Thread: Quick way of interpreting polynomials?

I'm studying asymptotes and wondering if there's a quicker way to determine where the graph is going.
Let me explain with an example:

$$f(x)=\frac{2x^3+x^2-1}{x^2-1}$$

So there are two vertical asymptotes x=-1 and x=1
To know how the curve is turning i check f(1.001), f(-1.001), f(-0.999), f(0.999).
Doing it by hand is pretty slow, is there a better way of knowing where the function goes when x approaches the asymptotes?

Originally Posted by TriForce

I'm studying asymptotes and wondering if there's a quicker way to determine where the graph is going.
Let me explain with an example:

$$f(x)=\frac{2x^3+x^2-1}{x^2-1}$$

So there are two vertical asymptotes x=-1 and x=1
To know how the curve is turning i check f(1.001), f(-1.001), f(-0.999), f(0.999).
Doing it by hand is pretty slow, is there a better way of knowing where the function goes when x approaches the asymptotes?

You have the critical points, so graph those. Now look for when the function is positive or negative. There are two critical points so the x-axis is broken into three pieces: x < -1, -1 < x < 1, and x < 1. For example, all we need to know is whether the function is negative or positive on the three intervals. So pick easy numbers. For x < -1 choose x = -100. Is the function positive or negative here? (It's negative.) Then do -1 < x < 1. Rinse and repeat.

You can often simplify the function a bit. In this case I'd write $\displaystyle \frac{2x^3 + x^2 - 1}{x^2 - 1} = \frac{2x^3}{x^2 - 1} + \frac{x^2 - 1}{x^2 - 1} = \frac{2x^3}{x^2 - 1} + 1$

It's a bit easier to work with in this case.

-Dan