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Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 4: Q. 15 (page 234)

In Problems $7 - 44$ , follow Steps $1$ through $7$ on page $228$ to analyze the graph of each function.
$H (x) = \frac{x^{3} - 1}{x^{2} - 9}$

Short Answer

Expert verified

The graph of the function $H (x) = \frac{x^{3} - 1}{x^{2} - 9}$ is,

Step by step solution

01

Step 1 Given function is,

$H (x) = \frac{x^{3} - 1}{x^{2} - 9}$

which may also written as,

$H (x) = \frac{(x - 1) (x^{2} + x + 1)}{(x + 3) (x - 3)}$

The domain of $H$ is $\{x | x 鈮� - 3, x 鈮� 3\}$

So, there will be domain in $0$ .

Substitute $x = 0$ in the given function.

$H (0) = \frac{0 - 1}{0 - 9} H (0) = \frac{1}{9}$

So, the obtained point is, $(0, \frac{1}{9})$ .

02

Step 2 The function H is in lower terms.

The real zeros of the denominator are the real solutions of the equation.

$(x + 3) (x - 3) = 0 x = 3, - 3$

The two lines $x = 3, x = - 3$ are two vertical asymptotes of the graph.

Graph these asymptotes using the dashed lines.

03

Step 3 In the given function degree of the numerator is greater than the degree of the denominator.

Use the method of long division to find the horizontal asymptote.

Therefore, $H (x) = x + \frac{9 x - 1}{(x + 3) (x - 3)}$ .

The quotient is $x$ . So, the line $y = x$ is an oblique asymptote of the graph. Indicate the line $y = x$ on the graph using a dashed line.

04

Step 4 Substitute Hx=x to determine whether the graph intersects at the asymptote y=x.

$x = \frac{x^{3} - 1}{x^{2} - 9} x (x^{2} - 9) = x^{3} - 1 x^{3} - 9 x = x^{3} - 1 - 9 x = - 1 x = \frac{1}{9}$

As $y = x$ , $y = \frac{1}{9}$ .

So, the point of intersection is, $(\frac{1}{9}, \frac{1}{9})$ .

05

Step 5 Now plot a graph by using the known values.

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