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

2 Answers
Nov 4, 2016

Answer:

An horizontal asimptote in #y=0# and a vertical one at #x=7x#

Explanation:

The function can be factorized at the numerator and denominator so that it becomes #f(x)=8(x-6)/((x-6)(x-7))=# that under the existence condition excludind #x=6# and #x=7# can be rewritten

#f(x)=8/(x-7)#. This function has got a vertical asymptote in #x=7# and an horizontal one at #y=0# graph{8/(x-7) [-11.67, 28.33, -9.68, 10.32]}

Nov 5, 2016

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]}