How do you find the vertical, horizontal and slant asymptotes of: #y = (3)/(x+2)#?

Question

Sadie Alexander · Accepted Answer

vertical asymptote x = -2
horizontal asymptote y = 0

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation set the denominator equal to zero. solve : x + 2 = 0 → x = - 2 is the asymptote Horizontal asymptotes occur as #lim_(xto+-oo),ytoc" (a constant)"# divide terms on numerator/denominator by x #(3/x)/(x/x+2/x)=(3/x)/(1+2/x)# as #xto+-oo,yto0/(1+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 0 , denominator-degree 1 ) Hence there are no slant asymptotes. graph{3/(x+2) [-10, 10, -5, 5]}

Ariana Alvarez · Answer

undefined To find the vertical, horizontal, and slant asymptotes of the function $y = \frac{3}{x+2}$, follow these steps:

### Vertical Asymptote
A vertical asymptote occurs where the denominator is zero and the function is undefined. Set the denominator equal to zero and solve for $x$:
$$x + 2 = 0$$
$$x = -2$$
Therefore, the vertical asymptote is at $x = -2$.

### Horizontal Asymptote
A horizontal asymptote is determined by the degree of the polynomial in the numerator and the denominator. For $y = \frac{3}{x+2}$, the degrees are 0 in the numerator (since 3 is a constant) and 1 in the denominator. When the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote is $y = 0$.

### Slant (Oblique) Asymptote
A slant or oblique asymptote occurs when the degree of the polynomial in the numerator is exactly one more than the degree of the polynomial in the denominator. Since this is not the case for $y = \frac{3}{x+2}$ (the numerator's degree is less than the denominator's), there is no slant asymptote for this function.

In summary:
- Vertical asymptote: $x = -2$
- Horizontal asymptote: $y = 0$
- There is no slant asymptote for this function.