How do you find the vertical, horizontal or slant asymptotes for #x/(x^2+ 5x +6)#?

Question

Elijah Allen · Accepted Answer

vertical asymptotes x = -3 , x = -2
horizontal asymptote y = 0

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation/s set the denominator equal to zero. solve : #x^2+5x+6=0rArr(x+2)(x+3)=0# #rArrx=-3,x=-2" are the asymptotes"# Horizontal asymptotes occur as #lim_(xto+-oo),f(x)toc" (a constant)"# divide terms on numerator/denominator by #x^2# #(x/x^2)/(x^2/x^2+(5x)/x^2+6/x^2)=(1/x)/(1+5/x+6/x^2)# as #xto+-oo,f(x)to0/(1+0+0)# #rArry=0" is the asymptote"# Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 1 , denominator-degree 2 ) Hence there are no slant asymptotes. graph{x/(x^2+5x+6) [-10, 10, -5, 5]}

Serenity Jones · Answer

undefined To find the vertical asymptotes, set the denominator equal to zero and solve for $x$. For the function $ \frac{x}{x^2 + 5x + 6} $, the denominator $x^2 + 5x + 6$ equals zero when $x = -2$ and $x = -3$. So, there are vertical asymptotes at $x = -2$ and $x = -3$.

To find horizontal asymptotes, 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 degree of the numerator is equal to the degree of the denominator, divide the leading coefficients to find the horizontal asymptote. If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

In this case, the degree of the numerator (1) is less than the degree of the denominator (2). So, the horizontal asymptote is at $y = 0$.

To find the slant (oblique) asymptote, if the degree of the numerator is exactly one more than the degree of the denominator, you can use polynomial long division to find the equation of the slant asymptote. In this case, since the degree of the numerator is less than the degree of the denominator, there is no slant asymptote.

Therefore, the vertical asymptotes are at $x = -2$ and $x = -3$, and the horizontal asymptote is at $y = 0$. There is no slant asymptote.