Curve sketching

There is no universal way of going about sketching the graph of a function. For complicated functions, sketching the graph may be impractical; in such cases, computer plots are more helpful. In this section we outline some points that are useful when considering the shape of graphs.

1. Is the function $f$ even, odd, periodic?
2. Does the graph cross the $x$ or $y$ axes?
3. Does $f$ have stationary points? If so, what is their nature?
4. Does the graph have asymptotes (lines or curves the function approaches as $x \rightarrow \pm \infty$ or $y \rightarrow \pm \infty)$ ?

The figure shows the function $y=f(x)$. It has 3 stationary points at $x=-4,-1,1$. The point $\left(-4, y_{1}\right)$ is a local and global minimum, $\left(1, y_{2}\right)$ is a local minimum and $\left(-1, y_{3}\right)$ is a local and global maximum.

5. If $f^{\prime}(x)>0$ for every $x$ on some interval, $\mathcal{I}$ then $f(x)$ is increasing on the interval. Similarly, if $f^{\prime}(x)<0$ for every $x$ on $\mathcal{I}$ then $f(x)$ is decreasing on the interval. If $f^{\prime}(x)=0$ for every $x$ on $\mathcal{I}$ then $f(x)$ is constant on the interval.
6. Is the function concave up or down? If $f^{\prime \prime}(x)>0\left(f^{\prime \prime}(x)<0\right)$ for all $x$ on some interval $\mathcal{I}$ then $f(x)$ is concave up (concave down) on $\mathcal{I}$.

Let us consider a function $f(x)=x^{3}-3 x$ where we want to extract information on its graph. We first note that $f(-x)=-f(x)$ and hence the function is odd. The function has 3 roots which occur at $f(x)=0$ given at $x=0$ and $x= \pm \sqrt{3}$. Further, there are two stationary points at $x= \pm 1$ where $f^{\prime}(x)=0$ and, by application of the second derivative test, $f^{\prime \prime}(-1)<0$ so $x=-1$ is a maximum while $f^{\prime \prime}(1)>0$ and so $x=1$ is a minimum. Finally $f(x)$ tends to $\pm \infty$ as $x \rightarrow \pm \infty$. This gives us enough information to sketch the graph of $f(x)$:

Rational functions

Consider now a function $f(x)$ given as

where $P$ and $Q$ are polynomial functions. The graph of $f$ has vertical asymptotes if $Q$ has real zeros, assuming that $P$ does not have zeros at the same points. For instance, consider the odd function

The graph of $f$ has vertical asymptotes at $x= \pm 1$. For the case where the degree of $P(x)$ is greater than the degree of $Q(x)$ (like in our example), long division gives us a quotient, $q(x)$ and a remainder, $r(x)$ such that

provided that the degree of $r(x)$ is strictly less than that of $Q(x)$. It follows that $r(x) / Q(x) \rightarrow 0$ as $x \rightarrow \pm \infty$ and so $f(x)$ and $q(x)$ are close to each other. In our example, the degree of $P$ is one more than the degree of $Q$ and the quotient is a linear function known as the slant asymptote. To determine $q(x)$ for our example, we rewrite $f(x)$ as

Using the binomial expansion for the term in the denominator we have

Recall that in the binomial expansion given by $(1+x)^{n}$ with $n$ not a positive integer, the resulting series does not terminate and it converges for $-1<x<1$. In our case, we have $\left(1-1 / x^{2}\right)^{-1}$ so we would expect convergence in $-1<x^{-2}<1$. Note that $f(x) \approx q(x)$ when $x$ is large (and $x^{-2}$ is small). Substituting the series expansion in:

as $x \rightarrow \pm \infty, f(x) \approx q(x)=x$. Further, a Maclaurin expansion around $x=0$ gives $f(x)=-x^{3}+\mathcal{O}\left(x^{5}\right)$; it follows that near $x=0$ the graph of $f$ looks like $-x^{3}$.

The graph of the function $y=f(x)$ where $f(x)$ is given by $f(x)=\frac{x^{3}}{x^{2}-1}$. There are two vertical asymptotes at $x= \pm 1$ shown with grey, dashed lines and a slant asymptote given by $y=x$ shown with a red, dashed line. Near $x=0$, the function behaves like $-x^{3}$.