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!
