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Section 8.4 – Graphing Rational Functions

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1 Section 8.4 – Graphing Rational Functions
EQ: How do I graph a rational function using the vertical and horizontal asymptotes?

2 A Rational Function An equation of the form𝑓 π‘₯ = π‘Ž(π‘₯) 𝑏(π‘₯) , where a(x) and b(x) are polynomial functions and b(x) β‰  0

3 Vertical and Horizontal Asymptotes
Vertical Asymptotes Whenever b(x) = 0 Horizontal Asymptotes (at most one) If the degree of π‘Ž π‘₯ is greater than the degree of 𝑏 π‘₯ , there is no horizontal asymptote If the degree of π‘Ž π‘₯ is less than the degree of 𝑏 π‘₯ , the horizontal asymptote is the line y = 0 If the degree of π‘Ž π‘₯ equals the degree of 𝑏 π‘₯ , the horizontal asymptote is the line 𝑦= π‘™π‘’π‘Žπ‘‘π‘–π‘›π‘” π‘π‘œπ‘’π‘“π‘“π‘–π‘π‘–π‘’π‘›π‘‘ π‘œπ‘“ π‘Ž(π‘₯) π‘™π‘’π‘Žπ‘‘π‘–π‘›π‘” π‘π‘œπ‘’π‘“π‘“π‘–π‘π‘–π‘’π‘›π‘‘ π‘œπ‘“ 𝑏(π‘₯)

4 Graph by finding the following
Vertical Asymptote(s), if any: Set the denominator = 0 and solve Horizontal Asymptote(s), if any. Compare the degree of the numerator (m) to the degree of the denominator (n) m < n: HA is y = 0 m = n: HA is (LC of numerator/LC of denominator) m > n: HA does not exist y – intercept: Let x = 0 and solve x – intercept(s): Set the numerator = 0 and solve for x Choose other x – values, 2 on each side of an asymptote

5 Example 1 What are the asymptotes of the function 𝑓 π‘₯ = π‘₯βˆ’3 π‘₯βˆ’2 ?
Vertical Asymptote: Set the denominator = 0 and solve. VA: x = 2 Horizontal Asymptote: compare the degree of the numberator (m) to the degree of the denominator (n). HA: 1π‘₯ 1 1π‘₯ 1 = 1 1 →𝑦=1

6 Example 2 What are the asymptotes of the function 𝑓 π‘₯ = 2π‘₯βˆ’7 π‘₯+4 ?
Vertical Asymptote: Set the denominator = 0 and solve. VA: x = -4 Horizontal Asymptote: compare the degree of the numberator (m) to the degree of the denominator (n). HA: 2π‘₯ 1 1π‘₯ 1 = 2 1 →𝑦=2

7 Example 3 Graph the function, then state the domain and range and any asymptote equations. a) 𝑓 π‘₯ = π‘₯ π‘₯ 2 βˆ’9 VA: Set the denominator = 0 and solve. π‘₯ 2 βˆ’9=(π‘₯βˆ’3)(π‘₯+3) π‘₯=βˆ’3, 3 HA: π‘₯ π‘₯ 2 β†’π‘š<𝑛→𝑦=0 *Use calculator to find other points, then connect the points Domain: all real, π‘₯β‰ βˆ’3, 3 Range: all real, 𝑦≠0

8 Example 3 Graph the function, then state the domain and range and any asymptote equations. b) 𝑓 π‘₯ = π‘₯ 2 π‘₯ 2 βˆ’1 VA: Set the denominator = 0 and solve. π‘₯ 2 βˆ’1=(π‘₯βˆ’1)(π‘₯+1) π‘₯=βˆ’1, 1 HA: π‘₯ 2 π‘₯ 2 β†’π‘š=𝑛→𝑦=1 *Use calculator to find other points, then connect the points Domain: all real, π‘₯β‰ βˆ’1, 1 Range: all real, 𝑦≠1

9 Point Discontinuity A point where a graph is undefined, looks like a hole in the graph

10 Point of discontinuity is at x = 3
Example 4 What is the point of discontinuity of the following functions? a) π‘₯ 2 βˆ’9 π‘₯βˆ’3 (π‘₯+3)(π‘₯βˆ’3) π‘₯βˆ’3 Point of discontinuity is at x = 3

11 Point of discontinuity is at x = -1
Example 4 What is the point of discontinuity of the following functions? b) 2 π‘₯ 2 +2 π‘₯+1 2(π‘₯+1)(π‘₯βˆ’1) π‘₯+1 Point of discontinuity is at x = -1

12 Point of discontinuity is at x = -5
Example 4 What is the point of discontinuity of the following functions? c) π‘₯ 2 +4π‘₯βˆ’5 π‘₯+5 (π‘₯+5)(π‘₯βˆ’1) π‘₯+5 Point of discontinuity is at x = -5

13 Point of discontinuity is at x = 2
Example 5 Graph 𝑓 π‘₯ = π‘₯ 2 βˆ’4 π‘₯βˆ’2 Hole at x = 2 (π‘₯+2)(π‘₯βˆ’2) π‘₯βˆ’2 Point of discontinuity is at x = 2 𝑓 π‘₯ =π‘₯+2

14 Point of discontinuity is at x = -4
Example 6 Graph 𝑓 π‘₯ = π‘₯ 2 βˆ’16 π‘₯+4 (π‘₯+4)(π‘₯βˆ’4) π‘₯+4 Point of discontinuity is at x = -4 𝑓 π‘₯ =π‘₯βˆ’4 Hole at x = -4


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