Question

State the horizontal asymptote of the rational function. For full credit, explain the reasoning you used to find the horizontal asymptote. f(x) =x+9/x^2+4x+2

Answer 1

i'm assuming that f(x) is actually :

\[(x+9)/(x^2 +4x +2)\]

Answer 2

the highest power in the numerator is 1
the highest power in the denominator is 2

since the highest power in the numerator is less than the highest power in the denominator, the horizontal asymptote is y = 0

Answer 3

The way I like to think of it is, when you are near zero you can go over horizontal (NOT VERTICAL!!) asymptotes. This means you care about the behavior at very large or very small x.

In this limit the highest powers in the numerator and denominator are all that contributes (since x^5 grows much after than x for example, plug in 10,000 and you'll see what I mean). This means at large (or small) x it acts as x/x^2 or 1/x but as you increase x that quantity goes towards zero(!!!!).

For example: starting at 10 we have 1/10 = .1, 100 implies .01, 1000 implies .001, etc. Therefore the horizontal asymptote is at y = 0. (as @robz8) said :P

Answer 4

And using negative x only changes the sign (doing -10 gives -.1, -100 gives -.01). The point is it still approaches zero.