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 $x^2-9=(x-3)(x+3)$ is a parabola with zeros at $x=\pm 3$:

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

Squaring gives a quartic with double zeros at $x=\pm 3$:

$$\epsffile{graphsx12.eps}$$

Dividing by $x$ will produce a simple pole at $x=0$ and also a sign change at negative $x$:

$$\epsffile{graphsx13.eps}$$

Function $y(x)$:

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has an $x$-extent $x\ne 0$ and a $y$-extend $-\infty < y < \infty$;

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is odd (symmetric with respect to the origin);

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has a relative maximum at -3 of finite curvature: $y\propto (x+3)^2$;

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has a relative minimum at 3 of finite curvature: $y\propto (x-3)^2$;

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has a vertical asymptote at $x=0$ with asymptotic behavior: $y \sim 81/x$ for $\vert x\vert\to 0$;

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behaves asymptotically as $y \sim x^3$ for $x\to\pm\infty$;

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is concave up for $x>0$, down for $x<0$. (Should really prove this, I guess.)

1.2.2 Using brute force

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

Hence

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intercepts with $x$-axis are at $x=\pm3$;

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no intercepts with the $y$ axis;

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$y$ is an odd function of $x$ (symmetric about the origin);

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for $x\downarrow 0$, $y \to \infty$ (vertical asymptote);

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for $x\uparrow 0$, $y \to -\infty$ (singularity is an odd, simple pole);

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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,

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$y'>0$ for $-\infty< x<-3$ ($y$ increases from $-\infty$);

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$y'=0$ for $x=-3$ (local maximum, $y=0$);

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$y'<0$ for $-3<x<0$ ($y$ decreases towards $-\infty$);

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$y'=-\infty$ for $x=0$ (singular point, vertical asymptote);

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$y'<0$ for $0<x<-3$ (decreases from $\infty$);

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$y'=0$ for $x=3$ (local minimum, $y=0$);

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$y'>0$ for $3<x<\infty$ (increases to $\infty$).

Also,

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$y'\to \infty$ when $x\to \pm \infty$ (no horizontal or oblique asymptotes);

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all derivatives exist, except at $x=0$, which has no point on the curve (no corners, cusps, infinite curvature, or other singular points);

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probably no inflection points.

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

Hence

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really no inflection points (since there is no point at $x=0$);

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cocave downward for $x<0$, upward for $x>0$.

$$\epsffile{graphsx13.eps}$$

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