Question

What are the asymptotes and removable discontinuities, if any, of `f(x)= (2x-2x^2)/(x(x+1)(x-1)`?

Answer 1

vertical asymptote at x = -1
horizontal asymptote at y = 0

Explanation:

factoring gives : `(-2x(x-1))/(x(x+1)(x-1)`

`rArr (-2cancel(x)cancel(x-1))/(cancel(x)(x+1)cancel(x-1))`

left with `( -2)/(x+1)`

removable discontinuities at x = 0 and x = 1

point discontinuities at (0,-2) and ( 1 , -1 )

vertical asymptote when denominator of rational function tends to zero. To find equation let denominator equal zero.

solve : x+ 1 = 0 → x = -1 is the equation

Horizontal asymptotes occur as `lim_(x→±∞) f(x) → 0`

If the degree of the numerator is less than the degree of the denominator , as is the case here , numerator degree 0 and denominator degree 1 then the equation is always y = 0.

thus equation is y = 0