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Objectives
Identify rational functions
Analyze graphs of rational
functions’
Given a function, draw the graph
Explain characteristics of the
graph
Given a variety of graphs, select
the rational function graph and
defend your choice
Introduction
• The following are great websites to
explore rational functions:
http://id.mind.net/~zona/mmts/functionI
nstitute/rationalFunctions/rationalFu
nctions.html
http://hh4.hollandhall.org/kluitwieler/Pa
ges/RationalFunctions/Welcome.htm
I. Definition
• A rational function can be written in the
form:
p(x)
f(x) =
q(x)
• p(x) and q(x) are polynomials where q(x)≠0
Examples of rational functions
2x  7
2x  5
2x  7
1
x
3x
x5
2x  7
x  5x  6
x2
2
Which of the following functions are rational?
Yes!
No!
Yes!
These rational functions make all
kinds of “funny looking” graphs…
Continuous or Discontinuous?
No!
Yes!
No!
No!
Yes!
No!
Graphs are discontinuous because
they have breaks in the graph.
• These breaks called “points of discontinuity”
are either holes or vertical asymptotes.
• What causes these points of discontinuity?
• First we must look at the domain of the function.
II. Domain
• The domain of a rational functions of x
includes all real numbers except x-values
that make the denominator zero.
Think…
What can you plug in for x? Or...
What can you NOT plug in for x?
II. Domain
• Example:
1
f ( x) 
x
• What value for x will make the
denominator zero?
II. Domain
1
• Example: f ( x ) 
x
• 0 will make the denominator zero.
• Domain of f(x) is:
{all real numbers except x=0.}
In order to find what makes the denominator
zero, you will need to factor.
x  4 x  45
f ( x)  2
x  2 x  15
2
( x  5)( x  9)
f ( x) 
( x  5)( x  3)
Since -5 and 3 will make the denominator
zero, these graphs will have points of
discontinuity at x = -5 and x = 3.
Find the points of discontinuity:
3  5x
f ( x) 
x( x  4)
x0
x  4
3
y 2
x  x  12
( x  4)( x  3)
x4
x  3
But how do you know if it is a hole or vertical
asymptote?
It will produce a hole in the line if it
cancels…
• Example:
 x  3 ( x  4)
 x  4
 x  3 ( x  4)
 x  4
Because 4 will make the denominator = 0, then
there is a point of discontinuity at x = 4.
However, since the (x - 4)’s cancel, then it is a
hole.
Since the (x-4)’s cancel out, we are really left with a
line that looks like the line y = x + 3, but with a
hole in it at x = 4.
 x  3 ( x  4)
y
 x  4
Notice how to enter it into calculator…
Graph won’t appear to have a hole at x = 4.
hole
III. Vertical Asymptote
• Abbreviated V.A.
• Occur on the graph of the function at the
zeros of the denominator
• Ex.
3 x
x2
2 would make the denominator = 0, and it
does not cancel, so there is a VA at x = 2.
III. Vertical Asymptote
• Graph
State whether the points of discontinuity are
holes or asymptotes.
x7
y
( x  1)( x  5)
x 2  3x
f ( x) 
x3
x  1
x ( x  3)
x3
Hole
x  5
x  3
Nothing Cancels
VA
h( x ) 
( x  6)( x  9)
( x  9)( x  9)( x  6)
Hole x  6
VA : x  9
IV. Horizontal Asymptote
• Abbreviated H.A.
• There are 3 different cases to
determine horizontal asymptotes:
Horizontal Asymptotes
1. If the degree of the numerator is less
than the degree of denominator, the
graph has the x - axis (y = 0) as a
horizontal asymptote.
If the degree of the denominator is
greater than the numerator, then .
If the degree is Bigger On Bottom
its 0. BOB0. (y = 0)
4x  9
f ( x)  3
x  12
2
4x  9
f ( x)  3
x  12
2
We can ignore the
(-9) and the (+ 12). They really do
not add anything to the graph when
you go to ±∞
2
4x
f ( x)  3
x
So, let’s look at this graphically
As the graph approaches ±∞, what
is the “height” of the graph?
2
4x
f ( x)  3
x
4x  9
3
x  12
2
4x  9
H .A  3
0
x  12
2
If the degree of the denominator is
greater than the numerator, then .
If the degree is
Bigger On Bottom its 0.
BOB0.
(y = 0)
Horizontal Asymptotes
• Case 1
2x
f ( x)  2
3x  1
Notice that the graph
crosses the H.A.
H.A.s only deal with
END BEHAVIOR.
Horizontal Asymptotes
2. If the degree of numerator equals degree
of denominator, the graph has the line
y=a/b, where a is the leading coefficient of
numerator and b is the leading coefficient
of denominator.
Degrees of both the numerator and
denominator are equal
• Then divide the leading coefficients.
That’s your horizontal asymptote.
EATS-D/C.
4 x  10 x  1
f ( x)  3
2
5x  2 x  x
3
4 x  10 x  1
f ( x)  3
2
5x  2 x  x
3
4 x  10 x  1
f ( x)  3
2
5x  2 x  x
3
4
y
5
Horizontal Asymptotes
• Case 2
2x 2
f ( x)  2
3x  1
y= a/b = 2/3
Horizontal Asymptotes
3. If the degree of numerator is greater
than the degree of denominator, the graph
has no horizontal asymptote.
Bigger On Top, there’s No horizontal
asymptote.
6x  x  7
f ( x) 
2
2x  8
4
BOTN
3
6x  x  7
f ( x) 
2
2x  8
4
3
6x  x  7
f ( x) 
2
2x  8
4
3
6x  x  7
f ( x) 
2
2x  8
4
3
Horizontal Asymptotes
• Case 3
2x3
f ( x)  2
3x  1
No H.A.
V. X - intercepts
• Occur where Numerator
equals zero
2x  2
f ( x)  2
3x  1
(set the numerator = to zero)
If numerator is a constant
(or imaginary), there is no
x-intercept
2
3
2x  2  0
3
x3  1
x  1
2x  2  0
2 x2  2
f ( x) 
3x  1
x  1
2
x  1
2
f ( x) 
3x  1
VI. y-intercepts
• Find by substituting zero in for x
3
• Ex.
2x  2
f ( x)  2
3x  1
2(0)  2 2
f (0) 
 2
3(0)  1 1
(0,2) is the y-intercept
Guidelines for Graphing Rational
Functions
1. Find and plot the y-intercept
2. Find the zeros of the numerator and plot x-intercepts
3. Find zeros of denominator. Sketch V.A.
4. Find and sketch the horizontal asymptotes
5. Plot a few points around intercepts and asymptotes
6. Use smooth curves to complete the graph
Example 1
• y-intercept
(0,-3/2)
3
-3
=
0-2
2
3
f ( x) 
x2
Example 1
3
f ( x) 
x2
• x-intercept – num. is 3, there are no x-int.
Example 1
3
f ( x) 
x2
• Vertical asymptotes: x - 2=0 so x = 2
Example 1
• Horizontal asymptotes is y = 0
3
f ( x) 
x2
3
f ( x) 
x2
Example 1
• Additional Points
x
-4
1
3
5
f(x)
-0.5
-3
3
1
Example 1
• Smooth Curves
3
f ( x) 
x2
x
f ( x)  2
x x2
Example 2
• y-intercept
(0,0)
0
0-0-2
=
0
2
= 0
Example 2
• x-intercept
(0,0)
Numerator is zero when x=0
f ( x) 
x
x2  x  2
f ( x) 
Example 2
• Vertical asymptotes
X2 - x-2 = (x + 1)(x - 2)
X = -1
X=2
x
x2  x  2
Example 2
• Horizontal asymptotes
Y=0
x
f ( x)  2
x x2
f ( x) 
Example 2
• Additional Points
x
-3
f(x)
-0.3
-0.5
x
x2  x  2
1
3
0.4 -0.5 0.75
Example 2
• Smooth Curves
x
f ( x)  2
x x2
Guess what time it is
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Homework
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