1.2 Example

From [1, p. 128, 13a]

Asked: Draw the graph of

$$xy=\left(x^2-9\right)^2$$

1.2.1 Using reasoning

$$xy=\left(x^2-9\right)^2$$

Instead of starting to crunch numbers, look at the pieces first:

Factor $x^2-9$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(x-3)(x+3)$ is a parabola with zeros at $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\pm 3$:

$$\begin{array}{c} \epsffile{graphsx11.eps} \end{array}$$

Squaring gives a quartic with double zeros at $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\pm 3$:

$$\epsffile{graphsx12.eps}$$

Dividing by $x$ will produce a simple pole at $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 and also a sign change at negative $x$:

$$\epsffile{graphsx13.eps}$$

Function $y(x)$:

• has an $x$-extent $x\ne 0$ and a $y$-extend $-\infty$ $\raisebox{.3pt}{$<$}$ $y$ $\raisebox{.3pt}{$<$}$ $\infty$;
• is odd (symmetric with respect to the origin);
• has a relative maximum at -3 of finite curvature: $y\propto (x+3)^2$;
• has a relative minimum at 3 of finite curvature: $y\propto (x-3)^2$;
• has a vertical asymptote at $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 with asymptotic behavior: $y \sim 81/x$ for $\vert x\vert\to 0$;
• behaves asymptotically as $y \sim x^3$ for $x\to\pm\infty$;
• is concave up for $x$ $\raisebox{.3pt}{$>$}$ 0, down for $x$ $\raisebox{.3pt}{$<$}$ 0. (Should really prove this, I guess.)

1.2.2 Using brute force

$$y=\frac{\left(x^2-9\right)^2}{x}$$

Hence

• intercepts with $x$-axis are at $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\pm3$;
• no intercepts with the $y$ axis;
• $y$ is an odd function of $x$ (symmetric about the origin);
• for $x\downarrow 0$, $y \to \infty$ (vertical asymptote);
• for $x\uparrow 0$, $y \to -\infty$ (singularity is an odd, simple pole);
• for $x\to \pm \infty$, $y \sim x^3 \to \pm \infty$.

$$y' \equiv\frac{{\rm d}y}{{\rm d}x}=\frac{\left(x^2-9\right)\left(3x^2+9\right)}{x^2}$$

Hence,

• $y'$ $\raisebox{.3pt}{$>$}$ 0 for $-\infty$ $\raisebox{.3pt}{$<$}$ $x$ $\raisebox{.3pt}{$<$}$ $-3$ ($y$ increases from $-\infty$);
• $y'$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 for $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-3$ (local maximum, $y$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0);
• $y'$ $\raisebox{.3pt}{$<$}$ 0 for $-3$ $\raisebox{.3pt}{$<$}$ $x$ $\raisebox{.3pt}{$<$}$ 0 ($y$ decreases towards $-\infty$);
• $y'$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-\infty$ for $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 (singular point, vertical asymptote);
• $y'$ $\raisebox{.3pt}{$<$}$ 0 for 0 $\raisebox{.3pt}{$<$}$ $x$ $\raisebox{.3pt}{$<$}$ $-3$ (decreases from $\infty$);
• $y'$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 for $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ 3 (local minimum, $y$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0);
• $y'$ $\raisebox{.3pt}{$>$}$ 0 for 3 $\raisebox{.3pt}{$<$}$ $x$ $\raisebox{.3pt}{$<$}$ $\infty$ (increases to $\infty$).

Also,

• $y'\to \infty$ when $x\to \pm \infty$ (no horizontal or oblique asymptotes);
• all derivatives exist, except at $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, which has no point on the curve (no corners, cusps, infinite curvature, or other singular points);
• probably no inflection points.

$$y'' = \frac{6x^4+162}{x^3}$$

Hence

• really no inflection points (since there is no point at $x$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0);
• cocave downward for $x$ $\raisebox{.3pt}{$<$}$ 0, upward for $x$ $\raisebox{.3pt}{$>$}$ 0.

$$\epsffile{graphsx13.eps}$$

Hence the $x$- and $y$-extends are as before.