Question 1163607
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There is an oblique (linear) asymptote only if the degree of the numerator is 1 more than the degree of the denominator.<br>
So there is never any need to "keep doing polynomial division" to find an asymptote.  If there is an oblique asymptote, a single division will find it.<br>
Example 1: {{{(x^2+2x)/(x+1) = (x+1)-1/(x+1)}}}<br>
A graph of the rational function and the oblique asymptote, y=x+1:<br>
{{{graph(400,400,-5,5,-10,10,(x^2+2x)/(x+1),x+1)}}}<br>
Example 2: {{{(x^3+3x^2+3x)/(x^2+2x+1) = (x+1)-1/(x^2+2x+1)}}}<br>
A graph of the rational function and the oblique asymptote, y=x+1:<br>
{{{graph(400,400,-5,5,-20,20,(x^3+3x^2+3x)/(x^2+2x+1),x+1)}}}<br>
If the degree of the numerator is more than 1 greater than the degree of the denominator, then you get an asymptote which is not oblique (linear).<br>
For example, if the degree of the numerator is 2 more than the degree of the denominator, the asymptote is degree 2 (quadratic -- a parabola).<br>
Example 3: {{{(x^3+3x^2+3x)/(x+1) = (x^2+2x+1)-1/(x+1)}}}<br>
A graph of the rational function and the quadratic asymptote:<br>
{{{graph(400,400,-5,5,-20,20,(x^3+3x^2+3x)/(x+1),x^2+2x+1)}}}<br>
In general, if the degree of the numerator is n more than the degree of the denominator, the asymptote will be of degree n.<br>
Example 4: numerator degree 4, denominator degree 1; asymptote degree 3: {{{(x^4+4x^3+6x^2+4x)/(x+1) = (x^3+3x^2+3x+1)-1/(x+1)}}}<br>
A graph....<br>
{{{graph(400,400,-5,5,-20,20,(x^4+4x^3+6x^2+4x)/(x+1),(x^3+3x^2+3x+1))}}}<br>