I have been asked to sketch the graph $y =\frac{x^3+1}{x^2}$ I need to be able to show all point of interest, stationary point, asymptotes and where it meets the $x$ and $y$ axis. And be able to show the working.
 A: We give a partial answer that does not consider derivatives. Note that our function simplifies to $x+\frac{1}{x^2}$. 
For $x$ close to $0$, our function is very large positive. Lightly draw $y$ very big positive on both sides of the line $x=0$ (the $y$-axis). The curve is asymptotic to the $y$-axis. 
For $x$ large positive, $x+\frac{1}{x^2}$ is large, and very close to $x$. Draw the line $y=x$. For $x$ large, $y=x+\frac{1}{x^2}$ is a tiny bit above the line $y=x$. Draw it doing that. 
Similarly, for $x$ large negative, $y=x+\frac{1}{x^2}$ is a tiny bit above the line $y=x$. That line is an (oblique) asymptote to the curve. Draw. 
Once we have sketched in what happens near $0$ and when $|x|$ is large, we have a pretty good idea of the basic behaviour of the function. You can probably visually fill in what the curve should look like. 
For finer detail, we can now turn to the derivative, and the second derivative. Little mechanical errors, and sign errors, are quite possible at this stage. The preliminary sketch we made before computing may serve to detect such errors. 
A: 
first find domain ,assymptote ,roots(optional)
then calculate f'
then determine the sign of f'
and go on
A: The function is defined everywhere except at $0$, where it has a vertical asymptote. It is continuous on its whole domain. It has an oblique asymptote (in both directions, being a rational function), because
$$
\lim_{x\to\infty}\frac{f(x)}{x}=\dots,
\qquad
\lim_{x\to\infty}f(x)-\dots x=\dots
$$
It can't intersect the $y$-axis and intersects the $x$-axis only at $\dots$.
The derivative is
$$
f'(x)=\left(x+\frac{1}{x^2}\right)'=1-\frac{2}{x^3}=\frac{x^3-2}{x^3}
$$
that's quite easy to study.
The second derivative is $f''(x)=6/x^4$ that's everywhere positive.
