warm-up 3/27
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Warm-Up 3/27. 1. G. Rigor: You will learn how to analyze, graph and solve equations of rational functions. Relevance: You will be able to use graphs and equations of rational functions to solve real world problems. . 2-5a Rational Functions. - PowerPoint PPT PresentationTRANSCRIPT
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1.Warm-Up 3/27
G
π₯=βπ2π
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Rigor:You will learn how to analyze, graph and solve
equations of rational functions.
Relevance:You will be able to use graphs and equations of rational functions to solve real world problems.
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2-5a Rational Functions
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A rational function is the quotient of two polynomial functions.
An asymptote is a line or curve that a graph approaches.
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If a factor is removable, then there is a hole at that x value.
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Example 1a: Find the domain of the function and the equations of the vertical or horizontal asymptotes, if any.π (π₯ )= π₯+4
π₯β3
Step 1 Find the domain. π₯β3β 0 π₯β 3 (ββ ,3)βͺ(3 ,β )
Step 2 Find the asymptotes, if any.(π₯β3 )is not removable , so π₯=3a vertical asymptote .
Degree of numerator equals the degree of the denominator, so is the horizontal asymptote.
Check
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Example 1b: Find the domain of the function and the equations of the vertical or horizontal asymptotes, if any.π (π₯ )=8 π₯2+5
4 π₯2+1
Step 1 Find the domain. 4 π₯2+1β 0 π₯2β β 14 (ββ ,β)
Step 2 Find the asymptotes, if any.Since domain is all realnumbers ,there are novertical asymptotes .
Degree of numerator equals the degree of the denominator, so therefore is the horizontal asymptote.
Check
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Example 2a: Find the domain, the vertical or horizontal asymptotes and intercepts. Then graph the function.π (π₯ )= 6
π₯+3
Step 1 Find the domain. π₯+3β 0 π₯β β3 (ββ ,β3)βͺ (β3 ,β)
Step 2 Find the asymptotes, if any.(π₯+3 ) is not removable , so π₯=β3 avertical asymptote .
Degree of numerator less than the degree of the denominator, so is the horizontal asymptote.Step 3 There are no x-intercepts and (0, 2) is the y-intercept.
Step 4
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Example 2b: Find the domain, the vertical or horizontal asymptotes and intercepts. Then graph the function.π (π₯ )= π₯2β7 π₯+10
π₯β3
Step 1 Find the domain. π₯β3β 0 π₯β 3 (ββ ,3)βͺ (3 ,β)
Step 2 Find the asymptotes, if any.(π₯β3 ) is not removable , so π₯=3a vertical asymptote .
Degree of numerator greater than the degree of the denominator, so no horizontal asymptote.Step 3 x-intercepts (5, 0) & (2, 0) and (0,) is the y-intercept.
Step 4 x yβ 5 β 8.75
β 1 β 4.5
1 β 2
4 β 2
7 2.5
π (π₯ )=(π₯β5)(π₯β2)(π₯β3)
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Example 3: Find the domain, the vertical or horizontal asymptotes and intercepts. Then graph the function.π (π₯ )=3 π₯2β3
π₯2β9
Step 1 Find the domain. π₯2β9β 0 π₯β Β±3 {π₯|π₯β Β±3 ,π₯ββ }
Step 2 Find the asymptotes, if any.(π₯β3 ) , (π₯+3 )are not removable , so π₯=Β±3 vertical asymptotes .
Degree of numerator is equal to the degree of the denominator, so π¦ = 3 is the horizontal asymptote.
Step 3 x-intercepts (β 1, 0) & (1, 0) and (0,) is the y-intercept.
Step 4 x yβ 7 3.6
β 5 4.5
β 2 β 1.8
2 β 1.8
5 4.5
7 3.6
π (π₯ )=3 (π₯β1)(π₯+1)(π₯β3)(π₯+3)
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An oblique asymptote is a slant line that occurs when the degree of the numerator is exactly one more than the degree of the denominator.
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Example 4: Find the domain, any asymptotes and intercepts. Then graph the function.π (π₯ )= 2 π₯3
π₯2+π₯β12
Step 1 Find the domain. π₯β β4 π₯β 3 {π₯|π₯β β4 ,3 ,π₯ββ }Step 2 Find the asymptotes, if any.
(π₯β3 ) , (π₯+4 )are not removable , so π₯=β4 ,3 vertical asymptotes .Degree of numerator is exactly one more than the degree of the denominator, so No H. A. & π¦ = 2x β 2 is the oblique asymptote.
Step 3 x- & y-intercept (0, 0).
Step 4x y
β 7 β 22.87
β 6 β 24
β 5 β 31.25
β 3 9
1 β 0.2
4 16
5 13.889
6 14.4
π (π₯ )= 2π₯3
(π₯β3)(π₯+4)
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Example 5: Find the domain, any asymptotes, holes and intercepts. Then graph the function.h (π₯ )= π₯2β4
π₯2β2 π₯β8
Step 1 Find the domain. π₯β 4 π₯β β2 {π₯|π₯β β2 ,4 ,π₯ββ }Step 2 Find the asymptotes, if any.
(π₯β4 )is not removable , so π₯=4 vertical asymptotes .Degree of numerator is equal to the degree of the denominator, so π¦ = 1 is the horizontal asymptote.
Step 3 x-intercept (2, 0) and (0,) is the y-intercept.
Step 4x y
β 2 hole
β 1 .6
0 .5
1 .3333
2 0
3 β 1
4 V.A.
5 3
6 2
h (π₯ )=(π₯β2)(π₯+2)(π₯β4 )(π₯+2)
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ββ1math!
2-5a Assignment: TX p138, 4-28 EOE
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1.Warm-Up 3/26
A
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ββ1math!
2-5a Assignment: TX p138, 4-28 EOE