Find the asymptotes of the following curve: f x = x 2 x 2 - 1

Given curve is f x = x 2 x 2 - 1 . Finding where the curve is undefined, that is when x 2 - 1 = 0 ⇒ x = ± 1 Now, lim x → 1 + x 2 x 2 - 1 = ∞ and lim x → 1 - x 2 x 2 - 1 = ∞ lim x → - 1 + x 2 x 2 - 1 = ∞ and lim x → - 1 - x 2 x 2 - 1 = ∞ Hence, the required vertical asymptotes are x = - 1 and x = 1 . Now, lim x → + ∞ x 2 x 2 - 1 = lim x → + ∞ 1 1 - 1 x 2 = 1 lim x → - ∞ x 2 x 2 - 1 = lim x → - ∞ 1 1 - 1 x 2 = 1 Hence, y = 1 is a horizontal asymptote. There is no oblique asymptote because the degree of the numerator is less than or equal to the degree of the denominator.

Find the asymptotes of the following curve: f x = x 2 x + 1

Given curve is f x = x 2 x + 1 For Horizontal asymptote: lim x → + ∞ x 2 x + 1 = ∞ lim x → - ∞ x 2 x + 1 = - ∞ So no horizontal asymptotes. For Vertical asymptotes: Finding where the curve is undefined, that is when x + 1 = 0 ⇒ x = - 1 lim x → - 1 + x 2 x + 1 = ∞ lim x → - 1 - x 2 x + 1 = - ∞ So x = - 1 is a vertical asymptote. For Slant asymptotes: Since the polynomial in numerator has a higher degree than polynomial in the denominator, we have a slant asymptote. Slant asymptote occurs for a curve if lim x → ∞ f x - y = 0 , where f x is the curve and y is asymptote. ⇒ lim x → ∞ x 2 x + 1 - y = 0 ⇒ lim x → ∞ x 2 - 1 + 1 x + 1 - y = 0 ⇒ lim x → ∞ x + 1 x - 1 x + 1 + 1 x + 1 - y = 0 ⇒ lim x → ∞ x - 1 - y + lim x → ∞ 1 x + 1 = 0 ⇒ lim x → ∞ x - 1 - y = 0 This happens if, y = x - 1 So, slant asymptote is y = x - 1 . So, given curve has a vertical asymptote x = - 1 and slant asymptote y = x - 1 .

Find the asymptotes of the following curve: f x = 3 x x 2 + 2

Given curve is f x = 3 x x 2 + 2 . For Vertical asymptote: Solve x 2 + 2 = 0 ⇒ x 2 + 2 = 0 x 2 = - 2 so, curve is always defined So there are no vertical asymptotes. For Horizontal asymptote: lim x → + ∞ 3 x x 2 + 2 = lim x → + ∞ 3 1 + 2 x 2 = ± 3 lim x → - ∞ 3 x x 2 + 2 = lim x → - ∞ 3 1 + 2 x 2 = ± 3 So y = 3 and y = - 3 are horizontal asymptotes. For Slant asymptote: The Polynomial in numerator does not have a higher degree than the polynomial in the denominator. So no slant asymptotes.

Find the asymptotes of the following curve: f x = x 2 - 6 x - 1 x + 3

Given f x = x 2 - 6 x - 1 x + 3 For Horizontal asymptote: lim x → + ∞ x 2 - 6 x - 1 x + 3 = lim x → + ∞ x - 6 x - 1 x 1 + 3 x = ∞ lim x → - ∞ x 2 - 6 x - 1 x + 3 = lim x → - ∞ x - 6 x - 1 x 1 + 3 x = - ∞ So no horizontal asymptotes. For Vertical asymptote: x + 3 = 0 ⇒ x = - 3 lim x → - 3 + x 2 - 6 x - 1 x + 3 = ∞ lim x → - 3 - x 2 - 6 x - 1 x + 3 = - ∞ So x = - 3 is a vertical asymptote. For Slant asymptote: The polynomial in the numerator is a higher degree than the polynomial in the denominator. To find the slant asymptote you must divide the numerator by the denominator using either long division or synthetic division. x + 3 x 2 - 6 x - 1 x - 9 x 2 + 3 x - 9 x - 1 - 9 x - 27 26 By performing synthetic division we get f x = x - 9 + 26 x + 3 So y = x - 9 is a slant asymptote. So, given curve has a vertical asymptote x = - 3 and slant asymptote y = x - 9 .