# How do you find the Vertical, Horizontal, and Oblique Asymptote given #(8x-48)/(x^2-13x+42)#?

##### 2 Answers

#### Answer:

An horizontal asimptote in

#### Explanation:

The function can be factorized at the numerator and denominator so that it becomes

#### Answer:

vertical asymptote at x = 7

horizontal asymptote at y = 0

#### Explanation:

The denominator of the function cannot be zero as this would make the function undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

#(8x-48)/(x^2-13x+42)=(8cancel((x-6)))/(cancel((x-6))(x-7))=8/(x-7)# There is an excluded value at x = 6.

solve:

#x-7=0rArrx=7" is the asymptote"# Horizontal asymptotes occur as

#lim_(xto+-oo),f(x)toc" ( a constant)"# divide terms on numerator/denominator by x.

#f(x)=(8/x)/(x/x-7/x)=(8/x)/(1-7/x)# as

#xto+-oo,f(x)to0/(1-0)#

#rArry=0" is the asymptote"# Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here hence there are no oblique asymptotes.

graph{(8)/(x-7) [-20, 20, -10, 10]}