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

1 Answer
Nov 27, 2017

Answer:

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
enter image source here
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