# How do you find the asymptotes for (t^3 - t^2 - 4t + 4) /(t^2 + t - 2)?

Feb 11, 2016

There isn't an asymptote.

#### Explanation:

The way asymptotes work is that there is a value that $x$, or in this case $t$, cannot equal when in the denominator, or else it would be dividing by zero. What this looks like when graphed is a value that the line will get closer and closer to but never touch.

The way to find an asymptote is to factor everything, cross out what divides out, and then solve for $t$ in the denominator.

So first, we factor. ${t}^{2} + t - 2$ can be factored to $\left(t - 1\right) \left(t + 2\right)$. The top one is a little harder, but through synthetic division I found that $\left(x - 1\right)$ is a factor, which leaves $\left({t}^{2} - 4\right)$, which we can simplify with the "difference of squares" to become $\left(t - 2\right) \left(t + 2\right)$. All together we now have $\frac{\left(t - 1\right) \left(t + 2\right) \left(t - 2\right)}{\left(t - 1\right) \left(t + 2\right)}$.

Now we move on to the second step, dividing out factors that are the same in the numerator as the denominator. $\frac{\cancel{t - 1} \cancel{t + 2} \left(t - 2\right)}{\cancel{t - 1} \cancel{t + 2}}$. We are left with $\frac{t - 2}{1}$, or just $t - 2$.

Now, remember that an asymptote is the value that $x$ cannot equal or else it would be dividing by zero. But in this case, $x$ has no such constraints because $x$ isn't even in the denominator. So, there is no asymptote as it is now a linear equation. If you don't believe me, let's graph our original equation.
graph{y=((x-1)(x^2-4))/(x^2+x-2)}

There is clearly no asymptote, and we can see that from the graph and from the algebra we did.

Nice job, thanks for sticking with me through this problem. Good work!