Rational Functions and Their Graphs

Contributed by:
Sharp Tutor
To determine horizontal or oblique asymptotes, compare the degrees of the numerator and denominator.
1. If the degree of the top < the bottom, horizontal asymptote along the x-axis (y = 0).
2. If the degree of the top = bottom, horizontal asymptote at y = leading coefficient of the top over the leading coefficient of the bottom.
3. If the degree of the top > the bottom, oblique asymptote is found by long division.
1. 8.3 RATIONAL
FUNCTIONS AND
THEIR GRAPHS
2. SUMMARY OF HOW TO FIND ASYMPTOTES
Vertical Asymptotes are the values that are NOT in the domain. To find them,
set the denominator = 0 and solve.
“WHAT VALUES CAN I NOT PUT IN THE DENOMINATOR????”
To determine horizontal or oblique asymptotes, compare the degrees of the
numerator and denominator.
1. If the degree of the top < the bottom, horizontal asymptote along the x
axis (y = 0)
2. If the degree of the top = bottom, horizontal asymptote at y = leading
coefficient of top over leading coefficient of bottom
3. If the degree of the top > the bottom, oblique asymptote found by long
division.
3. Finding Asymptotes
VERTICAL ASYMPTOTES
There will be a vertical asymptote at any “illegal” x value, so
anywhere that would make the denominator = 0
So there are vertical
2
x  2x  5 asymptotes at x = 4 and x = -1.
R x   2
 x x 43xx140
Let’s set the bottom = 0 and
factor and solve to find where
the vertical asymptote(s) should
be.
4. HORIZONTAL ASYMPTOTES
We compare the degrees of the polynomial in the numerator and the
polynomial in the denominator to tell us about horizontal asymptotes.
1<2
degree of top = 1
If the degree of the numerator is less than
1 the degree of the denominator, (remember
the x axis
2x  5 degree
is a horizontal
is the highest
asymptote.
powerThis
on any
is along
x term)
R x   2 the line
x axis
y =is0.a horizontal asymptote.
x  3x  4
degree of bottom = 2
5. HORIZONTAL ASYMPTOTES
The leading coefficient is the
number in front of the highest If the degree of the numerator is equal to the
powered x term. degree of the denominator, then there is a
horizontal asymptote at:
y = leading coefficient of top
degree of top = 2
leading coefficient of bottom
2
2x  4x  5
R x   2
1 x  3x  4
degree of bottom = 2
horizontal asymptote at:
2
y  2
1
6. OBLIQUE ASYMPTOTES - Slanted
If the degree of the numerator is greater
than the degree of the denominator, then
there is not a horizontal asymptote, but an
oblique one. The equation is found by
degree of top = 3 doing long division and the quotient is the
equation of the oblique asymptote ignoring
the remainder.
3 2
x  2 x  3x  5
R x   2
x  3x  4
degree of bottom = 2
x  5  a remainder
x 2  3x  4 x 3  2 x 2  3x  5 Oblique asymptote at y = x
+5
7. STRATEGY FOR GRAPHING A RATIONAL
1. Graph your asymptotes
2. Plot points to the left and right of each
asymptote to see the curve
8. SKETCH THE GRAPH
2x  3
f ( x) 
5 x  10
9. 2x  3
f ( x) 
5 x  10
The vertical asymptote is x = -2
The horizontal asymptote is y = 2/5
10. 2x  3
f ( x) 
5 x  10
10
8
6
4
2
-10 -8 -6 -4 -2 2 4 6 8 10
-2
-4
-6
-8
-10
11. OF: 1
g(x) 
x1
Vertical asymptotes at??
x=1
Horizontal asymptote at??
y=0
12. OF: 2
f (x) 
x
Vertical asymptotes at??
x=0
Horizontal asymptote at??
y=0
13. OF:  4
h(x) 
x
Vertical asymptotes at??
x=0
Horizontal asymptote at??
y=0
14. THE Vertical asymptotes at?? x=1
GRAPH Horizontal asymptote at?? y=0
OF: 1
y 2
x 3
Hopefully you remember,
y = 1/x graph and it’s asymptotes:
Vertical asymptote: x = 0
Horizontal asymptote: y = 0
15. OR…
We have the function:
1
y 2
x 3
But what if we simplified this and combined like terms:
1 2(x  3) Now looking at this:
y 
x 3 x 3 Vertical Asymptotes??
x = -3
1  2x  6
y
x 3 Horizontal asymptotes??
y = -2
 2x  5
y
x 3
16. OF: 2
x  3x
h(x) 
x
x(x  3)
h(x) 
x
Hole at??
x=0
17. FIND THE ASYMPTOTES
OF EACH FUNCTION: Vertical Asymptote:
x 2 3x 4
y  
2
x  3x  4 x x x
x=0
y 4
x y x  3  Slant Asymptote:
 x
y=x+3
2
x  3x  
28 Hole at x = 4
y 3
 x  11x 2  28x Vertical Asymptote:
x = 0 and x = 7
(x  7)(x  4)
y
x(x  7)(x  4) Horizontal Asymptote:
 y=0
18. WHAT MAKES A FUNCTION
Continuous functions are predictable…
1) No breaks in the graph
A limit must exist at every x-value or
the graph will break.
2) No holes or jumps
The function cannot have undefined
points or vertical asymptotes.
19. Key Point:
Continuous functions
can be drawn with a
single, unbroken
pencil stroke.
20. CONTINUITY OF
POLYNOMIAL AND
RATIONAL FUNCTIONS
A polynomial function is continuous at every real
number.
A rational function is continuous at every real
number in its domain.
21. Discontinuity: a
point at which a
function is not
22. Two Types of Discontinuities
1) Removable (hole in the graph)
2) Non-removable (break or vertical
asymptote)
A discontinuity is called removable if a function can
be made continuous by defining (or redefining) a
point.
23. TWO TYPES OF
24. Find the intervals on which these function
are continuous.
x2 Point of discontinuity:
f ( x)  2
x  3x  10 x  2 0 Removable
discontinuity
x2 x  2

( x  2)( x  5) Vertical Asymptote:
1 x  5 0 Non-removable
 x 5 discontinuity
( x  5)
25. DISCONTINUITY
x2
f ( x)  2
x  3x  10
Continuous on: ( ,  2)  ( 2, 5)  (5, )
26.  2 x, x 2
f ( x)  2
 x  4 x  1, x  2
lim (  2 x)  4
x 2
lim ( x 2  4 x  1) 3
x 2
Continuous on:
f (2)  4
( , 2]  (2, )
27. Determine the value(s) of x at which
the function is discontinuous. Describe
the discontinuity as removable or non-
removable. x 2  1
(A) (B) 2
x  10 x  9
f ( x)  2
f ( x)  2
x  5x  6 x  81
2
(C) 2
x  4x  5 (D) x 4
f ( x)  2 f ( x)  2
x  25 x  2x  8
28. 2
x 1
(A) f ( x)  2
x  5x  6
( x  1)( x  1)

( x  6)( x  1)
x  1 Removable discontinuity
x 6 Non-removable discontinuity
29. 2
x  10 x  9
(B) f ( x) 
x 2  81
( x  9)( x  1)

( x  9)( x  9)
x  9 Removable discontinuity
x 9 Non-removable discontinuity
30. 2
x  4x  5
(C) f ( x)  2
x  25
( x  5)( x  1)

( x  5)( x  5)
x 5 Removable discontinuity
x  5 Non-removable discontinuity
31. x2  4
f ( x)  2
x  2x  8
( x  2)( x  2)

( x  4)( x  2)
x  2 Removable discontinuity
x 4 Non-removable discontinuity
32. Continuous functions have no breaks, no holes, and no
If you can evaluate any limit on the function using only
the substitution method, then the function is