Question

Hey guys, i need help with slant asymptotes asap <3
i got (t^3-t^2-12t)/(t^2-9) and i dk how to do it ;-;

Answer 1

one word. long division

Answer 2

that was 2 words ;) but yeah, find your remainder

Answer 3

or rather your quotient, the remainder will go to zero as x gets large ....

Answer 4

In general, an asymptote to \(f(x)\) is a function \(p(x)\) such that
\[\lim_{x\to\infty}\bigg(f(x)-p(x)\bigg)=0\]To view this in the light of your particular problem, \(f(x)\) is a rational function, ie. a ratio of polynomials \(\dfrac{A(x)}{B(x)}\), and you're looking to find \(p(x)\).

Since \(\deg A(x)=\deg B(x)+1\), long division would give you a polynomial quotient term and a remainder term, so that you can write
\[f(x)=\frac{A(x)}{B(x)}=C(x)+\frac{D(x)}{E(x)}\]where \(\deg E(x)>\deg D(x)\). This last part is important, because as \(x\to\infty\), the remainder term will disappear, leaving you with
\[\lim_{x\to\infty}\bigg(f(x)-p(x)\bigg)=\lim_{x\to\infty}\bigg(C(x)+\frac{D(x)}{E(x)}-p(x)\bigg)=\lim_{x\to\infty}\bigg(C(x)-p(x)\bigg)\]You want this limit to be zero, and the easiest way for that to happen is if \(C(x)=p(x)\). That is, the quotient term *is* the asymptote.

Answer 5

@SithsAndGiggles
nice explanation.
can we say degree E(x) is one greater than D(x)
or in general D(x) / E(x) will have form c / (x - a)

Answer 6

Not necessarily. Consider this example:
\[\frac{x^2}{x^2+1}=\frac{x^2+1-1}{x^2+1}=1-\frac{1}{x^2+1}\]so in this case, \(\deg D+2=\deg E\)