How do you find the vertical, horizontal or slant asymptotes for $\frac{4 x^{2} + 11}{x^{2} + 8 x - 9}$ $(4x^2+11) /( x^2+8x−9)$ ?

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Answer 1 · 2016-12-27T08:26:26

The vertical asymptotes are $x = - 9$ $x=-9$ and $x = 1$ $x=1$
No slant asymptote.
The horizontal asymptote is $y = 4$ $y=4$

Let's factorise the denominator

$x^{2} + 8 x - 9 = (x - 1) (x + 9)$ $x^2+8x-9=(x-1)(x+9)$

Let, $f (x) = \frac{4 x^{2} + 11}{x^{2} + 8 x - 9}$ $f(x)=(4x^2+11)/(x^2+8x-9)$

The domain of $f (x)$ $f(x)$ is $D_f(x)=RR-{-9,1}$

As we cannot divide by $0$ , $x!=-9$ and $x!=1$

So, the vertical asymptotes are $x=-9$ and $x=1$

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

To find the horizontal asymptote, we compute the limits of $f(x)$ as $x->+-oo$

$lim_(x->+-oo)f(x)=lim_(x-+-oo)4x^2/x^2=4$

The horizontal asymptote is $y=4$

graph{(y-(4x^2+11)/(x^2+8x-9))(y-4)(x-1)(x+9)=0 [-52, 52, -26, 26]}

How do you find the vertical, horizontal or slant asymptotes for (4x^2+11) /( x^2+8x−9)4x2+11x2+8x−9(4x^2+11) /( x^2+8x−9)?