Question

What are the asymptotes and removable discontinuities, if any, of `f(x)= (x^3 -16x) / (4x^2 - 4x )`?

Answer 1

Oblique asymptotes `f(x) =x/4` and `f(x) = -x/4`. Discontinuity at `x=1` and removable discontinuity at `x=0`

Explanation:

Factor both the numerator and denominator
`f(x) = (x(x^2 - 16))/(4x(x-1)`
The bracketed term in the numerator is the difference of two squares and can therefore be factored
`f(x) = (x(x-4)(x+4))/(4x(x-1))`
Discontinuities exist wherever the denominator is zero, which will happen when `x=0` or when `x=1`. The first of these is a removable discontinuity because the single `x` will cancel out from numerator and denominator.
`f(x) = ((x-4)(x+4))/(4(x-1))`
As `x` gets larger positively the function will approach `f(x) = x/4` and as it gets larger negatively it will approach `f(x) = -x/4`