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

##### 1 Answer
Nov 27, 2017

V.A. $x = 0$,$x = 1$
H.A. non
S.A. $x + 1$
HOLE. $\left(0 , 4\right)$

#### 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 $x = 0 \mathmr{and} x = 1$

. 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)

$\frac{{x}^{3} - 4 x}{{x}^{2} - x}$

factor x from numerator and denominator
rewrite 4 as ${2}^{2}$

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

factor

$\frac{x \left(x + 2\right) \left(x - 2\right)}{x \left(x - 1\right)}$
the common factor is x