We need
(u/v)'=(u'v-uv')/(v^2)
Let's calculate the first derivative
f(x)=x^2/(x-2)^2
u=x^2, =>, u'=2x
v=(x-2)^2, =>, v'=2(x-2)
f'(x)=(2x*(x-2)^2-x^2*(2(x-2)))/(x-2)^4
=((x-2)(2x(x-2)-2x^2))/(x-2)^4
=(2x^2-4x-2x^2)/(x-2)^3
=(-4x)/(x-2)^3
f'(x)=0, when x=0
We build our first chart
color(white)(aaaa)xcolor(white)(aaaa)-oocolor(white)(aaaaaaaa)0color(white)(aaaaaaaaaa)2color(white)(aaaaaa)+oo
color(white)(aaaa)-4xcolor(white)(aaaaaa)+color(white)(aaaa)0color(white)(aaaa)-color(white)(aaaa)||color(white)(aaa)-
color(white)(aaaa)(x-2)^3color(white)(aaa)-color(white)(aaaa)0color(white)(aaaa)-color(white)(aaaa)||color(white)(aaa)+
color(white)(aaaa)f'(x)color(white)(aaaaa)-color(white)(aaaa)0color(white)(aaaa)+color(white)(aaaa)||color(white)(aaa)-
color(white)(aaaa)f(x)color(white)(aaaaaa)↘color(white)(aaa)0color(white)(aaaa)↗color(white)(aaa)||color(white)(aaa)↘
The critical point is when x=0
The intervals of increasing are x in (0,2) and decreasing are x in (-oo,0) uu (2,+oo)
Now, we calculate the second derivative
u=-4x, =>, u'=-4
v=(x-2)^3, =>, v'=3(x-2)^2
So,
f''(x)=(-4(x-2)^3+4x*3(x-2)^2)/(x-2)^6
=(-4(x-2)^2((x-2)-3x))/(x-2)^6
=-4(x-2-3x)/(x-2)^4
=-4(-2x-2)/(x-2)^4
=(8(x+1))/(x-2)^4
Therefore,
f''(x) =0 when x=-1, this is the inflexion point
The chart is as follows
color(white)(aaaa)Icolor(white)(aaaaaa)(-oo,-1)color(white)(aaaa)(-1,2)color(white)(aaaa)(2,+oo)
color(white)(aaaa)f''(x)color(white)(aaaaaaa)-color(white)(aaaaaaaaa)+color(white)(aaaaaaaa)+
color(white)(aaaa)f''(x)color(white)(aaaaaaa)nncolor(white)(aaaaaaaaa)uucolor(white)(aaaaaaaa)uu
The curve is concave is up when x in (-1,2) uu (2,+oo)
The curve is concave down when x in (-oo,-1)
Also,
f''(0)>0, this is a minimum
graph{x^2/(x-2)^2 [-11.36, 11.14, -1.58, 9.67]}