Following tools might help you to draw the graph for the given function,
$f(x) = \frac{x^3 – 1}{x^2 – 1}$
Here, $f(x)$ is undefined for $\pm 1.$ Hence domain for the given function is $(-\infty,-1) \cup (-1,1) \cup (1,\infty)$
Now, we try to find out for what interval, function is increasing or decreasing and for this, sign of the derivative of $f(x)$ would help us.
$f’(x) = \frac{x(x+2)}{(x+1)^2}$ here denominator is always $>0$ for the given domain. And so, sign of $f’(x)$ is decided by $x(x+2).$ Here, $f’(x)$ is negative for interval $(-2,0)$ and positive for $(-\infty,-2) \cup (0,\infty)$ (You can verify it by making parabola for $x(x+2)$ and check when it is above x axis and when it is below.
It means $f(x)$ increases for the interval $(-\infty,-2) \cup (0,\infty)$ and decreases for the interval $(-2,0)$
Now, try to find out asymptotes.
$\lim_{x \rightarrow -1^+} \frac{x^3 - 1}{x^2 -1} = \infty$ and
$\lim_{x \rightarrow -1^-} \frac{x^3 - 1}{x^2 -1} = -\infty$
$\lim_{x \rightarrow \infty} \frac{x^3 - 1}{x^2 -1} = \infty$
$\lim_{x \rightarrow -\infty} \frac{x^3 - 1}{x^2 -1} = -\infty$
Now, here, $f’(x) = \frac{x(x+2)}{(x+1)^2}$ which has roots for $x=0,-2$ and so, local minima is for $x=0$ and local maxima is for $x=-2$ and $f(0)=1$ and $f(-2) = -3$ and hence, range for $f(x)$ is $(-\infty, -3] \cup [1,\infty)$
You might check concavity also and can try other things also but if you follow the above points, you plot the given function $f(x)$ which you can verify here
