How do you find vertical, horizontal and oblique asymptotes for $f (x) = \frac{x^{2} - 9}{x - 4}$ $f(x) = (x^2 - 9) / (x - 4)$ ?

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Answer 1 · 2017-01-19T14:34:42

The vertical asymptote is $x = 4$ $x=4$
The oblique asymptote is $y = x + 4$ $y=x+4$
No horizontal asymptote.

The domain of $f (x)$ $f(x)$ is $D_f(x)=RR-{4}$

As you cannot divide by $0$ , $x!=4$

The vertical asymptote is $x=4$

As the degree of the numerator is $>$ than the denominator, there is an oblique asymptote.

Let's do a long division

$color(white)(aaaa)$ $x^2$ $color(white)(aaaaaa)$ $-9$ $color(white)(aaaa)$ $∣$ $color(red)(x-4)$

$color(white)(aaaa)$ $x^2-4x$ $color(white)(aaaaaaaa)$ $∣$ $color(blue)(x+4)$

$color(white)(aaaaa)$ $0+4x-9$

$color(white)(aaaaaaa)$ $+4x-16$

$color(white)(aaaaaaaaa)$ $+0+7$

Therefore,

$f(x)=(x^2-9)/(x-4)=color(blue)(x+4)+7/(x-4)$

$lim_(x->-oo)(f(x)-(x+4))=lim_(x->-oo)7/x=0^-$

$lim_(x->+oo)(f(x)-(x+4))=lim_(x->+oo)7/x=0^+$

The oblique asymptote is $y=x+4$

graph{(y-(x^2-9)/(x-4))(y-x-4)=0 [-52, 52, -26.03, 26]}

How do you find vertical, horizontal and oblique asymptotes for f(x) = (x^2 - 9) / (x - 4)f(x)=x2−9x−4 f(x) = (x^2 - 9) / (x - 4)?