1. ## function question

This is the function:
f(x)=(x^2-Bx+4)/(x-A)

given:
x=2 critical point
y=x+5 is an oblique asymptote

I need to find A and B and decide if x=2 is min or max.

i got to f'(x)= (x^2-2Ax-4+BA)/{(x-A)^2}
but I'm not even sure that I need a derivative here.

can someone explain what would be the way to solve this?
Please, if you can, show your work so that I can understand it.

THANKS!

2. Indeed, the derivative is $\displaystyle f'(x)=\frac{x^2-2Ax+AB-4}{(x-A)^2}$.
If x is a critical point then $f'(2)=0\Rightarrow A(B-4)=0$. (1)

The oblique asymptote is $y=mx+n$ where $\displaystyle m=\lim_{x\to\infty}\frac{f(x)}{x}$ and $\displaystyle n=\lim_{x\to\infty}[f(x)-mx]$.

Now, $\displaystyle m=\lim_{x\to\infty}\frac{x^2-BX+4}{x^2-Ax}=1$.
$\displaystyle n=\lim_{x\to\infty}\left(\frac{x^2-Bx+4}{x-A}-x\right)=\lim_{x\to\infty}\frac{(A-B)x+4}{x-A}=A-B=5$ (2)

From (1) and (2) we get:
I) $A=0,B=-5\Rightarrow x=2$ point of minimum
II) $A=9,B=4\Rightarrow x=2$ point of maximum.

I still don't understand the second part.
How did you decide what A goes with what B
and how do you know what's max and whats min?
If you could clear this up for me that would be great
thanks!

4. For $A=0,B=-5\Rightarrow \displaystyle f'(x)=\frac{x^2-4}{x^2}$
The roots of f'(x) are $x_1=-2,x_2=2$.
Then $f'(x)<0,\forall x\in(0,2)$, so f is decreasing and
$f'(x)>0,\forall x\in(2,\infty)$, so f is increasing.
Then x=2 is a point of minimum.

For $A=9,B=4\Rightarrow \displaystyle f'(x)=\frac{x^2-18x+32}{(x-9)^2}$.
The roots of f'(x) are $x_1=2,x_2=16$.
$f'(x)>0,\forall x\in(-\infty,2)$, so f is increasing and
$f'(x)<0,\forall x\in(2,9)$, so f is decreasing.
Then x=2 is a point of maximum.

5. thanks for all your help