The oblique asymptote always occurs in a rational function. It occurs when there is no horizontal asymptote.

2) In a rational function $f(x)=\frac{p(x)}{q(x)}$, the highest exponent of the polynomial p(x) (numerator) is 1 more than the highest exponent of the polynomial q(x) denominator.Then the function has an oblique asymptote.

Degree of p(x) > Degree of q(x) (by 1)

3) An oblique asymptote can be found out either by long division or by synthetic division.

1) Find the oblique asymptote of the rational function $f(x)=\frac{x^2-x-6}{x+1}$

$y=\frac{x^2-x-6}{x+1}$

Here p(x) = $x^{2}$-x-6 with degree 2.

q(x) = x+1 with degree 1.

Degree of p(x) > Degree of q(x) (by 1) so there is a oblique asymptote

Now we will divide p(x) by q(x) using long division. $y=(x-2)+\frac{-4}{x^2-x-6}$

So the oblique asymptote : y=x-2

2) Find the slant asymptote of the rational function $f(x)=\frac{x^2+3x+2}{x-2}$

$y=\frac{x^2+3x+2}{x-2}$

Here p(x) = $x^{2}$+3x+2 with degree 2.

q(x) = x-2 with degree 1.

Degree of p(x) > Degree of q(x) (by 1) so there is a oblique asymptote

Now we will divide p(x) by q(x) using synthetic division.

x-2= 0

∴ x= 0 $y=(x+5)+\frac{12}{x^2+3x+2}$

So the oblique asymptote : y=x+5

3) Find the oblique asymptote of the rational function $f(x)=\frac{3x^2+3x+1}{x+2}$

$y=\frac{3x^2+3x+1}{x+2}$

Here p(x) = $3x^{2}$+3x+1 with degree 2.

q(x) = x+2 with degree 1.

Degree of p(x) > Degree of q(x) (by 1) so there is a oblique asymptote

By doing the long division we will get

$y=(3x-3)+\frac{7}{3x^2+3x+1}$

So the oblique asymptote : y=3x-3

4)Find the oblique asymptote of the rational function $f(x)=\frac{(x+1)(x-4)}{x}$

$y=\frac{(x+1)(x-4)}{x}$

If we multiply the numerator we will get $x^{2}$-3x-4

Here p(x) = $x^{2}$-3x-4 with degree 2.

q(x) = x with degree 1.

Degree of p(x) > Degree of q(x) (by 1) so there is a oblique asymptote

By doing the long division we will get

$y=(x-3)+\frac{-4}{x^2-3x-4}$

So the oblique asymptote : y=x-3

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