Question

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

Answer 1

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]}

Answer 2

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]}