Lesson 3-7

Example 1
Determine the asymptotes for the graph of f(x) = .

Since f(-1) is undefined, there may be a vertical asymptote at x = -1. To verify that x = -1 is indeed a vertical asymptote, you have to make sure that f(x)   or f(x)  - as x  -1 from either the left or the right.
The values in the table suggest that f(x)   as x -1 from the left, so there is a vertical asymptote at x = -1.	x	f(x)	
	-1.1	14	
	-1.01	104	
	-1.001	1004	
	-1.0001	10,004	

Two different methods may be used to find the horizontal asymptote.
Method 1

Let f(x) = y and solve for x in terms of y. Then find where the function is undefined for values of y.

	y	= 
	y(x + 1)	= 4x + 3
	xy + y	= 4x + 3
	xy - 4x	= 3 - y
	x(y - 4)	= 3 - y
	x	= 
	Method 2

First divide the numerator and denominator by the highest power of x.

y	= 
y	= 
y	= 	
The rational expression  is undefined
for y = 4. thus, the horizontal asymptote is
the line y = 4.	As the value of x increases positively or negatively, the values of  and  approach zero. therefore, the value of the entire expression approaches  or 4. So, the line y = 4 is the horizontal asymptote.	

Method 2 is preferable when the degree of the numerator is greater than that of the denominator.

The graph of this function verifies that	
the lines of y = 4 and x = -1 are asymptotes.

	

Example 2
Use the parent graph f(x) =  to graph each function. Describe the transformation(s) that take place. Identify the new location of each asymptote.

a.	g(x) = 
To graph g(x) = , translate the parent graph 2 units to the right. The new vertical asymptote is x = 2. The horizontal asymptote, y = 0, remains unchanged.

		
b.	h(x) = -
To graph h(x) = -, reflect the parent graph over the
x-axis, and compress the result horizontally by a factor of 3. This does not affect the vertical asymptote at x = 0. The horizontal asymptote, y = 0, is also unchanged.

c.	k(x) = 
To graph k(x) = , stretch the parent graph vertically by a factor of 5 and translate the parent graph 4 units to the right. The new vertical asymptote is x = 4. The horizontal asymptote, y = 0, is unchanged.
		
d.	m(x) =  - 2
To graph m(x) =  - 2, stretch the result vertically by a factor of 3, and translate the result 1 unit to the right and 2 units down. The new vertical asymptote is x = 1. The horizontal asymptote changes from y = 0 to y = -2.		

Example 3
GEOMETRY  A regular triangular prism with a constant volume of 80 cubic centimeters has a surface area A(x) =  where x is the height of the prism.

a.	Write the function as a transformation of the parent f(x) = .

b.	Graph the function and identify the asymptotes. Interpret their meaning in terms of the problem.

a.	In order to write A(x) = as a transformation of f(x) = , you must perform the indicated division of the linear functions in the numerator and denominator.

Therefore, A(x) = 160 + 12x.

b.	The graph of A(x) is a vertical stretch of	
the parent function by a factor of 160. The
vertical asymptote x = 0. There is no
horizontal asymptote.

For the purposes of our model, the domain
of A(x) must be restricted to x > 0 since
surface area cannot be negative. As x
approaches 0 centimeters from the righ, the
surface area approaches infinity because in
order to maintain a constant volume of
80 cubic centimeters, the bases would have to grow without out bound if the length of the prism became infinitely small. As x approaches infinity, the surface area also approaches infinity.

Example 4
Determine the slant asymptote for f(x) = .

First, use division to rewrite the function.

As x  ,   0. So, the graph of f(x) will	
approach that of y = 3x + 5. This means that the
line y = 3x + 5 is a slant asymptote for the graph
of f(x). Note that x = 1 is a vertical asymptote.

The graph of this function verifies that the line
y = 3x + 5 is a slant asymptote.

Example 5
Graph y= .
y	= 
	= , x ≠ 2

Since x - 2 is a common factor that does not remain in the denominator after the division, there is point discontinuity at x = 2. Because y increases or decreases without bound close to x = 0 and
x = -1, there are vertical asymptotes at x = 0 and x = -1. There is also a horizontal asymptote at
y = 0.

The graph is the graph of y = ,	
except for point discontinuity at (-1, 1).