Is it possible to use repeated synthetic division (rather than long division) to find a slant asymptote for a rational function such as $\displaystyle \frac{2x^3 + 3x^2 + 5x + 7}{(x-1)(x-3)}$? It appears to work, but I am not sure that it is valid to ignore the remainder term from the first synthetic division.
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when you are finding the slant asymptote you are ignoring terms that are small compared to the asymptote. it is ok to use synthetic division. in your example if you divide first by $x-1$ you will have $${2x^3 + 3x^2 + 5x + 7 \over(x-1)} = 2x^2 + 5x + 10 + {17 \over (x-1)}$$ and when you synthetically divide $2x^2 + 5x + 10$ by $x-3$ you have $$ {2x^2 +5x + 10 \over (x - 3)} = 2x + 11 + {43 \over (x - 3)}$$ putting the two together we get $${2x^3 + 3x^2 + 5x + 7 \over(x-1)(x-3)} = 2x + 11 + {43 \over (x - 3)} + {17 \over (x-1)(x-3)}$$ you can see that slant asymptote is $y = 2x + 11$ and that we are ignoring the remainder terms as being small when $x$ is large. therefore you could ignore the remainder $17$ and $43$ where you are dividing.
abel
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It's OK, for a slant asymptote to exist degree of numerator must exceed degree of denominator by $1$. In this case the quotient will be a polynomial $f(x)=ax+b$, and $y=f(x)$ is the asymptote.
The remainder in the first division is completely immaterial, as you can see by substituting an arbitrary constant $c$ instead of $7$ in the numerator.
Stefan Hansen
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P. Hasdeu
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