Single Variable Calculus, Chapter 4, 4.5, Section 4.5, Problem 10

Use the guidelines of curve sketching to sketch the curve. $\displaystyle y = \frac{x^2 - 4}{x^2 - 2x}$

The guidelines of Curve Sketching
By factoring, we can results $\displaystyle y = \frac{x^2 - 4}{x^2 - 2x} = \frac{(x+2)\cancel{(x-2)}}{x \cancel{(x-2)} } = \frac{x+2}{2}$ It means that there is a removal discontinuity at $x = -2$

A. Domain.
We know that $f(x)$ is a rational function that is defined everywhere except for the value of $x$ that would make its denominator equal to zero. In this case we have $x = 0$ and $x = -2$ (removable). Therefore the domain is $(-\infty, -2) \bigcup (-2,0) \bigcup(0,\infty)$


B. Intercepts.
Solving for $y$-intercept, when $x = 0$
$\displaystyle y = \frac{0+2}{0} =$ is undefined since 0 is not included in the domain
Solving for $x$-intercept, when $y = 0$
$\displaystyle 0 = \frac{x+2}{x}$

We have, $x = -2$. However, the function has a removal discontinuity at $x = -2$. It shows that we don't have intercept.


C. Symmetry.
The function is not symmetric to both $y$-axis and origin by using symmetry test.

D. Asymptotes.
For vertical asymptotes, we equate the denominator to 0, that is $x = 0$. For horizontal asymptotes, since we have equal coefficients of $x$ with numerator and denominator, we obtain $\displaystyle y = \frac{1}{1} = 1$

E. Intervals of Increase or Decrease.
If we take the derivative of $f(x)$, by using Quotient Rule...
$\displaystyle f'(x) = \frac{x(1) - (x + 2)(1)}{x^2} = \frac{-2}{x^2}$
When $f'(x) = 0$,
$\displaystyle 0 = \frac{-2}{x^2}$
The critical numbers do not exist.
If we divide the interval by its domain, we can determine the intervals of increase or decrease

$\begin{array}{|c|c|c|} \hline\\ \text{Interval} & f'(x) & f\\ \hline\\ x < -2 & - & \text{decreasing on } (-\infty, -2)\\ \hline\\ -2 < x < 0 & - & \text{decreasing on } (-2 , 0 )\\ \hline\\ x > 0 & - & \text{decreasing on } (0,\infty)\\ \hline \end{array}$



F. Local Maximum and Minimum Values.
Since $f'(x)$ doesn't change sign, we can say that the function has no local maximum and minimum.

G. Concavity and Points of Inflection.

$\begin{equation} \begin{aligned} \text{if } f'(x) &= \frac{-2}{x^2} , \text{ then}\\ \\ f''(x) &= \frac{x^2(0)-(-2)(2x)}{(x^2)^2} = \frac{4}{3}\\ \\ \\ \text{when } f''(x) &= 0\\ \\ 0 &= \frac{4}{x^3} \end{aligned} \end{equation}$

It means that we have no inflection points.
If we divide the interval within the domain, we can determine the concavity as...

$\begin{array}{|c|c|c|} \hline\\ \text{Interval} & f''(x) & \text{Concavity}\\ \hline\\ x < -2 & - & \text{Downward}\\ \hline\\ -2 < x < 0 & - & \text{Downward}\\ \hline\\ x > 0 & + & \text{Upward}\\ \hline \end{array}$




H. Sketch the Graph.