Functions can ‘get big’ in different ways.
When they ‘get big’ by looking more and more like a ‘slanted’ line (i.e.,
not horizontal and not vertical),
then the function is said to have a slant asymptote.
Slant asymptotes are also called oblique asymptotes.
As discussed in Introduction to Asymptotes,
an asymptote is a
curve (usually a line)
that another curve gets arbitrarily close to
as $\,x\,$ approaches $\,+\infty\,$, $\,\infty\,$, or a finite number.
Here's how this general definition is ‘specialized’ to get a
slant asymptote:
A ‘slant asymptote’ is a nonvertical, nonhorizontal line
that another curve gets arbitrarily close to
as $\,x\,$ approaches $\,+\infty\,$ or $\,\infty\,$.
What's the key to a slant asymptote situation?
The outputs are behaving more and more like $\,y = mx + b\,$ for $\,m\ne 0\,$
as inputs get arbitrarily large (big and positive, or big and negative).
This section gives a precise discussion of slant asymptotes,
as well as
conditions under which a rational function has a
slant asymptote.

The red line is a slant asymptote
for the blue curve.
As $\,x\rightarrow\infty\,$,
the blue curve approaches the red line.

Conditions under which
$\,y = mx + b\,$ ($\,m\ne 0\,$)
is a Slant Asymptote for a Function $\,f\,$
Let $\,f\,$ be a nonlinear function.
That is, $\,f(x)\,$ is
not of the form $\,f(x) = cx + d\,$ for any real numbers $\,c\,$ and $\,d\,$.
Let $\,m \ne 0\,$.
The line $\,y = mx + b\,$ is a
slant asymptote for $\,f\,$
if and only if at least one of
the following conditions holds:
 as $\,x\rightarrow\infty\,$, $\,f(x)  (mx + b) \rightarrow 0\,$
 as $\,x\rightarrow \infty\,$, $\,f(x)  (mx + b) \rightarrow 0\,$
That is, the vertical distance between the line $\,y = mx + b\,$ and the graph of $\,f\,$ approaches zero as $\,x\rightarrow\infty\,$
or $\,x\rightarrow \infty\,$ (or both).

A function can have at most two slant asymptotes.
It can approach one line as $\,x\rightarrow\infty\,$, and a different line as
$\,x\rightarrow \infty\,$.

A function can have a slant asymptote and a horizontal asymptote.
For example, it might approach a horizontal line as $\,x\rightarrow\infty\,$,
and a slanted line as $\,x\rightarrow \infty\,$.
Example:
A NonRational Function with a Slant Asymptote
Let $\,f(x) = 3x + \text{e}^{x}\,$ (the blue curve at right).
As $\,x\rightarrow \infty\,$, we have $\,\text{e}^{x}\rightarrow 0\,$, so the outputs from $\,f\,$ are looking more and more like $\,3x\,$.
More precisely, as $\,x\rightarrow \infty\,$, we have:
$$
\begin{gather}
f(x)  3x \ = \ (3x + \text{e}^{x})  3x \ = \ \text{e}^{x}\cr
\text{and}\cr
\text{e}^{x}\rightarrow 0
\end{gather}
$$
Thus, $\,y = 3x\,$ (the red line at right) is a slant asymptote for $\,f\,$.


Example:
A Rational Function with a Slant Asymptote;
degree of numerator is exactly one greater than degree of denominator
Let $\displaystyle\,R(x) = \frac{x^2  4x  5}{x  3}\,$.
When $\,x\,$ is big, the outputs from $\,R(x)\,$ looks like $\,\frac{\text{big}}{\text{big}}\,$, which is not very useful.
As usual, we will rename the function to better understand what happens when $\,x\,$ is big:
The key is to do a long division:
 $x$  $$  $1$   
$x3$  $x^2$  $$  $4x$  $$  $5$ 
 $(x^2$  $$  $3x$  $)$  
 
   $x$  $$  $5$ 
  $($  $x$  $+$  $3)$ 
  
     $8$ 
Thus,
$$
R(x) = \frac{x^2  4x  5}{x  3} \ \ =\ \ x  1 \ \ \ + \overbrace{\left(\frac{8}{x3}\right)}^{\text{this part tends to zero}}
$$
Note that as $\,x\rightarrow\pm\infty\,$, $\displaystyle\,\frac{8}{x3}\rightarrow 0\,$.
Therefore, as inputs get big, the function is looking more and more like
the line $\,y = x  1\,$.
Thus, $\,y = x  1\,$ is a slant asymptote for the function $\,R\,$.
The graph of the function $\,\color{blue}{R}\,$ (in blue), together with its slant asymptote
$\,\color{red}{y = x  1}\,$ (in red), is shown at right.
When we were doing the long division of $\,x^2  4x  5\,$ by $\,x  3\,$ up above,
what caused the $\,x = x^1\,$ to appear in the quotient $\,x  1\,$?
It was the fact that the degree of the numerator is one more than the degree of the denominator!
This is the key ingredient for determining if a rational function has a slant asymptote.
(There are a couple other technical requirements, which are stated below.)


In summary, we have:
Conditions under which
a Rational Function has a Slant Asymptote
Let $\,N(x)\,$ and $\,D(x)\,$ be polynomials, so that
$\displaystyle\,R(x) = \frac{N(x)}{D(x)}\,$ is a rational function.
The function $\,R\,$ has a slant asymptote when the following conditions are
met:

$\deg{N(x)} = \deg{D(x)} + 1$
(the degree of the numerator is exactly one more than the degree of the denominator)

$\deg{N(x)} \ge 2$
(the numerator is at least quadratic)
 when dividing $\,D(x)\,$ into $\,N(x)\,$, the remainder is not zero

Why do we require that $\deg{N(x)}\ge 2\,$?
Notice that $\displaystyle \,y = \frac{x1}{3}\,$ is a rational function, since both numerator and denominator are polynomials.
The degree of the numerator is $\,1\,$.
The degree of the denominator is $\,0\,$.
Thus, the degree of the numerator
is exactly one more than the degree of the denominator.
However, this is just the linear function $\,y = \frac 13x  \frac 13\,$, which does not have a slant asymptote.

Why do we require the remainder to be nonzero?
Notice that $\displaystyle \,y = \frac{x^21}{x1}\,$ is a rational function, and the degree of the numerator is exactly one more than the degree of the denominator.
However, in this case $\,x  1\,$ goes into $\,x^2  1\,$ evenly (the remainder is zero):
$$ y = \frac{x^21}{x1} = \frac{(x1)(x+1)}{x1} = x + 1\,,\ \ \text{for $\,x\ne 1\,$}
$$
This function graphs as a line that is
punctured at $\,x = 1\,$. It does not have a slant asymptote.