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

Question

Serenity Johnson · Accepted Answer

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

. 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

Anna Allen · Answer

undefined To identify the asymptotes, holes, and intercepts for the function $ f(x) = \frac{x^3 - 4x}{x^2 - x} $, follow these steps:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $ x $. Any solutions will be vertical asymptotes, except where they are canceled out by factors in the numerator.
   $ x^2 - x = 0 $ when $ x = 0 $ and $ x = 1 $.
   However, at $ x = 0 $, the factor in the numerator is also zero, so it cancels out the vertical asymptote. Thus, the only vertical asymptote is $ x = 1 $.

2. Horizontal asymptotes: If the degree of the numerator is greater than or equal to the degree of the denominator, there are no horizontal asymptotes. In this case, the degree of the numerator (3) is greater than the degree of the denominator (2), so there are no horizontal asymptotes.

3. Holes: Check if there are any common factors between the numerator and the denominator that can be canceled out. If there are, these points will be holes in the graph.
   Factoring the numerator: $ x(x^2 - 4) $
   Factoring the denominator: $ x(x - 1) $
   We can cancel out the common factor $ x $.
   So, there is a hole at $ x = 0 $.

4. $ x $-intercepts: Find the values of $ x $ for which $ f(x) = 0 $.
   Setting $ f(x) = 0 $, we get:
   $ \frac{x(x^2 - 4)}{x(x - 1)} = 0 $
   The $ x $-intercepts are where the numerator equals zero, which are $ x = 0 $ and $ x = ±2 $.

5. $ y $-intercept: Find the value of $ f(0) $.
   $ f(0) = \frac{(0)^3 - 4(0)}{(0)^2 - 0} = \frac{0 - 0}{0} = 0 $
   So, the $ y $-intercept is at the point (0, 0).

To summarize:
- Vertical asymptote: $ x = 1 $
- Horizontal asymptote: None
- Hole: At $ x = 0 $
- $ x $-intercepts: $ x = 0 $ and $ x = ±2 $
- $ y $-intercept: (0, 0)