Question

How do you find vertical, horizontal and oblique asymptotes for `f(x) =( 10x^2)/(x-4)`?

Answer 1

Vertical asymptote at `x = 4`, no horizontal asymptote and
slant asymptote is `y= 10x +40`

Explanation:

`f(x)= (10 x^2)/(x-4)`

The vertical asymptotes will occur at those values of `x` for which

the denominator is equal to zero. `x-4=0 or x=4`

Thus, the graph will have a vertical asymptote at `x = 4`

The degree of the numerator is `2` and the degree of the

denominator is `1`. Since the larger degree occurs in the

numerator, the graph will have no horizontal asymptote.

Since the degree of the numerator is greater by margin of `1`

we have a slant asymptote , found by quotient on long division.

`f(x)= (10 x^2)/(x-4)= (10 x +40) + 160/(x-4)`

Slant asymptote is `y= 10x +40`. [Ans]