What are the asymptote(s) and hole(s), if any, of  f(x) =3/x-(8x)/(x^2-3x) ?

Apr 8, 2018

Asymptotes: $x = 3 , x = 0 , y = 0$

Explanation:

$f \left(x\right) = \frac{3}{x} - \frac{8 x}{{x}^{2} - 3 x}$

f(x)=(3(x^2-3x)-8x*x)/(x(x^2-3x)

For the asymptotes, we look at the denominator.
Since the denominator cannot equal to $0$

ie $x \left({x}^{2} - 3 x\right) = 0$

${x}^{2} \left(x - 3\right) = 0$

therefore $x \ne 0 , 3$

For the y asymptotes, we use the limit as $x \to 0$

lim x->0 (3(x^2-3x)-8x*x)/(x(x^2-3x)

=$\lim x \to 0 \frac{3 {x}^{2} - 9 x - 8 {x}^{2}}{x \left({x}^{2} - 3 x\right)}$

=$\lim x \to 0 \frac{- 5 {x}^{2} - 9 x}{{x}^{3} - 3 {x}^{2}}$

=$\lim x \to 0 \frac{\left(- \frac{5}{x} - \frac{9}{x} ^ 2\right)}{1 - \frac{3}{x}}$

=$0$

therefore $y \ne 0$