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Published byΞΞ΅Ο ΞΊΞ±Ξ»Ξ―ΟΞ½ ΞΞΏΟ Ξ½ΟΞΏΟ ΟΞΉΟΟΞ·Ο Modified over 5 years ago
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Section 8.4 β Graphing Rational Functions
EQ: How do I graph a rational function using the vertical and horizontal asymptotes?
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A Rational Function An equation of the formπ π₯ = π(π₯) π(π₯) , where a(x) and b(x) are polynomial functions and b(x) β 0
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Vertical and Horizontal Asymptotes
Vertical Asymptotes Whenever b(x) = 0 Horizontal Asymptotes (at most one) If the degree of π π₯ is greater than the degree of π π₯ , there is no horizontal asymptote If the degree of π π₯ is less than the degree of π π₯ , the horizontal asymptote is the line y = 0 If the degree of π π₯ equals the degree of π π₯ , the horizontal asymptote is the line π¦= πππππππ πππππππππππ‘ ππ π(π₯) πππππππ πππππππππππ‘ ππ π(π₯)
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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 β intercept: Let x = 0 and solve x β intercept(s): Set the numerator = 0 and solve for x Choose other x β values, 2 on each side of an asymptote
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Example 1 What are the asymptotes of the function π π₯ = π₯β3 π₯β2 ?
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 βπ¦=1
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Example 2 What are the asymptotes of the function π π₯ = 2π₯β7 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 βπ¦=2
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Example 3 Graph the function, then state the domain and range and any asymptote equations. a) π π₯ = π₯ π₯ 2 β9 VA: Set the denominator = 0 and solve. π₯ 2 β9=(π₯β3)(π₯+3) π₯=β3, 3 HA: π₯ π₯ 2 βπ<πβπ¦=0 *Use calculator to find other points, then connect the points Domain: all real, π₯β β3, 3 Range: all real, π¦β 0
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Example 3 Graph the function, then state the domain and range and any asymptote equations. b) π π₯ = π₯ 2 π₯ 2 β1 VA: Set the denominator = 0 and solve. π₯ 2 β1=(π₯β1)(π₯+1) π₯=β1, 1 HA: π₯ 2 π₯ 2 βπ=πβπ¦=1 *Use calculator to find other points, then connect the points Domain: all real, π₯β β1, 1 Range: all real, π¦β 1
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Point Discontinuity A point where a graph is undefined, looks like a hole in the graph
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Point of discontinuity is at x = 3
Example 4 What is the point of discontinuity of the following functions? a) π₯ 2 β9 π₯β3 (π₯+3)(π₯β3) π₯β3 Point of discontinuity is at x = 3
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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₯β1) π₯+1 Point of discontinuity is at x = -1
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Point of discontinuity is at x = -5
Example 4 What is the point of discontinuity of the following functions? c) π₯ 2 +4π₯β5 π₯+5 (π₯+5)(π₯β1) π₯+5 Point of discontinuity is at x = -5
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Point of discontinuity is at x = 2
Example 5 Graph π π₯ = π₯ 2 β4 π₯β2 Hole at x = 2 (π₯+2)(π₯β2) π₯β2 Point of discontinuity is at x = 2 π π₯ =π₯+2
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Point of discontinuity is at x = -4
Example 6 Graph π π₯ = π₯ 2 β16 π₯+4 (π₯+4)(π₯β4) π₯+4 Point of discontinuity is at x = -4 π π₯ =π₯β4 Hole at x = -4
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