How do you find vertical, horizontal and oblique asymptotes for $(x^2-x+1)/(x-3)$ ?

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Answer 1 · 2016-12-15T19:09:29

The vertical asymptote is $x=3$
No horizontal asymptote.
The oblique asymptote is $y=x+2$

As you cannot divide by $0$ , $x!=3$

Therefore,

The vertical asymptote is $x=3$

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

Let's do a long division

$color(white)(aaaa)$ $x^2-x+1$ $color(white)(aaaa)$ $∣$ $x-3$

$color(white)(aaaa)$ $x^2-3x$ $color(white)(aaaaaaa)$ $∣$ $x+2$

$color(white)(aaaaa)$ $0+2x+1$

$color(white)(aaaaaaa)$ $+2x-6$

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

Therefore,

$(x^2-x+1)/(x-3)=x+2+7/(x-3)$

The oblique asymptote is $y=x+2$

There is no horizontal asymptote. When $x->+-oo$ ,

$f(x)->+-oo$

graph{(y-(x^2-x+1)/(x-3))(y-x-2)=0 [-41.1, 41.1, -20.56, 20.56]}

How do you find vertical, horizontal and oblique asymptotes for (x^2-x+1)/(x-3)(x^2-x+1)/(x-3)?