How do you find the Vertical, Horizontal, and Oblique Asymptote given $\frac{(x - 3) (9 x + 4)}{x^{2} - 4}$ $((x-3)(9x+4))/(x^2-4)$ ?

Question

1257 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-26T06:36:40

The vertical asymptotes are $x = 2$ $x=2$ and $x = - 2$ $x=-2$
The horizontal asymptote is $y = 9$ $y=9$
No slant asymptote

Let $f (x) = \frac{(x - 3) (9 x + 4)}{x^{2} - 4}$ $f(x)=((x-3)(9x+4))/(x^2-4)$

Let's factorise the denominator

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

As you cannot divide by $0$ $0$ , $x \neq 2$ $x!=2$ and $x \neq - 2$ $x!=-2$

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

As the degree of the numerator is $=$ $=$ the degree of the denominator, there is no slant asymptote

$lim_(x->+-oo)f(x)=lim_(x->+-oo)(9x^2)/x^2=9$

The horizontal asymptote is $y=9$

graph{(y-((x-3)(9x+4)/(x^2-4)))(y-9)=0 [-41.1, 41.1, -20.55, 20.57]}

How do you find the Vertical, Horizontal, and Oblique Asymptote given ((x-3)(9x+4))/(x^2-4)(x−3)(9x+4)x2−4((x-3)(9x+4))/(x^2-4)?