How do you find vertical, horizontal and oblique asymptotes for # f(x)= (5x-15)/(2x+4)#?

Question

Theodore Amidon · Accepted Answer

vertical asymptote x = -2
horizontal asymptote #y = 5/2 # Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation set the denominator equal to zero. solve : 2x + 4 = 0 → x = -2 is the asymptote. Horizontal asymptotes occur as #lim_(x to +-oo) f(x) to 0 # divide terms on numerator/denominator by x #((5x)/x - 15/x)/((2x)/x + 4/x) = (5-15/x)/(2+4/x) # as # xto+-oo , y to (5-0)/(2+0)# # rArr y = 5/2 " is the asymptote " # Oblique asymptotes occur when the degree of the numerator > degree of denominator. This is not the case here hence there are no oblique asymptotes. graph{(5x-15)/(2x+4) [-14.24, 14.24, -7.11, 7.13]}

Anna Allen · Answer

undefined To find the vertical asymptote, set the denominator equal to zero and solve for x. For this function, the vertical asymptote occurs at x = -2.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at y = 0. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote. If the degree of the numerator is greater, there is no horizontal asymptote. In this case, since the degree of the numerator (1) is less than the degree of the denominator (1), the horizontal asymptote is at y = 5/2.

For oblique asymptotes, perform long division or polynomial division on the numerator and denominator. If the result is a polynomial plus a proper fraction (where the degree of the numerator is less than the degree of the denominator), the oblique asymptote is the polynomial part of the quotient. If the degrees are equal or the numerator's degree is greater, there is no oblique asymptote. In this case, since the degree of the numerator (1) is less than the degree of the denominator (1), there is no oblique asymptote.