Question

r(x)=(2x^3+11x^2+5x-1)/(x^2+3x+5) need help with finding out the oblique asymptote anyone.

Answer 1

Just going to have to carry out the long division here...can you show your process so far so we can see where you're going off?

Answer 2

Oh offline already...nvm

Answer 3

An oblique asymptote for \(r(x)\) is any polynomial \(p(x)\) that satisfies
\[\lim_{x\to\infty}(r(x)-p(x))=0\]Given a rational function \(r(x)=\dfrac{a(x)}{b(x)}\) with the degree of \(a(x)\) greater than \(b(x)\), you can use long division as suggested above to write
\[r(x)=\frac{a(x)}{b(x)}=c(x)+\frac{d(x)}{b(x)}\]where \(c\) is the quotient of \(\dfrac{a}{b}\) and \(d\) is the remainder. By definition, the degree of \(d\) will be smaller than that of \(b\), so as \(x\to\infty\) the remainder term will disappear and you're left with the requirement that
\[\lim_{x\to\infty}(c(x)-p(x))=0\]The easiest way for this to be guaranteed is by choosing \(c(x)=p(x)\).

Long division is a good approach, but for the sake of variety I'll suggest the expanded synthetic division algorithm for division involving denominators of degree greater than \(1\).
\[\begin{array}{cc|cc|cc}
&&2&11&5&-1\\[1ex]
&-5&&&-10&-25\\[1ex]
-3&&&-6&-15\\[1ex]
\hline
&&2&5&-20&-26
\end{array}\]The result here translates to
\[r(x)=2x+5-\frac{20x+26}{x^2+3x+5}\]