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1. Warm-Up 3/27 G = 2

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 Presentation

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Page 1: Warm-Up 3/27

1.Warm-Up 3/27

G

𝑥=−𝑏2𝑎

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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.𝑓 (𝑥 )= 𝑥+4

𝑥−3

Step 1 Find the domain. 𝑥−3≠0 𝑥≠3 (−∞ ,3)∪(3 ,∞ )

Step 2 Find the asymptotes, if any.(𝑥−3 )is not removable , so 𝑥=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.𝑔 (𝑥 )=8 𝑥2+5

4 𝑥2+1

Step 1 Find the domain. 4 𝑥2+1≠0 𝑥2≠− 14 (−∞ ,∞)

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.𝑔 (𝑥 )= 6

𝑥+3

Step 1 Find the domain. 𝑥+3≠0 𝑥≠−3 (−∞ ,−3)∪ (−3 ,∞)

Step 2 Find the asymptotes, if any.(𝑥+3 ) is not removable , so 𝑥=−3 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.𝑘 (𝑥 )= 𝑥2−7 𝑥+10

𝑥−3

Step 1 Find the domain. 𝑥−3≠0 𝑥≠3 (−∞ ,3)∪ (3 ,∞)

Step 2 Find the asymptotes, if any.(𝑥−3 ) is not removable , so 𝑥=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 y– 5 – 8.75

– 1 – 4.5

1 – 2

4 – 2

7 2.5

𝑘 (𝑥 )=(𝑥−5)(𝑥−2)(𝑥−3)

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Example 3: Find the domain, the vertical or horizontal asymptotes and intercepts. Then graph the function.𝑓 (𝑥 )=3 𝑥2−3

𝑥2−9

Step 1 Find the domain. 𝑥2−9≠0 𝑥≠±3 {𝑥|𝑥≠±3 ,𝑥∈ℝ }

Step 2 Find the asymptotes, if any.(𝑥−3 ) , (𝑥+3 )are not removable , so 𝑥=±3 vertical asymptotes .

Degree of numerator is equal to the degree of the denominator, so 𝑦 = 3 is the horizontal asymptote.

Step 3 x-intercepts (– 1, 0) & (1, 0) and (0,) is the y-intercept.

Step 4 x y– 7 3.6

– 5 4.5

– 2 – 1.8

2 – 1.8

5 4.5

7 3.6

𝑓 (𝑥 )=3 (𝑥−1)(𝑥+1)(𝑥−3)(𝑥+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.𝑓 (𝑥 )= 2 𝑥3

𝑥2+𝑥−12

Step 1 Find the domain. 𝑥≠−4 𝑥≠3 {𝑥|𝑥≠−4 ,3 ,𝑥∈ℝ }Step 2 Find the asymptotes, if any.

(𝑥−3 ) , (𝑥+4 )are not removable , so 𝑥=−4 ,3 vertical asymptotes .Degree of numerator is exactly one more than the degree of the denominator, so No H. A. & 𝑦 = 2x – 2 is the oblique asymptote.

Step 3 x- & y-intercept (0, 0).

Step 4x y

– 7 – 22.87

– 6 – 24

– 5 – 31.25

– 3 9

1 – 0.2

4 16

5 13.889

6 14.4

𝑓 (𝑥 )= 2𝑥3

(𝑥−3)(𝑥+4)

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Example 5: Find the domain, any asymptotes, holes and intercepts. Then graph the function.h (𝑥 )= 𝑥2−4

𝑥2−2 𝑥−8

Step 1 Find the domain. 𝑥≠4 𝑥≠−2 {𝑥|𝑥≠−2 ,4 ,𝑥∈ℝ }Step 2 Find the asymptotes, if any.

(𝑥−4 )is not removable , so 𝑥=4 vertical asymptotes .Degree of numerator is equal to the degree of the denominator, so 𝑦 = 1 is the horizontal asymptote.

Step 3 x-intercept (2, 0) and (0,) is the y-intercept.

Step 4x y

– 2 hole

– 1 .6

0 .5

1 .3333

2 0

3 – 1

4 V.A.

5 3

6 2

h (𝑥 )=(𝑥−2)(𝑥+2)(𝑥−4 )(𝑥+2)

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√−1math!

2-5a Assignment: TX p138, 4-28 EOE

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1.Warm-Up 3/26

A

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√−1math!

2-5a Assignment: TX p138, 4-28 EOE