Question

Find the asymptotes of the following curves:

(i) f(x) = x²/(x² - 1)

(ii) f(x) = x²/(x + 1)

(iii) f(x) = 3x/√(x² + 2)

(iv) f(x) = (x² - 6x - 1)/(x + 3)

(v) f(x) = (x² + 6x - 4)/(3x - 6)

Answer 1

(i) f(x) = x²/(x² - 1)

y = 1 is a horizontal asymptote 

So the asymptotes are x = -1, x = +1, y = 1

(ii) Since the numerator is of higher degree than the denominator we have a slant asymptote to find that asymptote we have to divide the numerator by the denominator So the slant asymptote is y = x – 1

Thus, x = - 1 is a vertical asymptote

(iii) f(x) = 3x/√(x² + 2)

lim_{x → ∞⁺} 3x/√(x² + 2) = 3 lim_{x → ∞} x/√(x² + 2)

∴ y = 3 and y = -3 are the horizontal asymptotes and there is no slant asymptote 

(iv) Since the numerator is of highest degree than the denominator. We have a slant asymptote to find it we have to divide numerator by the denominator.

So the equation of asymptotes is y = x – 9 and x = -3

(v) Since the numerator is of highest degree than the denominator. We have a slant asymptote to find it we have to divide the numerator by the denominator.

So the equation of asymptote is y = (x/3) + (8/3)

and 3x - 6 = 0

x = 2