How do you find vertical, horizontal and oblique asymptotes for #(2x+4)/(x^2-3x-4)#?

Question

James Allen · Accepted Answer

vertical asymptotes x = -1 , x = 4
horizontal asymptote y = 0

When the denominator of a rational function tends to zero, vertical asymptotes occur. Let the denominator equal zero to find the equation or equations. solve: # x^2 - 3x - 4 = 0 → (x -4)(x + 1) = 0 # # rArr x = -1 , x = 4 " are the asymptotes" # Horizontal asymptotes occur as # lim_(xto+-oo) f(x) to 0 # The equation is always y = 0 when the degree of the numerator is less than the degree of the denominator, as it is in this instance. There are no oblique asymptotes in this case because oblique asymptotes arise when the degree of the numerator is greater than the degree of the denominator. The function's graph is shown here: graph{(2x+4)/(x^2-3x-4) [-10, 10, -5, 5]}

Greyson Brown · Answer

undefined To find the vertical asymptotes, set the denominator equal to zero and solve for x. For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is less than the degree of the denominator, there is a horizontal asymptote at y = 0. If the degrees are equal, divide the leading coefficients to find the horizontal asymptote. For oblique asymptotes, divide the numerator by the denominator using polynomial long division, and the quotient represents the oblique asymptote.