How do you find all the asymptotes for function $f(x) = (x+4)/(3x^2+5x-2)$ ?

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Answer 1 · 2016-11-12T16:04:55

The vertical asymptotes are $x=1/3$ and $x=-2$
The vertical asymptote is $y=0$
There are no slant asymptotes

Let's factorise the denominator $3x^2+5x-2 =(3x-1)(x+2)$

As we cannot divide by $0$ , the vertical asymptotes are $x=1/3$ and $x=-2$

The degree of the numerator is $<$ the degree of the denominator, so we don't have a slant asymptote.

$lim_(x->-oo)f(x)=lim_(x->-oo)x/(3x^2)=lim_(x->-oo)1/(3x)=0^(-)$

$lim_(x->+oo)f(x)=lim_(x->+oo)x/(3x^2)=lim_(x->+oo)1/(3x)=0^(+)$

So $y=0$ is a horizontal asymptote
graph{(x+4)/(3x^2+5x-2) [-11.05, 6.73, -6.2, 2.69]}

How do you find all the asymptotes for function f(x) = (x+4)/(3x^2+5x-2)f(x) = (x+4)/(3x^2+5x-2)?