Rational Functions and Their Characteristics (DAY 2):
Example 1: Determine the intercepts and the vertical and horizontal asymptotes for
3x  2
f  x 
. Then sketch what you know about the function so far.
x 1
To be able to sketch the curve in the boundaries created by the asymptotes, you need to
understand the behaviour of the graph as it approaches the asymptotes.
Behaviour the Function as it approaches the Vertical Asymptote (VA):
(Does the curve point toward positive or negative infinity on either side of the VA?)
You need to examine the sign of the function as x “approaches” the VA from the left side
( x  VA value  ) and the sign of the function as x “approaches” the VA from the right side
( x  VA value  ). To do this for the left side, you substitute a value smaller than the VA but
really close to it and determine whether the output would be positive or negative. Then for
the right side, you substitute a value larger than the VA but really close to it and
determine whether the output would be positive or negative. A positive output would imply
that the graph is pointing upwards towards positive infinity on that side of the VA and a
negative output would imply that the graph is pointing downwards towards negative infinity
on that side of the VA.
3x  2
(From ex 1) around the
x 1
vertical asymptote. Then add that information to your sketch of f  x 
Example 2: Examine the behaviour of the function f  x  
Behaviour of the Function as it approaches the Horizontal Asymptote:
(Is the function above or below the HA as x is a very large positive (as x   ) or as x is a
very large negative (as x   )?
When is the function above the HA? Solve: ___________________________
When is the function below the HA? Solve: ____________________________
Does the function cross the HA? Solve: ______________________________
As , x   is the function above or below the HA? You should substitute a very ________
_________ x – value and if the output is _____________________ the function above
the HA and if the output is _______________________ the function if below the HA.
As x   is the function above or below the HA? You should substitute a very _________
___________ x – value and if the output is _____________________ the function above
the HA and if the output is _______________________ the function if below the HA.
Example 3:
a) Does the function f  x  
3x  2
(from ex 1) cross its horizontal asymptote? (Justify)
x 1
b) As x   , is the function, f  x  , above or below the HA?
c) As x   , is the function, f  x  , above or below the HA?
d) Add the above information to your sketch of f  x  
3x  2
.
x 1
e) Given that curves usually “hug” the asymptotes, fill-in the possible sketch of the
function.
(Linear) Oblique Asymptotes:
For rational functions, linear oblique asymptotes occur when the degree of the numerator
is exactly one more than the degree of the denominator. The equation of the linear
oblique asymptote can be found by dividing the numerator by the denominator using long
division. (or synthetic division where appropriate)
2 x3  x 2  3
.
x2
a) Determine the oblique asymptote.
Example 4: Let f  x  
b) Determine the intercepts.
c) Determine the vertical asymptote(s) and examine the behaviour around the VA.
d) Use your analysis to sketch the graph of f  x  .
ASSIGNED EXERCISES:
Knowledge
1.
2.
Find the intercepts and the equation of the vertical and horizontal asymptotes of each
curve. Examine the behaviour around the VA and determine whether the function is above
or below the HA as x   .
(a) f ( x ) 
2x  3
5 x
(b)
g( x) 
(c) h( x ) 
x2  1
x2  1
(d)
k ( x)  1 
x
x 9
2
Find an equation of the oblique asymptote of each curve.
3x 2  4 x  5
(a) y 
x
6x 2
(c) y 
3x  2
3.
x
x 4
2
(b)
x3  4
h( x) 
x2
(d)
x 3  5x 2  3x  10
f ( x) 
x2  2
Find the vertical asymptote and the linear oblique asymptote of each curve and determine
the behaviour of the function around the VA. Use the information to help you sketch the
graph.
(a) y  2 x  3 
3
x 1
(b)
y
x2  4
x4
Application
4.
A piece of machinery depreciates in value, V, in dollars, over time, t, in months. The value
is given by
V (t )  5000 
2000t
(t  2) 2
(a) Find the value of the machinery after:
(i)
(iii)
1 month
1 year
(ii)
(iv)
6 months
10 years
(b) Would you expect to find a local maximum or minimum value in the interval [0,]?
Explain
(c) Find V(t) as t becomes extremely large.
(d) Will the machinery ever have a value of $0?
(e) In light of your result in part (d), does V(t) model the value of the machinery for all
time?
5.
For the function f ( x) 
x 2  4x  3
, use the domain, intercepts and vertical, horizontal
x2
and/or oblique asymptotes to sketch the graph.
A company that installs carpet charges $600 for any area less than or equal to 40 m2 and
an additional $20/m2for any area over 40 m2.
6.
(a) Find a piecewise function y = c(x) to represent the average cost, per square metre, to
install x square metres of carpet.
(b) Find the value of c(x) as x becomes extremely large
(c) Graph y = c(x) for x>0
(d) Would you call this company to carpet a very small area? Explain.
Communication/Thinking
7.
8.
(a)
Under what conditions does a rational function have a linear oblique asymptote?
(b)
Explain how to find the linear oblique asymptote of a rational function that satisfies
the conditions in part (a)
A telecommunications company’s sales for the last 20 years can be modelled by the
2n 2  2n  4
function S (n) 
, where S(n) represents annual sales, in millions of dollars,
17n  4
and n represents the number of years since the company’s founding.
(a)
Find S(n) as n becomes extremely large. Interpret this result
(b)
If the long-term average rate of inflation is 3.3%, what is the true growth of
the company?
Answers:
1.

 3 

 5 

VA : x = 5, as x  5 , f  x    , as x  5 , f  x   
3
2
(a) x – int :  , 0  y – int :  0,
HA : y = - 2, as x   f  x  is below, as x   f  x  is above.
(b) x – int & y – int:  0, 0 
VA :
x = 2, as x  2 , f  x    , as x  2 , f  x   
x = - 2, as x  2 , f  x    , as x  2 , f  x   
HA : y = 0, as x   f  x  is above, as x   f  x  is below.
(c) x – int : None, y – int :  0, 1
VA :
x = 1, as x  1 , f  x    , as x  1 , f  x   
x = - 1, as x  1 , f  x    , as x  1 , f  x   
HA : y = 1, as x   f  x  is above, as x   f  x  is above.
(d) x – int :  2.54,0 and  3.54,0 y – int :  0,1
VA :
x = 3, as x  3 , f  x    , as x  3 , f  x   
x = - 3, as x  3 , f  x    , as x  3 , f  x   
HA : y = 1, as x   f  x  is below, as x   f  x  is above.
2.
(a) y = 3x + 4
(b) y = x
(c) y = 2x + 4/3
(d) y = x + 5
3.
(a) VA: x = -1, as x  1 , f  x    , as x  1 , f  x    , OA: y = 2x + 3
(b) VA: x = 4, as x  4 , f  x    , as x  4 , f  x    , OA: y = x + 4
4.
(a)
(i) $4777.78 (ii) $3875
(iii) $3021.74 (iv) $3022.19
(b) max at t = 0
(c) 3000
5.
Graph
6.
(a) c( x ) 
(d)
no
600
20 x  200
forx  40 , c( x ) 
, x  40
x
x
(b) 20
(c) Graph
(d) No
8.
(e)
(a) S(n) = 
(b) 13.9%
no