Single Variable Calculus, Chapter 4, 4.5, Section 4.5, Problem 48

Use the guidelines of curve sketching to sketch the curve $\displaystyle y = \frac{x^2 + 12}{x - 2}$ then find the equation of slant asymptote.

A. Domain. We know that $f(x)$ is a rational function that is defined for all values of $x$ except for the values that would make its denominator equal to zero. In our case, it's $x = 2$. Therefore, the domain is $(- \infty, 2) \bigcup(2, \infty)$

B. Intercepts. Solving for $y$ intercept, when $x = 0$,

$\displaystyle y = \frac{0^2 + 12}{0 - 2} = -6$

Solving for $x$ intercept, when $y = 0$,


$
\begin{equation}
\begin{aligned}

0 =& \frac{x^2 + 12}{x - 2}
\\
\\
0 =& x^2 + 12

\end{aligned}
\end{equation}
$


$x$ intercept does not exist

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

D. Asymptote. For vertical asymptote, we set denominator equal to zero

that is, $x - 2 = 0$

Therefore, the vertical asymptote is $x = 2$.

For horizontal asymptote,

Since $\lim_{x \to \pm \infty} f(x) = \pm \infty$, the function has no horizontal asymptote

For slant asymptote, by using long division,







So we can rewrite $\displaystyle y = \frac{x^2 + 12}{x - 2}$ as $\displaystyle y = x + 2 + \frac{16}{x - 2}$

$\displaystyle \lim_{x \to \pm \infty} f(x) - (x + 2) = \frac{16}{x - 2} = \frac{\displaystyle \frac{16}{x}}{\displaystyle 1 - \frac{2}{x}} = 0$

Therefore, the equation of slant asymptote is $y = x + 2$.

E. Intervals of increase or decrease,

If $\displaystyle f(x) = \frac{x^2 + 12}{x - 2}$, then by using Quotient Rule,


$
\begin{equation}
\begin{aligned}

f'(x) =& \frac{(x - 2)(2x) - (x^2 + 12)(1)}{(x - 2)^2}
\\
\\
f'(x) =& \frac{2x^2 - 4x - x^2 - 12}{(x - 2)^2} = \frac{x^2 - 4x - 12}{(x - 2)^2}

\end{aligned}
\end{equation}
$


When $f'(x) = 0$


$
\begin{equation}
\begin{aligned}

0 =& \frac{x^2 - 4x - 12}{(x - 2)^2}
\\
\\
0 =& x^2 - 4x - 12

\end{aligned}
\end{equation}
$


By using Quadratic Formula, the critical numbers are

$x = 6$ and $x = -2$

Hence, the intervals of increase and decrease are..

$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f'(x) & f \\
x < -2 & + & \text{increasing on } (- \infty, -2) \\
-2 < x < 2 & - & \text{decreasing on } (-2, 2) \\
2 < x < 6 & - & \text{decreasing on (2, 6)} \\
x > 6 & + & \text{increasing on } (6, \infty) \\
\hline

\end{array}
$


F. Local Maximum and Minimum Values

Since $f'(x)$ changes from positive to negative at $x = -2, f(-2) = -4$ is a local maximum. On the other hand, since $f'(x)$ changes from negative to positive at $x = -6, f(6) = 13$ is a local minimum.

G. Concavity and inflection point

If $f'(x) = \displaystyle \frac{x^2 - 4x - 12}{(x - 2)^2}$, then by using Quotient Rule and Chain Rule,


$
\begin{equation}
\begin{aligned}

f''(x) =& \frac{(x - 2)^2 (2x - 4) - (x^2 - 4x - 12) (2(x - 2)(1))}{[(x - 2)^2]^2}
\\
\\
f''(x) =& \frac{(x -2 ) [(x - 2) (2x - 4) - 2 (x^2 - 4x - 12)]}{(x - 2)^4}
\\
\\
f''(x) =& \frac{\cancel{2x^2} - \cancel{4x} - \cancel{4x} + 8 - \cancel{2x^2} + \cancel{8x} + 24 }{(x - 2)^3}
\\
\\
f''(x) =& \frac{32}{(x - 2)^3}

\end{aligned}
\end{equation}
$


when $f''(x) = 0,$

$\displaystyle 0 = \frac{32}{(x - 2)^3}$

$f''(x)= 0$ does not exist, therefore the function has no inflection point.

So the concavity is..

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

H. Sketch the graph.