1 warm-up solve the following rational equation
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1
Warm-up• Solve the following rational equation.
xx
x
xx
2
4
1
4
122
: 1
: 4
Solution x
Extr x
2
Set Equation to ZERO
xx
x
xx
2
4
1
4
122
02
4
1
4
122
xx
x
xx
Next Slide
12 1 20
( 4) 4
x
x x x x
12 ( 1) 2( 4)0
( 4)
x x x
x x
3
Problem Continued
0)4(
)4(2)1(12
xx
xxx
08212 2 xxx
0432 xx
0432 xx
0)1)(4( xx
14 xorx
x = -4 Extraneous
x = 1
MUST CHECK ANSWERS
x = -4 does not work
Rational Function Discontinuities
Section 2-6
5
Objectives
• I can identify Graph Discontinuities– Vertical Asymptotes – Horizontal Asymptotes – Slant Asymptotes– Holes
• I can find “x” and “y” intercepts
6
Rational Functions
• A rational function is any ratio of two polynomials, where denominator cannot be ZERO!
• Examples:
1)(
x
xxf
103
1)(
2
xx
xxf
7
Asymptotes• Asymptotes are the boundary lines that a
rational function approaches, but never crosses.
• We draw these as Dashed Lines on our graphs.
• There are three types of asymptotes: – Vertical– Horizontal (Graph can cross these)– Slant
8
Vertical Asymptotes
• Vertical Asymptotes exist where the denominator would be zero.
• They are graphed as Vertical Dashed Lines
• There can be more than one!
• To find them, set the denominator equal to zero and solve for “x”
9
Example #1
• Find the vertical asymptotes for the following function:
1)(
x
xxf
•Set the denominator equal to zero
•x – 1 = 0, so x = 1
•This graph has a vertical asymptote at x = 1
10
1 2 63 4 5 7 8 9 10
4
3
2
7
5
6
8
9
x-axis
y-axis
0
1-2-6 -3-4-5-7-8-910
-4
-3
-2
-1
-7
-5
-6
-8
-9
0
-1
Vertical Asymptote at
X = 1
11
Other Examples:
• Find the vertical asymptotes for the following functions:
3
3)(
xxg
)5)(2(
1)(
xx
xxg
3: xVA
5:
2:
xVA
xVA
To find Vertical Asymptote(s)1) Set reduced denominator = 0
2) Solve for x = #.
3) Your answer is written as a line.
To find Vertical Asymptote(s)1) Set reduced denominator = 0
2) Solve for x = #.
3) Your answer is written as a line.
Horizontal Asymptotes
• Horizontal Asymptotes are also Dashed Lines drawn horizontally to represent another boundary.
• To find the horizontal asymptote you compare the degree of the numerator with the degree of the denominator
• See next slide:
Horizontal Asymptote (HA)
Given Rational Function:
Compare DEGREE of Numerator to Denominator
If N < D , then y = 0 is the HA
If N > D, then the graph has NO HA
If N = D, then the HA is
Numerator( )
Denominatorf x
N
D
LCy
LC
15
Example #1
• Find the horizontal asymptote for the following function:
1)(
x
xxf
•Since the degree of numerator is equal to degree of denominator (m = n)
•Then HA: y = 1/1 = 1
•This graph has a horizontal asymptote at y = 1
1 2 63 4 5 7 8 9 10
4
3
2
7
5
6
8
9
x-axis
y-axis
0
1-2-6 -3-4-5-7-8-910
-4
-3
-2
-1
-7
-5
-6
-8
-9
0
-1
Horizontal Asymptote at
y = 1
Other Examples:
• Find the horizontal asymptote for the following functions:
3
3)(
xxg
13
13)(
2
2
xx
xxg
5
1)(
3
x
xxg
0: yHA
3: yHA
NoneHA :
To find Horizontal Asymptote(s)
1) Compare DEGREE of numerator and denominator
Num BIGGER then NO HA
Num SMALLER then y = 0
Degree is SAME then
To find Horizontal Asymptote(s)
1) Compare DEGREE of numerator and denominator
Num BIGGER then NO HA
Num SMALLER then y = 0
Degree is SAME then Num
Den
LCy
LC
19
Slant Asymptotes (SA)
• Slant asymptotes exist when the degree of the numerator is one larger than the denominator.
• Cannot have both a HA and SA• To find the SA, divide the Numerator by the
Denominator. • The results is a line y = mx + b that is the
SA.
Example of SA
20
22 4 8( )
2
x xf x
x
2 4 8 -2
2
4
8
16
8
2 8y x
Remainder does not matter
To find Slant Asymptote(s)1) DEGREE of Numerator must be
ONE bigger than Denominator
2) Divide with Synthetic or Long Division
3) Don’t use the RemainderGet y = mx + b
To find Slant Asymptote(s)1) DEGREE of Numerator must be
ONE bigger than Denominator
2) Divide with Synthetic or Long Division
3) Don’t use the RemainderGet y = mx + b
22
Holes
• A hole exists when the same factor exists in both the numerator and denominator of the rational expression and the factor is eliminated when you reduce!
Example of Hole Discontinuity
23
( 4)( 1)( )
( 2)( 4)
x xf x
x x
Cancel LIKE factors
( 1)( )
( 2)
xf x
x
____, ____4
( 4 1) 5 5
( 4 2) 6 6
5
6
( 3)( 3)
( 3)( 1)
x xf x
x x
HOLES
To Find Holes1) Factor.2) Reduce. 3) A hole is formed when a factor is
eliminated from the denominator.4) Set eliminated factor = 0 and solve for
x.5) Find the y-value of the hole by
substituting the x-value into reduced form and solve for y.
6) Your answer is written as a point. (x, y)
To Find Holes1) Factor.2) Reduce. 3) A hole is formed when a factor is
eliminated from the denominator.4) Set eliminated factor = 0 and solve for
x.5) Find the y-value of the hole by
substituting the x-value into reduced form and solve for y.
6) Your answer is written as a point. (x, y)
To find x- intercept(s)
1) Set reduced numerator = 0
2) Solve for x.
3) Answer is written as a point.
(#, 0)
To find x- intercept(s)
1) Set reduced numerator = 0
2) Solve for x.
3) Answer is written as a point.
(#, 0)
To find y- intercept
1) Substitute 0 in for all x’s in reduced form.
2) Solve for y.3) Answer is a point. (0, #)
To find y- intercept
1) Substitute 0 in for all x’s in reduced form.
2) Solve for y.3) Answer is a point. (0, #)
27
Homework
• WS 5-3
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