How do you find the asymptote and holes for #y=(x-6)/(x^2+5x+6)#?

Question

Serenity Johnson · Accepted Answer

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

First step is to factorise the function. #rArr(x-6)/((x+2)(x+3))# Since there are no common factors on numerator/denominator this function has no holes. Vertical asymptotes occur as the denominator of a rational tends to zero. To find the equation/s set the denominator equal to zero. solve: (x+2)(x+3) = 0 → x = -3 , x= -2 #rArrx = -3" and "x=-2" are the asymptotes"# Horizontal asymptotes occur as #lim_(xto+-oo),yto0# When the degree of the numerator < degree of the denominator as is the case here (numerator-degree 1 , denominator-degree 2 ) then the equation of the asymptote is always y = 0 graph{(x-6)/(x^2+5x+6) [-10, 10, -5, 5]}

Elias Ackerman · Answer

undefined To find the asymptotes and holes for the rational function $y = \frac{x - 6}{x^2 + 5x + 6}$:

1. **Vertical Asymptotes**: Set the denominator equal to zero and solve for $x$. These values will give the vertical asymptotes, but check if they cancel out with any factors from the numerator to determine if there are any removable discontinuities (holes).

$$x^2 + 5x + 6 = 0$$
   $$x^2 + 3x + 2x + 6 = 0$$
   $$x(x + 3) + 2(x + 3) = 0$$
   $$(x + 2)(x + 3) = 0$$

Vertical asymptotes occur where the denominator is zero, so $x = -2$ and $x = -3$.

2. **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 x-axis (y = 0) is the horizontal asymptote. If the degree of the numerator is equal to the degree of the denominator, divide the leading coefficients to find the horizontal asymptote.

Degree of the numerator = 1
   Degree of the denominator = 2

Since the degree of the numerator is less than the degree of the denominator, there is a horizontal asymptote at y = 0.

3. **Holes**: Check if any factors from the numerator cancel out with factors from the denominator at the vertical asymptotes. If any factors cancel out, these points are holes in the graph.

Substitute the values of the vertical asymptotes into the function:

For $x = -2$:
   $$y = \frac{(-2) - 6}{((-2)^2 + 5(-2) + 6)}$$
   $$y = \frac{-8}{4 - 10 + 6}$$
   $$y = \frac{-8}{0}$$

For $x = -3$:
   $$y = \frac{(-3) - 6}{((-3)^2 + 5(-3) + 6)}$$
   $$y = \frac{-9}{9 - 15 + 6}$$
   $$y = \frac{-9}{0}$$

Both values result in division by zero, indicating holes at these points.

Therefore, the vertical asymptotes are $x = -2$ and $x = -3$, the horizontal asymptote is $y = 0$, and there are holes at $x = -2$ and $x = -3$.