What are the asymptote(s) and hole(s), if any, of $f (x) = (\frac{x - 3}{x + 2} \cdot x) \cdot (\frac{x^{2} - x}{x^{3} - 3 x^{2}})$ $f(x) =((x-3)/(x+2)x)((x^2-x)/(x^3-3x^2))$ ?

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Answer 1 · 2018-06-15T18:13:38

Vertical asymptote at $x = - 2$ $x=-2$

$f (x) = \frac{x (x - 3) (x^{2} - x)}{(x + 2) (x^{3} - 3 x^{2})}$ $f(x)={x(x-3)(x^2-x)}/{(x+2)(x^3-3x^2)}$

factor $(x^{2} - x)$ $(x^2-x)$ and $(x^{3} - 3 x^{2})$ $(x^3-3x^2)$ .

$f (x) = \frac{x^{2} (x - 3) (x - 1)}{x^{2} (x + 2) (x - 3)}$ $f(x)={x^2(x-3)(x-1)}/{x^2(x+2)(x-3)}$

Cancel likewise terms.

$f (x) = \frac{x - 1}{x + 2}$ $f(x)={x-1}/{x+2}$

Vertical asymptote at $x = - 2$ $x=-2$ as $f (x)$ $f(x)$ is not defined there.

What are the asymptote(s) and hole(s), if any, of f(x)=(x−3x+2⋅x)⋅(x2−xx3−3x2) f(x) =((x-3)/(x+2)*x)*((x^2-x)/(x^3-3x^2)) ?