Exercise 55 Page 969

Practice makes perfect

a We will graph the parent reciprocal function

f (x) = \frac{1}{x} .

To do so, we will first make a table of values to see the positive and negative values of

x .

$f (x) = \frac{1}{x}$
$x$	$\frac{1}{x}$	$(x, y)$
$- 4$	$\frac{1}{- 4}$	$(- 4, - 0.25)$
$- 1$	$\frac{1}{- 1}$	$(- 1, - 1)$
$- 0.25$	$\frac{1}{- 0 . 2 5}$	$(- 0.25, - 4)$
$0$	$\frac{1}{0}$	Undefined
$0.25$	$\frac{1}{0 . 2 5}$	$(0.25, 4)$
$1$	$\frac{1}{1}$	$(1, 1)$
$4$	$\frac{1}{4}$	$(4, 0.25)$

Notice that $f (x) = \frac{1}{x}$ is defined for all real numbers except $x = 0 .$ Therefore, there is no $y -$ intercept. Moreover, since the numerator is never $0,$ there is no $x -$ intercept. Knowing this, let's connect the points with a smooth curve.

Graph of f(x)=1/x. Points on the graph for x=-4,-1,-0.25,0.25,1,4 are shown.

b We will examine the graph of

g (x) = \frac{4}{x + 2}

by comparing the graph of

f (x) = \frac{1}{x} .

To do so, let's remember the general form of the reciprocal function family.

y = \frac{a}{x - h} + k

In this form, the value of

a

refers to stretches, shrinks, or reflections of the parent function

f (x) = \frac{1}{x} .

Moreover, the value of

h

represents the horizontal translation and

k

represents the vertical translation. With this in mind, let's examine

g (x) .

g (x) = ⇕ g (x) = \frac{4}{x + 2} \frac{4}{x - ( - 2 )} + 0

As we can see,

g (x)

is a stretch of the graph of

f (x)

by a factor of

4

and it is followed by a translation

2

units to the

l e f t .

Let's see them on the same coordinate plane.

c We are asked to find the horizontal asymptote of the graph of

g (x) .

Let's remember that in the general form of the reciprocal function,

y = \frac{a}{x - h} + k,

the line

y = k

represents the horizontal asymptote.

g (x) = \frac{4}{x - ( - 2 )} + 0

Therefore,

y = 0

is the horizontal asymptote of

g (x) .

d This time we will identify the vertical asymptote. One more time, we will recall that in the general form of the reciprocal function

y = \frac{a}{x - h} + k,

the line

x = h

represents the vertical asymptote.

g (x) = \frac{4}{x - ( - 2 )} + 0

Therefore,

x = - 2

is the vertical asymptote of

g (x) .

Notice that the function is undefined at

x = - 2

because it makes the denominator

0 .

x - (- 2) = 0 \Leftrightarrow x = - 2

This means that

- 2

is not included in the domain of the function. We can conclude that the domain of a rational function does not include the vertical asymptotes.

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