4 - (-4*0^2 + 8*0)(4*0-2)^3}/(0-2)^8 #

#= {(8)(-2)^4}/(2)^8 #, isn't zero such a lovely number to plug it?

#=128/256# all that for #1/2#

#1/2 >0# so #x=0# is a local minima.

To find the y value we need to plug it into the function.

#f(x)=0^2/{(0-2)^2}=0# The origin!