Chapter 3.7, Problem 10E

(a)

To determine

To complete the tables.

Answer to Problem 10E

The completed tables are as follows:

Table 1

$x$	$r\left(x\right)$
1.5	$r\left(x\right)=\frac{4\left(1.5\right)+1}{1.5-2}=-\frac{6+1}{0.5}=-\frac{7}{0.5}=-14$
1.9	$-86$
1.99	$-896$
1.999	$-8996$

Table 2

$x$	$r\left(x\right)$
2.5	$r\left(x\right)=\frac{4\left(2.5\right)+1}{2.5-2}=\frac{11}{0.5}=22$
2.1	$94$
2.01	$904$
2.001	$9004$

Table 3

$x$	$r\left(x\right)$
10	$r\left(x\right)=\frac{4\left(10\right)+1}{10-2}=\frac{41}{8}=5.125$
50	$4.1875$
100	$4.0918$
1000	$4.00902$

Table 4

$x$	$r\left(x\right)$
$-10$	$r\left(x\right)=\frac{4\left(-10\right)+1}{-10-2}=\frac{-39}{-12}=3.25$
$-50$	$3.8269$
$-100$	$3.9117$
$-1000$	$3.99102$

Explanation of Solution

Given information:

The rational function is $r\left(x\right)=\frac{3x^2+1}{{\left(x-2\right)}^2}$ .

Calculation:

To complete each table, substitute, each value of $x$ on $r\left(x\right)=\frac{3x^2+1}{{\left(x-2\right)}^2}$ to determine the value of $r\left(x\right)$ ,

Therefore, the tables are obtained as:

Table 1

$x$	$r\left(x\right)$
1.5	$r\left(x\right)=\frac{3{\left(1.5\right)}^2+1}{{\left(1.5-2\right)}^2}=\frac{3\left(2.25\right)+1}{{\left(-0.5\right)}^2}=\frac{7.75}{0.25}=31$
1.9	$1183$
1.99	$128803$
1.999	$12988003$

Table 2

$x$	$r\left(x\right)$
2.5	$79$
2.1	$1423$
2.01	$131203$
2.001	$13012003$

Table 3

$x$	$r\left(x\right)$
10	$4.703125$
50	$3.255642$
100	$3.123803$
1000	$3.012037$

Table 4

$x$	$r\left(x\right)$
$-10$	$2.090278$
$-50$	$2.774038$
$-100$	$2.883602$
$-1000$	$2.988037$

(b)

To determine

To describe the behaviour of the function near its vertical asymptote.

Answer to Problem 10E

The behaviour of the function near its vertical asymptote is as $x\to 2^{+}$ , $r\to \infty$ and as $x\to 2^{-}$ , $r\to \infty$ .

Explanation of Solution

Given information:

The rational function is $r\left(x\right)=\frac{3x^2+1}{{\left(x-2\right)}^2}$ .

The $x=2$ is a vertical asymptote because the denominator of $r$ is zero when $x=2$ .

Use table 1 and table 2 to identify the behaviour of the function for $x$ -values to the left and right of 2.

Table 1$r\to \infty$ as $x\to 2^{-}$

$x$	$r\left(x\right)$
1.5	$31$
1.9	$1183$
1.99	$128803$
1.999	$12988003$

Table 2$r\to \infty$ as $x\to 2^{+}$

$x$	$r\left(x\right)$
2.5	$79$
2.1	$1423$
2.01	$131203$
2.001	$13012003$

From the above table shown above, it can be concluded that as $x\to 2^{+}$ , $r\to \infty$ and as $x\to 2^{-}$ , $r\to \infty$ .

Hence,

The behaviour of the function near its vertical asymptote is as $x\to 2^{+}$ , $r\to \infty$ and as $x\to 2^{-}$ , $r\to \infty$ .

(c)

To determine

To find the horizontal asymptote.

Answer to Problem 10E

The horizontal asymptote is the line $y=3$ .

Explanation of Solution

Given information:

The rational function is $r\left(x\right)=\frac{3x^2+1}{{\left(x-2\right)}^2}$ .

The horizontal asymptote is the value that $y$ approaches as $x\to \pm \infty$

Table 3

$x$	$r\left(x\right)$
10	$4.703125$
50	$3.255642$
100	$3.123803$
1000	$3.012037$

Table 4

$x$	$r\left(x\right)$
$-10$	$2.090278$
$-50$	$2.774038$
$-100$	$2.883602$
$-1000$	$2.988037$

From the above table shown above, it can be concluded that as $x\to +\infty$ , $r\to 3^{+}$ and as $x\to -\infty$ , $r\to 3^{-}$ .

Hence,

The horizontal asymptote is the line $y=3$ .