# How do you find the Vertical, Horizontal, and Oblique Asymptote given #g(x)=( x+3) / (x(x-5))#?

##### 1 Answer

#### Answer:

vertical asymptotes x = 0 , x = 5

horizontal asymptote y = 0

#### Explanation:

The denominator of g(x) cannot be zero as this is undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

solve: x(x -5) = 0 → x = 0 , x = 5 are the asymptotes

Horizontal asymptotes occur as

#lim_(xto+-oo),g(x)toc" (a constant)"#

#g(x)=(x+3)/(x^2-5x)# divide terms on numerator/denominator by highest power of x , that is

# x^2#

#(x/x^2+3/x^2)/(x^2/x^2-(5x)/x^2)=(1/x+3/x^2)/(1-5/x)# as

#xto+-oo,g(x)to(0+0)/(1-0)#

#rArry=0" is the asymptote"# Oblique asymptotes occur when the degree of the numerator > degree of denominator. This is not the case here (numerator-degree 1 , denominator-degree 2) Hence there are no oblique asymptotes.

graph{(x+3)/(x(x-5)) [-10, 10, -5, 5]}