Exercise 1: computing asymptotes - Emory University
Since the denominator vanishes for x = 1 we may have a vertical asymptote If we check, we get lim x 1+ f(x) = +1 lim x 1 f(x) = 1 thus x = 1 is a vertical asymptote There are no horizontal asymptote, since lim x1 f(x) = 1 Let’s check if there are oblique asymptotes lim x1 f(x) x = lim x1 x2 + 1 x2 + x = 1 So we have an oblique
Lesson 23 Exercises, pages 114–121
oblique asymptote Determine: x2 (4 x) Write: 2x2 (x 4) 4 10 0 4 16 1 4 16 The quotient is x 4; so the equation of the oblique asymptote is: y x 4 There are no common factors The degree of the numerator is 1 more than the degree of the denominator, so there is an oblique asymptote Determine: ( 2x 3x 1) (x 2) 2 23 1 4 14 27 13
Lesson 32 Exercises, pages 151–158 Exercises
Lesson 3 2 Exercises, pages 151–158 i) The vertical asymptote has equation x ˜ ˚1 The oblique asymptote has slope ˚1 and y-intercept 1, so its equation is y ˜ ˚x ˛ 1 ii) The domain is: x 3˚1 i) The vertical asymptote has equation x ˜ 2 The oblique asymptote has slope 2 and y-intercept ˚1, so its equation is y ˜ 2x ˚ 1 ii) The
Solved Problems on Limits at Infinity, Asymptotes and
is an oblique asymptote at infinity and negative infinity ----- Snezhana Gocheva-Ilieva, Plovdiv University ----- 21/24 ; Representation of the graphics of the
Lesson 33 Exercises, pages 167–173 Exercises
The vertical asymptote has equation: x ˜ 2 There is also an oblique asymptote Determine: (2x2 ˚ 3x ˛ 1) ˝ (x ˚ 2) The quotient is 2x ˛ 1; so the equation of the oblique asymptote is y ˜ 2x ˛ 1 Draw broken lines for the asymptotes Close to the vertical asymptote: When x ˜ 0, y ˜ ˚0 5 When y ˜ 0, 2x2 ˚ 3x ˛ 1 ˜ 0 (2x ˚ 1)(x
Vertical and Horizontal Asymptotes
Oblique Asymptotes and Point Discontinuity An oblique asymptote is an asymptote that is neither horizontal nor vertical In some cases, graphs of rational functions may have point discontinuity, which looks like a hole in the graph That is because the function is undefined at that point Oblique Asymptotes If f(x) = − a(x), b(x)
Math 1A - So you think you can slant (asymptote)?
Hence, the slant asymptote to f at 1is: y = x+2 (which is the same answer we found above) This procedure is also good to show a function cannot have a slant asymptote Problem Show that f(x) = x+ p x does not have a slant asymptote at 1 We’ll do a proof by contradiction Suppose f has a slant asymptote y = ax + b Then we must have: a = lim
Asymptote Practice Problems And Answers
Oblique Asymptote or Slant Asymptote Some curves have asymptotes that are oblique, that is, neither horizontal nor vertical If then the line y = mx + b is called the oblique or slant asymptote because the vertical distances between the curve y = f(x) and the line y = mx + b approaches 0
College Algebra Worksheet (8) Multiple Choice Questions on
a the graph of f has a vertical asymptote b the graph of f has a hole on the x axis c f(x) = 0 d the graph of f has a hole at (3,2) 3 Which of the following functions has an oblique asymptote? a f(x) = x5 +1 x4 +3x2 +2 b f(x) = x2 +1 x3 −x2 − 1 c f(x) = 4x2 +x+1 x2 d f(x) = x5 x2 −1 4 Find the equation of the oblique
RATIONAL FUNCTIONS
slant or oblique asymptotes If the degree of the numerator (n), is exactly one more than the degree of the denominator (d), then the graph will have an oblique asymptote To find an oblique asymptote, it is necessary to divide the numerator by the denominator You can use polynomial long division, or the CAS (on a CAS use propFrac command)
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