Graping y=x ^ 2 / (2 – x) usin derivatives .Plz Help!

Question

Graping y=x ^ 2 / (2 –  x) usin derivatives .Plz Help!

anonymous · Answer

Thanks Ishan. :)

anonymous · Answer

I found the vertcal asymptote to be x=2, no horz asympte, and y=-x oblique asymptote

anonymous · Answer

I lso found (0,0) to be a pont on
 the graph

anonymous · Answer

Hmm I haven't done this before, I will see if I can get it. 

$$y' = \frac{2x(2-x)+x^2}{(2-x)^2}$$
$$y'=0\implies \frac{4x-x^2}{(2-x)^2} = 0$$Yes, vertical asymptote at x=2. And minima at 0.

anonymous · Answer

The problem is how can we have a asmpytote of y=-x,when 0,0 id a point on the graph

anonymous · Answer

Never mind the last sentence

anonymous · Answer

actually y=-x-2 is an oblique asymptote

anonymous · Answer

asymptotes can be crossed... only the vertical asymptotes can never be crossed.

anonymous · Answer

I have found everything,including the second derivative ,concavity and increading/decreasing points

anonymous · Answer

but cant graph it

anonymous · Answer

Can you help me graph it

anonymous · Answer

|dw:1336410136774:dw|From (0,2). Second derivative $\large\frac{-8}{(2-x)^3}$.

anonymous · Answer

:o Where is my drawing??? :(

anonymous · Answer

:(
 
I got a graph from (-2,2] as f''(x) < 0.
|dw:1336410866104:dw|