Asymptotes the Derivative Is Dy = − , Which Simpliﬁes to Dx X Ex (2X 1) Dy = −

Home , Asymptote

curve sketching curve sketching Critical Points MCV4U: Calculus & Vectors Recap ex Determine any critical values for y = . √x

x x 1 e √x e 2√x Vertical, Horizontal and Oblique Asymptotes The derivative is dy = − , which simpliﬁes to dx x ex (2x 1) dy = − . J. Garvin dx 2√x3 Since ex = 0, the only critical point occurs when 2x 1 = 0, 1 6 − or x = 2 . e1/2 1 √ When x = 2 , the critical value is y = = 2e. 1 2 q

J. Garvin— Vertical, Horizontal and Oblique Asymptotes Slide 1/22 Slide 2/22

curve sketching curve sketching Vertical Asymptotes Vertical Asymptotes While polynomial functions do not have any vertical Example asymptotes, they often occur in rational functions. 2x Determine any vertical asymptotes for f (x) = . For rational functions involving polynomials, there will be a x2 2x 3 − − vertical asymptote at x = k if x k is a factor of the − denominator, provided it is not also a factor of the numerator. Factoring the denominator, we can rewrite the function as 2x If x k is a factor of both the numerator and the f (x) = . − (x 3)(x + 1) denominator, it is a point discontinuity (hole) instead. − Therefore, there are vertical asymptotes at x = 3 and Other functions may contain vertical asymptotes at values x = 1. where the denominator equates to zero. For example, − y = tan x has vertical asymptotes for all values of x where sin x cos x = 0, since tan x = cos x .

J. Garvin— Vertical, Horizontal and Oblique Asymptotes J. Garvin— Vertical, Horizontal and Oblique Asymptotes Slide 3/22 Slide 4/22

curve sketching curve sketching Vertical Asymptotes Vertical Asymptotes A graph of y shows how the function approaches the Example asymptotes at x = 3 and x = 1. − Determine any vertical asymptotes for x 1 f (x) = − . x3 x2 + 2x 2 − − Factor the denominator by grouping.

x3 x2 + 2x 2 = x2(x 1) + 2(x 1) − − − − = (x2 + 2)(x 1) − 1 Thus, we can rewrite the function as f (x) = , x = 1. x2 + 2 6 1 1 Since f (1) = 3 , there is a point discontinuity at 1, 3 .

J. Garvin— Vertical, Horizontal and Oblique Asymptotes J. Garvin— Vertical, Horizontal and Oblique Asymptotes Slide 5/22 Slide 6/22 curve sketching curve sketching Vertical Asymptotes Horizontal Asymptotes Since x2 + 2 2, the denominator is never equal to zero. A horizontal asymptote is a value that a function approaches ≥ Thus, there are no vertical asymptotes. as x or x . → ∞ → −∞ Formally, if lim f (x) = L, then y = L is a horizontal x asymptote for→±∞f (x). In previous courses, you have probably determined the equations of horizontal asymptotes by dividing polynomial terms by the highest power, then observing what happens as either x or x . → ∞ → −∞ This is the same technique used here, but formalized using limit notation.

J. Garvin— Vertical, Horizontal and Oblique Asymptotes J. Garvin— Vertical, Horizontal and Oblique Asymptotes Slide 7/22 Slide 8/22

curve sketching curve sketching Horizontal Asymptotes Horizontal Asymptotes Example Evaluate the limits as x and x . → ∞ → −∞ 2 4x 3x 3 3 Determine any horizontal asymptotes for y = − . 4 4 0 4 4 0 2x2 + 5 lim − x = − lim − x = − x 5 2 + 0 x 5 2 + 0 →∞ 2 + x2 ! →−∞ 2 + x2 ! Factor the highest power, x2, from each term. = 2 = 2

4x2 3x x2 4 3 In this case, the line y = 2 is a horizontal asymptote as − = − x 2x2 + 5 x2 2 + 5 x and as x . x2 → ∞ → −∞ 4 3 This is not always the case. Some functions may have = − x 5 diﬀerent asymptotes as x gets very large, or very small. 2 + x2

J. Garvin— Vertical, Horizontal and Oblique Asymptotes J. Garvin— Vertical, Horizontal and Oblique Asymptotes Slide 9/22 Slide 10/22

curve sketching curve sketching Horizontal Asymptotes Horizontal Asymptotes A graph of y shows the function approaching y = 2 as Example x and as x . x + 2 → ∞ → −∞ Determine any horizontal asymptotes for y = . √9x2 + 1

The square root complicates matters here, so we need to approach it with an analytical perspective. As x , 9x2 will have a much greater effect than the +1 → ∞ in the denominator. When x is sufficiently large, the 1 will be almost insignificant. Therefore, the denominator may be approximated by √9x2, or 3x, so we can use x as the highest power. x 1 + 2 1 + 2 This gives us y = x , or y = x . √9x2+1 √9x2+1 x x x J. Garvin— Vertical, Horizontal and Oblique Asymptotes J. Garvin— Vertical, Horizontal and Oblique Asymptotes Slide 11/22 Slide 12/22 curve sketching curve sketching Horizontal Asymptotes Horizontal Asymptotes The square root continues to be an issue in the denominator, Finally, we can evaluate the limits as x and x . → ∞ → −∞ so we need to simplify further. 1 + 2 1 + 0 1 + 2 1 + 0 Since x is positive when x , we can use the fact that lim x = lim x = x x 2 → ∞ 1 √9 + 0 1 √9 + 0 x = √x . Similarly, x is negative when x , so →∞ 9 + 2 →−∞ 9 + 2 x − x − √ 2 → −∞ x = x . q = 1 q = 1 − 3 − 3 1 + 2 1 + 2 1 + 2 1 + 2 x = x x = x Therefore, there are two horizontal asymptotes for the √9x2+1 2 √9x2+1 2 9x +1 9x +1 function. As x , there is a horizontal asymptote at √x2 x2 √x2 x2 → ∞ − − y = 1 , and as x , there is one at y = 1 . q1 + 2 q1 + 2 3 → −∞ − 3 = x = x 9 + 1 9 + 1 x2 − x2 q q

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curve sketching curve sketching Horizontal Asymptotes Oblique Asymptotes A graph of y shows both horizontal asymptotes. An oblique asymptote, or slant asymptote, is a linear asymptote that is neither vertical nor horizontal. For rational functions involving polynomials, oblique asymptotes occur when the numerator has a degree one greater than the denominator. x2 + 3x For example, the rational function y = has an x 2 oblique asymptote with equation y = x + 5,− whereas the x rational function y = does not have an oblique x2 4 asymptote. − Rational functions involving polynomials may contain a horizontal asymptote or an oblique asymptote, but not both. The equation of an oblique asymptote may be found using long or synthetic division. J. Garvin— Vertical, Horizontal and Oblique Asymptotes J. Garvin— Vertical, Horizontal and Oblique Asymptotes Slide 15/22 Slide 16/22

curve sketching curve sketching Oblique Asymptotes Oblique Asymptotes Example A graph of f (x) is below. Determine the equation of any oblique asymptotes for x2 + 5x 4 f (x) = − . x + 3

Use x = 3 and synthetic division to ﬁnd the quotient, − which is the equation of the asymptote.

1 5 4 − 3 3 6 − − − 1 2 10 −

Therefore, f (x) has an oblique asymptote with equation y = x + 2. J. Garvin— Vertical, Horizontal and Oblique Asymptotes J. Garvin— Vertical, Horizontal and Oblique Asymptotes Slide 17/22 Slide 18/22 curve sketching curve sketching Other Asymptotes Other Asymptotes While the asymptotes covered in this course are limited to Example vertical, horizontal and oblique, an asymptote may assume Determine the equation of the parabolic asymptote to the shape of any function. x3 + 4x2 2x + 1 y = − . For example, a rational function with a numerator of order 3 x 2 − and a denominator of order 1 will have a parabolic asymptote, while a quintic numerator and a quadratic Use synthetic division to determine the equation. denominator will result in a cubic asymptote. 1 4 2 1 In general, a rational function with a polynomial or order n − and a denominator of order m, with n > m, will have a 2 2 12 20 polynomial asymptote with order n m. − 1 6 10 21

The equation of the parabolic asymptote is y = x2 + 6x + 10, or y = (x + 3)2 + 1.

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curve sketching curve sketching Other Asymptotes Questions? A graph of y shows how the function approaches the asymptote.

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