How do you find the asymptotes for #Q(x) = (2x^2)/ (x^2 - 5x - 6)#?

Question

Theodore Amidon · Accepted Answer

vertical asymptotes x = -1 , x = 6
horizontal asymptote y = 2

Vertical asymptotes occur when the denominator of a rational function tends to zero. To find the equation/s let the denominator equal zero. solve : #x^2 - 5x - 6 = 0 → (x-6)(x+1) = 0# # rArr x = -1 , x = 6" are the asymptotes " # Horizontal asymptotes occur as # lim_(x→±∞) f(x) → 0 # divide all terms on numerator/denominator by # x^2 # # ((2x^2)/x^2)/(x^2/x^2 - (5x)/x^2 - 6/x^2)= 2/(1-5/x -6/x^2) # As x → ∞ , # 5/x" and " 6/x^2 → 0 # #rArr y = 2/1 = 2 " is the asymptote " # Here is the graph of the function. graph{(2x^2)/(x^2-5x-6) [-20, 20, -10, 10]}

Cora Alexander · Answer

undefined To find the asymptotes of the function $ Q(x) = \frac{2x^2}{x^2 - 5x - 6} $:

1. Identify the vertical asymptotes by finding the values of $ x $ for which the denominator becomes zero. These are the values that make the function undefined.
2. Identify the horizontal asymptotes by examining the behavior of the function as $ x $ approaches positive or negative infinity. If the degrees of the numerator and the denominator are equal, the horizontal asymptote is the ratio of the leading coefficients.
3. If there are any slant (oblique) asymptotes, determine them by performing polynomial long division if the degree of the numerator is one greater than the degree of the denominator.

Once you have found the vertical, horizontal, and slant asymptotes, you have identified all the asymptotes of the function.