Question

How do you find all the asymptotes for \[\dfrac{{2{x^3} + 11{x^2} + 5x - 1}}{{{x^2} + 6x + 5}}?\]

Answer 1

Hint: As we know that asymptote of a curve is the lien formed by the movement of the curve and the line moving continuously towards zero. We know that if the degree of the denominator is greater than the degree of the numerator, horizontal asymptote is at $ y = 0 $ . In this question, to find the value of the vertical, horizontal or slant asymptotes we see the degree of the denominator and equate to zero.

Complete step-by-step answer:
First we will find the value of the vertical asymptotes: This occurs when the rational function contains a zero denominator. So we can write it as $ {x^2} + 4 = 0 $ .
We will solve it now, $ {x^2} + 6x + 5 = {x^2} + 5x + x + 5 $ .
It can be further written as $ x(x + 5) + 1(x + 5) = (x + 5)(x + 1) $ .
So the vertical asymptote is $ x = - 5, - 1 $ .
As we can see that there are two solutions so there are two vertical asymptotes of this function.
In the horizontal asymptotes the value occurs if the numerator of a rational function has degree less than or equal to the degree of the denominator. But here this is not the case,
Let us consider the rational function $ R = \dfrac{{a{x^n}}}{{b{x^m}}} $ , where $ n $ is the degree of the numerator and m is the degree of the denominator.
We know that if $ n > m $ , then there is no horizontal asymptote.
Hence there are two vertical asymptotes and zero horizontal asymptote.

Note: We should note that for finding the slant asymptotes we have to use the concept that it occurs when the degree of the denominator of a rational function is one less than the degree of the numerator.. It is noted that if the degree of the denominator is $ > $ degree of the numerator, then the horizontal asymptote is $ y = 0 $ .