How do you find the asymptotes for #(2x^2 - x - 38) / (x^2 - 4)#?

Question

James Allen · Accepted Answer

vertical asymptotes at x = ± 2

horizontal asymptote at y = 2 As the denominator of a rational function tends to 0 there will be a vertical asymptote. solve #(x^2 - 4) =0 # #(x-2) = 0 or (x+2) = 0# hence vertical asymptotes at # x = ± 2# [horizontal asymptotes occur when # lim_(x→±∞) f(x) → 0#] when the degree of the numerator and denominator are equal the equation can be found by taking the ratio of leading coefficients. In this question they are equal , both of degree 2 and so # y = 2/1 rArr y = 2 # The graph shows the asymptotes. graph{(2x^2-x-38)/(x^2-4) [-10, 10, -5, 5]}

Cora Alexander · Answer

undefined To find the asymptotes of the function $ \frac{2x^2 - x - 38}{x^2 - 4} $, we need to identify the vertical and horizontal asymptotes.

Vertical asymptotes occur where the denominator of the rational function is equal to zero and the numerator is not zero. So, set the denominator $ x^2 - 4 $ equal to zero and solve for $ x $.

$ x^2 - 4 = 0 $  
$ (x + 2)(x - 2) = 0 $  
$ x = -2, 2 $

Therefore, the vertical asymptotes are $ x = -2 $ and $ x = 2 $.

To find horizontal asymptotes, we consider the behavior of the function as $ x $ approaches positive or negative infinity. Horizontal asymptotes exist if the degree of the numerator is less than or equal to the degree of the denominator.

The degree of the numerator is 2, and the degree of the denominator is also 2. Therefore, we need to divide the leading term of the numerator by the leading term of the denominator.

Leading term of the numerator: $ 2x^2 $  
Leading term of the denominator: $ x^2 $

$ \frac{2x^2}{x^2} = 2 $

Thus, the horizontal asymptote is $ y = 2 $.

Therefore, the asymptotes for the given function are $ x = -2 $, $ x = 2 $, and $ y = 2 $.