How do you find the vertical, horizontal or slant asymptotes for #R(x) = (3x) / (x^2 - 9)#?

Question

Lily Anderson · Accepted Answer

vertical asymptotes at x = ± 3
horizontal asymptote at y = 0

Vertical asymptotes occur when the denominator of a rational function tends to zero. To find the equation let the denominator equal zero. solve : # x^2 - 9 = 0 → (x-3)(x+3 ) = 0 → x = ± 3# Horizontal asymptotes occur as #lim_(x→±∞) f(x) → 0# If the degree of the numerator is less than the degree of the denominator , as in this question, numerator degree 1 , denominator degree 2. Then the equation is always y = 0. Here is the graph of the function to illustrate. graph{3x/(x^2-9) [-10, 10, -5, 5]}

Madelyn Anderson · Answer

undefined To find the vertical asymptotes of the function R(x) = (3x) / (x^2 - 9), you identify the values of x where the denominator becomes zero, as division by zero is undefined. In this case, the denominator x^2 - 9 becomes zero when x = ±3. Therefore, the vertical asymptotes are x = 3 and x = -3.

To find the horizontal asymptote, you compare the degrees of the numerator and denominator. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at y = 0. In this function, the degree of the numerator is 1 (since it's just x), and the degree of the denominator is 2. Therefore, there is a horizontal asymptote at y = 0.

To find any slant asymptotes, you perform polynomial long division or synthetic division. Since the degree of the numerator is less than the degree of the denominator, there are no slant asymptotes in this case.