Horizontal and Vertical Asymptotes from section 2.6 Definition: In either of the cases f(x) = L or f(x) = L we say that the x x horizontal line y = L is a horizontal asymptote of the function f. Note: For a rational function R(x) = P (x) Q(x) (P and Q being polynomials), if the degrees of the numerator and denominator are equal, then there exists a horizontal asymptote; the following example describes this.
Example: Find the horizontal asymptotes of the function y = 2x3 2 3x 3 + x 2 + x + 1 Solution: So, the line y = 2 3 x ± 2x 3 2 3x 3 + x 2 + x + 1 = x ± is the only horizontal asymptote. 2 2 x 3 3 + 1 x + 1 x 2 + 1 x 3 = 2 3 Page 2
Definition: In either of the cases f(x) = extreme ( or + ) x a OR f(x) = extreme x a + we say that the vertical line x = a is a vertical asymptote of the function f. Note: The vertical asymptotes of a function R(x) = P (x) Q(x) (not necessarily a rational function) are amongst the lines x = a where a is a root of the denominator. To decide which one of these is a vertical asymptote you should evaluate the its x a + R(x) and x a R(x) and if one of them equals either of ± then we have vertical asymptote at that point a. Here are two examples: Page 3
Example: Find the vertical and horizontal asymptotes of the function y = x2 16 x 2 + x 6 Solution: Horizontal asymptotes: x 6 x 2 + x 6 = x x 2 ( 1 16 x 2 ) x 2 ( 1 + 1 x 6 x 2 ) = x 1 16 x 2 1 + 1 x 6 x 2 = 1 The it would result in the same answer, so the line y = 1 is the only horizontal x asymptote. Vertical asymptotes: x 2 + x 6 = 0 x = 3, 2 6 x ( 3) + x 2 + x 6 = 6 x ( 3) + (x + 3)(x 2) = ( 7) (0 + )( 5) = so the line x = 3 is a vertical asymptote. 6 x 2 + x 2 + x 6 = 6 x 2 + (x + 3)(x 2) = ( 12) (5)(0 + ) = so the line x = 2 is a vertical asymptote. Page 4
Example: Find the vertical asymptotes of the function y = x2 2x 3 x 2 + 3x + 2 Solution: The roots of the denominator are x = 2, 1 x 2 2x 3 x ( 2) + x 2 + 3x + 2 = x 2 2x 3 x ( 2) + (x + 2)(x + 1) = 5 (0 + )( 1) = So, the line x = 2 is a vertical asymptote. x 2 2x 3 x ( 1) + x 2 + 3x + 2 = (x + 1)(x 3) x ( 1) + (x + 2)(x + 1) = (x 3) x ( 1) + (x + 2) = 4 1 = 4 And the it x ( 1) x2 2x 3 would result in the same value of 4. Therefore, although x 2 +3x+2 the point x = 1 is a root of the denominator however we do not have a vertical asymptote there. So remember that a root of the denominator is not necessarily a vertical asymptote Page 5
for vertical asymptotes Step1 : Find the candidates by setting the denominator equal to zero Step2 : Check whether each candidate is a good candidate or a bad candidate for horizontal asymptotes { Take both its f(x) and x f(x) to find the horizontal asymptotes Page 6
Example: Find the vertical asymptote(s) of the function y = 2x 2 + 3x + 1 Solution: x = b ± b 2 4ac 2a The roots of the denominator are: = 3 ± 9 8 4 = 3 ± 1 4 1 = 1 2 Now we check which one of these is a good candidate: Checking the line x = 1 x ( 1) + 2x 2 + 3x + 1 = ()(x + 1) x ( 1) + (2x + 1)(x + 1) = () x ( 1) + (2x + 1) = 2 1 = 2 x ( 1) 2x 2 + 3x + 1 = ()(x + 1) x ( 1) (2x + 1)(x + 1) = () x ( 1) (2x + 1) = 2 1 = 2 None of these two left-hand and right-hand equal either of ±, so the line x = 1 is not a vertical asymptote. Checking the line x = 1 2 x ( 1 2x 2 + 3x + 1 = ()(x + 1) 2 )+ x ( 1 (2x + 1)(x + 1) = 3 2 2 )+ x ( 1) + 0 + = So, the line x = 1 2 is a vertical asymptote of the graph of the function. Page 7
Example: Find the horizontal asymptote(s) of the function y = 2x 2 + 3x + 1 Solution: To search for horizontal asymptotes we must take the its and. 2x 2 + 3x + 1 = So, the line y = 1 2 2x 2 + 3x + 1 = x 2 ( 1 1 x 2 ) x 2 ( 2 + 3 x + 1 x 2 ) = is a horizontal asymptote. x 2 ( 1 1 x 2 ) x 2 ( 2 + 3 x + 1 x 2 ) = 1 1 x 2 2 + 3 x + 1 x 2 = 1 0 2 + 0 + 0 = 1 2 1 1 x 2 2 + 3 x + 1 x 2 = 1 0 2 + 0 + 0 = 1 2 So, the line y = 1 2 is a horizontal asymptote. But this line had been found before. So, in summary, we have a horizontal asymptote y = 1 2 and no vertical asymptote. Page 8
Example: Find the vertical asymptote(s) of the function y = x 2 + 3x 4 Solution: x = b ± b 2 4ac 2a The roots of the denominator are: = 3 ± 9 + 16 2 = 3 ± 5 2 4 = 1 Now we check which one of these is a good candidate: Checking the line x = 4 x ( 4) + x 2 + 3x 4 = x ( 4) + (x + 4)() = 1 x ( 4) + (x + 4) = 1 0 + = + So, the line x = 4 is a vertical asymptote. Now checking the line x = 1 x 1 x 2 + 3x 4 = x 1 (x + 4)() = 1 x 1 (x + 4) = 1 5 x 1 + x 2 + 3x 4 = x 1 + (x + 4)() = 1 x 1 + (x + 4) = 1 5 None of these two left-hand and right-hand equal either of ±, so the line x = 1 is not a vertical asymptote. Page 9
Example: Find the horizontal asymptote(s) of the function y = x 2 + 3x 4 Solution: To search for horizontal asymptotes we must take the its and. 0 x 2 + 3x 4 = x ( ) 1 1 x x ( 2 1 + 3 ) = 4 x x 2 1 1 x x ( 1 + 3 x 4 x 2 ) = 1 ( )(1) = So, the line y = 0 is a horizontal asymptote. x 2 + 3x 4 = ) x ( 1 1 x x ( 2 1 + 3 ) = 4 x x 2 1 1 x x ( 1 + 3 x 4 x 2 ) = 1 ( )(1) = 0 So, the line y = 0 is a horizontal asymptote. But this line had been found before. So, in summary, we have a horizontal asymptote y = 0 and a vertical asymptote x = 4. Note To have a horizontal asymptote the degree of the denominator must be equal or larger than the degree of the numerator. Page 10