Question

How do you find the vertical, horizontal and slant asymptotes of: `y=(8x-48)/(x^2-13x+42)`?

Answer 1

vertical asymptote x = 7
horizontal asymptote y = 0

Explanation:

The first step here is to factorise and simplify y.

`(8x-48)/(x^2-13x+42)=(8cancel((x-6)))/((x-7)cancel((x-6)))=8/(x-7)`

The denominator of this rational function cannot be zero as this would lead to division by zero which is undefined.By setting the denominator equal to zero and solving for x we can find the value that x cannot be and if the numerator is also non-zero for this value of x then it must be a vertical asymptote.

solve : x - 7 = 0 → x = 7 is the asymptote

Horizontal asymptotes occur as

`lim_(xto+-oo),ytoc" (a constant)"`

divide terms on numerator/denominator by x

`(8/x)/(x/x-7/x)=(8/x)/(1-7/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{8/(x-7) [-20, 20, -10, 10]}