Transcript Document

SFM Productions Presents:
Another semi-chilly non-snow day
in your Pre-Calculus journey!
2.6
Rational Functions
Homework for section 2.6/2.7
p190
p201
# 21-29, 41-45, 57, 61, 65
#49, 51 (this is something we did
at the beginning of the year)
Rational Functions: functions expressed as a
ratio.
N x 
f (x) 
D x 
N for numerator
D for denominator
What is the Domain of a Rational Function?
All x-values except those that
make the denominator zero.
As
1
f (x) 
x
x  
f (x)  0
5
Y
As
x  
4
f (x)  0
3
2
1
X
-5
-4
-3
-2
-1
0
1
2
3
4
5
As
x  0 
-1
f (x)  
-2
-3
-4
-5
As
x  0 
f ( x )  
10
9
8
7
6
5
4
3
2
1
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1-10
-2
-3
-4
-5
-6
-7
-8
-9
-10
Y
25
Y
20
15
10
5
X
1 2 3 4 5 6 7 8 9 10
X
-25
-20
-15
-10
-5
0
5
10
15
20
25
-5
-10
-15
-20
-25
All these graphs are the same…it just looks like
the graph disappears…in actuality, the graph
keeps getting closer and closer and closer and
closer and closer and closer to both the x and
y axes.
As
1
f (x)  2
x
f (x)  0
5
Y
As
4
2
1
X
-4
-3
-2
-1
0
-1
-2
-3
-4
-5
x  0 
f (x)  
3
-5
x  
1
2
3
4
5
NOTE: It is
only when very
far away from
the origin that
the graph
approaches an
asymptote…
5 x
f (x) 
3x2  1
5
Y
4
3
2
1
X
-10
-8
-6
-4
-2
0
-1
-2
-3
-4
-5
2
4
6
8
10
You CAN cross
the asymptote at
numbers that are
not far from
x=0.
2x 1
f (x) 
x 1
10
Asymptotes do
not have to be
just on either the
x or y axis…
Y
9
8
7
6
5
4
3
2
1
X
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0
-1
-2
-3
-4
-5
-6
-7
-8
-9
-10
1
2
3
4
5
6
7
8
9 10
But how are the
asymptotes
determined?......
Vertical Asymptote
m
VA when D(x) = 0
Horizontal Asymptote
3 different possibilities for HA
The degree of N(x) is less than that of D(x).
y=0
The degree of N(x) is equal to that of D(x).
y=a/b
The degree of N(x) is more than that of D(x).
no HA
2x
f (x) 
3x2  1
5
Y
4
3
2
1
X
-5
-4
-3
-2
-1
0
1
2
3
4
5
-1
-2
-3
-4
-5
HA is y = 0
2 x2
f (x) 
3x2  1
5
Y
4
3
HA is:
2
y = 2/3
1
X
-5
-4
-3
-2
-1
0
-1
-2
-3
-4
-5
1
2
3
4
5
2 x3
f (x) 
3x2  1
5
HA is:
Non existent
Y
4
3
However, there is a
slant asymptote:
2
1
X
-5
-4
-3
-2
-1
0
-1
-2
-3
-4
-5
1
2
3
4
5
The equation of
the slant
asymptote is equal
to the quotient of
the long division.
The slant asymptote is present only if the
degree of N is exactly 1 more than the degree
of D.
The equation of the slant asymptote is the
quotient of the long division of D into N.
So, what is the point of all this?......
It’s so you can sketch the graph of a
rational function without the use of a
calculator……muah, ha ha ha
Rules of
for graphing Rational Functions
1.
Set x = 0.
2.
Find the zeros of function by setting
N(x) = 0. Plot the x-intercepts.
Find the zeros of the denominator by
setting D(x) = 0. Sketch the VA.
Find and sketch any HA by comparing the
degree of N(x) and D(x).
3.
4.
5.
6.
Plot the y-intercepts.
Plot at least 1 point on either side of
x-intercepts and VA.
Draw nice smooth curves and then say
ahhhhhh…
f (x) 
x2
x 2
2x 8
10
Y
8
6
4
2
X
-10
-8
-6
-4
-2
0
-2
-4
-6
-8
-10
2
4
6
8
10
f (x) 
x2
2
x  3
10

Y
8
6
4
2
X
-10
-8
-6
-4
-2
0
-2
-4
-6
-8
-10
2
4
6
8
10
5
f (x) 
x
 4
x 2  x  12
10
Y
8
6
4
2
X
-10
-8
-6
-4
-2
0
-2
-4
-6
-8
-10
2
4
6
8
10
x 2  16
f (x) 
x 4
10
*: what is unique about this
problem?
Y
8
6
4
2
X
-10
-8
-6
-4
-2
0
-2
-4
-6
-8
-10
2
4
6
8
10
Can you work backwards???
10
Y
8
6
4
2
X
-10
-8
-6
-4
-2
0
-2
-4
-6
-8
-10
2
4
6
8
10
Go!
Do!