Calculus question with oblique asymptote involved This question is let $$f(x) = \frac{ax^2+bx-c}{x-4}.$$ Determine the values of $a,b,$ and $c$ so that $f(x)$ has an oblique asymptote of $y=3x+1$ and has local minimum at $x=-3$.
I know this requires two equations. The first one requires taking derivative of $f(x)$ and subbing local minimum. But how do I create second equation involving the oblique asymptote.
 A: We have
$$\frac{ax^2+bx-c}{x-4}=\overbrace{ax+b+4a}^\text{oblique asymptote}+\overbrace{\frac{16a+4b-c}{x-4}}^{\text{remainder}}$$ and therefore have $a=3$ and $b=-11.$
To calculate $c$, you mentioned that you have to take the derivative and sub in local minimum.  Hopefully you can take it from here.
A: Be aware that, given a $f(x)$ :

*

*one thing is to find its "asymptotic behaviour" as $x \to \infty$, which means to find a $g(x)$  such that
$$
\mathop {\lim }\limits_{x \to \infty } {{f(x)} \over {g(x)}} = 1
$$
and we say that $f$ and $g$ are asymptotically equivalent;


*another thing is to find the function $g(x)$ which is asymptotically equal to $f(x)$, i.e. such that
$$
\mathop {\lim }\limits_{x \to \infty } \left( {f(x) - g(x)} \right) = 0
$$
Now,  when we talk to find a horizontal or oblique asymptote to a curve we mean to find a line
which is asyptotic to the curve in the second meaning, i.e. that the distance goes to $0$.
In the first way (J. Wang's answer) you can only determine the slope of the line, but not the intercept because
$$
\mathop {\lim }\limits_{x \to \infty } {{f(x)} \over {mx}}
 = 0\quad  \Rightarrow \quad \mathop {\lim }\limits_{x \to \infty } {{f(x)} \over {mx + q}} = 0
$$
Therefore in your case we have
$$
\eqalign{
  & f(x) = {{ax^2  + bx - c} \over {x - 4}} =   \cr 
  &  = {{a\left( {x - 4} \right)^2  + b\left( {x - 4} \right) + 8ax - 16a + 4b - c} \over {x - 4}} =   \cr 
  &  = {{a\left( {x - 4} \right)^2  + b\left( {x - 4} \right) + 8a\left( {x - 4} \right) + 16a + 4b - c}
 \over {x - 4}} =   \cr 
  &  = a\left( {x - 4} \right) + b + 8a + {{16a + 4b - c} \over {x - 4}} \cr} 
$$
So for the asymptote we shall have
$$
ax + b + 4a = 3x + 1\quad  \Rightarrow \quad a = 3,\quad b =  - 11
$$
as already indicated by A.Chin.
Then the function becomes
$$
f(x) = {{3x^2  - 11x - c} \over {x - 4}} = 3x + 1 - {{c - 4} \over {x - 4}}
$$
It is clear that the function has the required oblique asymptote, as well as a vertical asymptote at $x=4$, where we have:
$$
\mathop {\lim }\limits_{x \to 4 + } f(x) = - {\rm sgn}(c - 4) \cdot \infty \quad
 \mathop {\lim }\limits_{x \to 4 - } f(x) =  {\rm sgn}(c - 4) \cdot \infty 
$$
This means that for $c < 4$ you will  have a minimum at $4 < x$ and a maximum at $x < 4$, while
for $4 < c$ you do not have any local extremum.
You can easily check this analytically.
Therefore a minimum at $x=-3$ is unattainable.
