How do you find the vertical, horizontal or slant asymptotes for $g(x)=(x+7)/(x^2-4)$ ?

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Answer 1 · 2016-12-31T10:58:35

The vertical asymptotes are $x=-2$ and $x=2$
No slant asymptote.
The horizontal asymptote is $y=0$

We need

$a^2-b^2=(a+b)(a-b)$

Let's factorise the denominator

$x^2-4=(x+2)(x-2)$

Then,

$g(x)=(x+7)/(x^2-4)=(x+7)/((x+2)(x-2))$

The domain of $g(x)$ is $D_g(x)=RR-{-2,2}$

As we cannot divide by $0$ , $x!=-2$ and $x!=2$

The vertical asymptotes are $x=-2$ and $x=2$

The degree of the numerator is $<$ than the degree of the denominator, there is no slant asymptote.

To find the horizontal asymptotes, we calculate $lim g(x)$ as $x->+-oo$

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

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

The horizontal asymptote is $y=0$

graph{(y-(x+7)/(x^2-4))(y)=0 [-11.25, 11.26, -5.62, 5.62]}

How do you find the vertical, horizontal or slant asymptotes for g(x)=(x+7)/(x^2-4)g(x)=(x+7)/(x^2-4)?