How do you find the asymptotes for $f(x) = (x^3 + x^2 - 6x) /( 4x^2 - 8x - 12)$ ?

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Answer 1 · 2016-01-26T13:56:50

Asymptotes are $x=3$ , $x=-1$ and $x/4$

Start by finding the factors for both numerator and denominator.
$f(x) = (x^3+x^2-6x)/(4x^2-8x-12)$

$f(x) = (x(x^2+x - 6))/(4(x^2-2x-3))$

$f(x) = (x(x+3)(x-2))/(4(x-3)(x+1))$

This confirms that there are no removable asymptotes as there are no factors that cancel out.

The denominator approaches zero, and therefore the function value $->oo$ as $x->3$ or $x->-1$ so there are vertical asymptotes at $x=3$ and $x=-1$

As $x->oo$ the function $f(x) -> (xcancel(x^2+x - 6))/(4cancel(x^2-2x-3)) -> x/4$

So there is a slant asymptote of $x/4$

How do you find the asymptotes for f(x) = (x^3 + x^2 - 6x) /( 4x^2 - 8x - 12)f(x) = (x^3 + x^2 - 6x) /( 4x^2 - 8x - 12)?