How do you find vertical, horizontal and oblique asymptotes for #y = (4x^3 + x^2 + x + 5 )/( x^2 + 3x)#?

Question

How do you find vertical, horizontal and oblique asymptotes for #y = (4x^3 + x^2 + x + 5 )/(  x^2 + 3x)#?

James Allen · Accepted Answer

Vertical asymptotes are #x=0# and #x=-3# and oblique asymptote is #y=4x#.

To find the asymptotes for function #(4x^3+x^2+x+5)/(x^2+3x)#, let us first start with vertical asymptotes, which are given by putting denominator equal to zero or #x^2+3x=0# i.e. #x(x+3)=0# and hence #x=-3# and #x=0# are two vertical asymptotes. As the highest degree of numerator is #3# and is just one higher than that of denominator i.e. #2#, we would have obique asymptote given by #y=4x^3/x^2=4x# and there is no horizontal asymptote. Hence, Vertical asymptotes are #x=0# and #x=-3# and oblique asymptote is #y=4x#. The graph is as shown below. graph{(4x^3+x^2+x+5)/(x^2+3x) [-40, 40, -20, 20]}

Lincoln Anderson · Answer

undefined To find the vertical asymptotes, set the denominator equal to zero and solve for x. Any value of x that makes the denominator zero will result in a vertical asymptote.

For the given function y = (4x^3 + x^2 + x + 5) / (x^2 + 3x), the denominator is x^2 + 3x. Setting this equal to zero gives us x(x + 3) = 0. Solving for x, we find that x = 0 and x = -3. These are the vertical asymptotes.

To find the horizontal asymptotes, 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 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 than the degree of the denominator, there is no horizontal asymptote.

For the given function, the degree of the numerator is 3 and the degree of the denominator is 2. Since the degree of the numerator is greater, there is no horizontal asymptote.

To find oblique asymptotes, perform polynomial long division or synthetic division to divide the numerator by the denominator. The quotient obtained represents the equation of the oblique asymptote.

In this case, performing long division or synthetic division, we get a quotient of 4x - 3 with a remainder of (18x + 15) / (x^2 + 3x). As x approaches positive or negative infinity, the remainder term tends to zero, leaving y = 4x - 3 as the equation of the oblique asymptote.