Question

How do you identify all asymptotes or holes and intercepts for `f(x)=(x^3-4x)/(x^2-x)`?

Answer 1

V.A. `x=0`,`x=1`
H.A. non
S.A. `x+1`
HOLE. `(0,4)`

Explanation:

. For a function to have V.A. the function needs to have undefined points (zeros of denominator)
In this function, the zeros of the denominator are 0 and 1 therefore the vertical asymptotes are `x=0 and x=1`

. A graph will have a horizontal asymptote if the degree of the denominator is greater than the degree of the numerator
In this function, the degree of nominator is 3 and the degree of numerator is 2
`3>2` so there is no Horizontal asymptote

. Since the degree is one greater in the numerator, I know that I will have a slant asymptote.
Use polynomial long division to get the Slant/Oblique asymptote

S.A. : `x+1`

.There is a hole at (0,4)

`(x^3-4x)/(x^2-x)`

factor x from numerator and denominator
rewrite 4 as `2^2`

`(x(x^2-2^2))/(x(x-1)`

factor

`(x(x+2)(x-2))/(x(x-1))`
the common factor is x