How do you identify all asymptotes or holes for $f (x) = \frac{- 2 x^{3} - 12 x^{2} - 16 x}{x^{3} + x^{2} - 2 x}$ $f(x)=(-2x^3-12x^2-16x)/(x^3+x^2-2x)$ ?

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Answer 1 · 2016-12-13T10:49:31

The vertical asymptote is $x = 1$ $x=1$
The holes are when $x = 0$ $x=0$ and $x = - 2$ $x=-2$

The horizontal asymptote is $y = - 2$ $y=-2$

Let's factorise the numerator

$- 2 x^{3} - 12 x^{2} - 16 x = - 2 x (x^{2} + 6 x + 8)$ $-2x^3-12x^2-16x=-2x(x^2+6x+8)$
$= - 2 x (x + 2) (x + 4)$ $=-2x(x+2)(x+4)$

Lets factorise the denominator

$x^{3} + x^{2} - 2 x = x (x^{2} + x - 2) = x (x + 2) (x - 1)$ $x^3+x^2-2x=x(x^2+x-2)=x(x+2)(x-1)$

So,

$f (x) = \frac{- 2 x^{3} - 12 x^{2} - 16 x}{x^{3} + x^{2} - 2 x}$ $f(x)=(-2x^3-12x^2-16x)/(x^3+x^2-2x)$

$=(-2cancelxcancel(x+2)(x+4))/(cancelxcancel(x+2)(x-1))$

$=(-2(x+4))/(x-1)$

There are holes when $x=0$ and $x=-2$

As you cannot divide by $0$ , $x!=1$

So, the vertical asymptote is $x=1$

Now, we calculate the limits as $x->oo$

$lim_(x->+-oo)f(x)=lim_(x->+-oo)(-2x)/x=lim_(x->+-oo)-2=-2$

The horizontal asymptote is $y=-2$

graph{(y-(-2(x+4))/(x-1))=0 [-41.1, 41.1, -20.56, 20.56]}

How do you identify all asymptotes or holes for f(x)=(-2x^3-12x^2-16x)/(x^3+x^2-2x)f(x)=−2x3−12x2−16xx3+x2−2xf(x)=(-2x^3-12x^2-16x)/(x^3+x^2-2x)?