How do you find vertical, horizontal and oblique asymptotes for $(x^2-5x+6)/(x-4)$ ?

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Answer 1 · 2017-03-07T13:11:08

The vertical asymptote is $x=4$
The oblique asymptote is $y=x-1$
No horizontal asymptote

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

The vertical asymptote is $x=4$

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

Let $f(x)=(x^2-5x+6)/(x-4)$

Let's do a long division

$color(white)(aaaa)$ $x^2-5x+6$ $color(white)(aaaa)$ $|$ $x-4$

$color(white)(aaaa)$ $x^2-4x$ $color(white)(aaaaaaaa)$ $|$ $x-1$

$color(white)(aaaaa)$ $0-x+6$

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

$color(white)(aaaaaaa)$ $-0+2$

Therefore,

$f(x)=(x-1)+2/(x-4)$

$lim_(x->-oo)(f(x)-(x-1))=lim_(x->-oo)2/(x-4)=0^-$

$lim_(x->+oo)(f(x)-(x-1))=lim_(x->+oo)2/(x-4)=0^+$

The oblique asymptote is $y=x-1$

graph{(y-(x^2-5x+6)/(x-4))(y-x+1)(y-100(x-4))=0 [-18.3, 17.74, -6.74, 11.28]}

How do you find vertical, horizontal and oblique asymptotes for (x^2-5x+6)/(x-4)(x^2-5x+6)/(x-4)?