Chapter 3.7, Problem 9E

(a)

To determine

To complete the tables.

Answer to Problem 9E

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 $r\left(x\right)=\frac{3x-10}{{\left(x-2\right)}^2}$ .

Calculation:

To complete each table, substitute, each value of $x$ on $r\left(x\right)=\frac{3x-10}{{\left(x-2\right)}^2}$ to determine the value of $r\left(x\right)$ ,

Therefore, the tables are obtained as:

Table 1

$x$	$r\left(x\right)$
1.5	$r\left(x\right)=\frac{3\left(1.5\right)-10}{{\left(1.5-2\right)}^2}=\frac{4.5-10}{{\left(-0.5\right)}^2}=\frac{-5.5}{0.25}=-22$
1.9	$-430$
1.99	$-40300$
1.999	$-4003000$

Table 2

$x$	$r\left(x\right)$
2.5	$r\left(x\right)=\frac{3\left(2.5\right)-10}{{\left(2.5-2\right)}^2}=\frac{7.5-10}{{\left(0.5\right)}^2}=\frac{-2.5}{0.25}=-10$
2.1	$-370$
2.01	$-39700$
2.001	$-3997000$

Table 3

$x$	$r\left(x\right)$
10	$r\left(x\right)=\frac{3\left(10\right)-10}{{\left(10-2\right)}^2}=\frac{20}{{\left(8\right)}^2}=\frac{20}{64}=0.3125$
50	$0.06076$
100	$0.030196$
1000	$0.003002$

Table 4

$x$	$r\left(x\right)$
$-10$	$r\left(x\right)=\frac{3\left(-10\right)-10}{{\left(-10-2\right)}^2}=\frac{-30-10}{{\left(-12\right)}^2}=\frac{-40}{144}=0.27778$
$-50$	$-0.05917$
$-100$	$-0.0298$
$-1000$	$-0.003$

(b)

To determine

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

Answer to Problem 9E

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 $r\left(x\right)=\frac{3x-10}{{\left(x-2\right)}^2}$ .

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	$r\left(x\right)=\frac{3\left(1.5\right)-10}{{\left(1.5-2\right)}^2}=\frac{4.5-10}{{\left(-0.5\right)}^2}=\frac{-5.5}{0.25}=-22$
1.9	$-430$
1.99	$-40300$
1.999	$-4003000$

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

$x$	$r\left(x\right)$
2.5	$r\left(x\right)=\frac{3\left(2.5\right)-10}{{\left(2.5-2\right)}^2}=\frac{7.5-10}{{\left(0.5\right)}^2}=\frac{-2.5}{0.25}=-10$
2.1	$-370$
2.01	$-39700$
2.001	$-3997000$

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 9E

The horizontal asymptote is the line $y=0$ .

Explanation of Solution

Given information:

The rational function is $r\left(x\right)=\frac{3x-10}{{\left(x-2\right)}^2}$ .

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

Table 3

$x$	$r\left(x\right)$
10	$r\left(x\right)=\frac{3\left(10\right)-10}{{\left(10-2\right)}^2}=\frac{20}{{\left(8\right)}^2}=\frac{20}{64}=0.3125$
50	$0.06076$
100	$0.030196$
1000	$0.003002$

Table 4

$x$	$r\left(x\right)$
$-10$	$r\left(x\right)=\frac{3\left(-10\right)-10}{{\left(-10-2\right)}^2}=\frac{-30-10}{{\left(-12\right)}^2}=\frac{-40}{144}=0.27778$
$-50$	$-0.05917$
$-100$	$-0.0298$
$-1000$	$-0.003$

From the above table shown above, it can be concluded that as $x\to +\infty$ , $r\to 0^{+}$ and as $x\to -\infty$ , $r\to 0^{-}$ .

Hence,

The horizontal asymptote is the line $y=0$ .