Domain, Intercepts and Asymptotes

Given a function \(f(x)\text{,}\) there are several important features that we can determine from that expression before examining its derivatives.

The domain of the function — take note of values where \(f\) does not exist. If the function is rational, look for where the denominator is zero. Similarly be careful to look for roots of negative numbers or other possible sources of discontinuities.
Intercepts — examine where the function crosses the \(x\)-axis and the \(y\)-axis by solving \(f(x)=0\) and computing \(f(0)\text{.}\)
Vertical asymptotes — look for values of \(x\) at which \(f(x)\) blows up. If \(f(x)\) approaches either \(+\infty\) or \(-\infty\) as \(x\) approaches \(a\) (or possibly as \(x\) approaches \(a\) from one side) then \(x=a\) is a vertical asymptote to \(y=f(x)\text{.}\) When \(f(x)\) is a rational function (written so that common factors are cancelled), then \(y=f(x)\) has vertical asymptotes at the zeroes of the denominator.
Horizontal asymptotes — examine the limits of \(f(x)\) as \(x\to+\infty\) and \(x\to-\infty\text{.}\) Often \(f(x)\) will tend to \(+\infty\) or to \(-\infty\) or to a finite limit \(L\text{.}\) If, for example, \(\lim\limits_{x\rightarrow+\infty}f(x)=L\text{,}\) then \(y=L\) is a horizontal asymptote to \(y=f(x)\) as \(x\rightarrow\infty\text{.}\)

Consider the function

\begin{align*} f(x) &= \frac{x+1}{(x+3)(x-2)} \end{align*}

We see that it is defined on all real numbers except \(x=-3,+2\text{.}\)
Since \(f(0)=-1/6\) and \(f(x)=0\) only when \(x=-1\text{,}\) the graph has \(y\)-intercept \((0,-1/6)\) and \(x\)-intercept \((-1,0)\text{.}\)
Since the function is rational and its denominator is zero at \(x=-3,+2\) it will have vertical asymptotes at \(x=-3,+2\text{.}\) To determine the shape around those asymptotes we need to examine the limits
\begin{align*} \lim_{x\to -3} f(x) && \lim_{x\to2} f(x) \end{align*}
Notice that when \(x\) is close to \(-3\text{,}\) the factors \((x+1)\) and \((x-2)\) are both negative, so the sign of \(f(x) = \frac{x+1}{x-2} \cdot \frac{1}{x+3}\) is the same as the sign of \(x+3\text{.}\) Hence
\begin{align*} \lim_{x\to -3^+} f(x) &= +\infty & \lim_{x\to -3^-} f(x) &= -\infty \end{align*}
A similar analysis when \(x\) is near \(2\) gives
\begin{align*} \lim_{x\to 2^+} f(x) &= +\infty & \lim_{x\to 2^-} f(x) &= -\infty \end{align*}
Finally since the numerator has degree 1 and the denominator has degree 2, we see that as \(x \to \pm \infty\text{,}\) \(f(x) \to 0\text{.}\) So \(y=0\) is a horizontal asymptote.
Since we know the behaviour around the asymptotes and we know the locations of the intercepts (as shown in the left graph below), we can then join up the pieces and smooth them out to get the a good sketch of this function (below right).