1. ## Asymptote Question...

I have found the slant asymptote of $\displaystyle \frac{x^4-6x^3-11x^2+60x+100}{x^3-9x^2+27x-27}$ to be $\displaystyle y=x+3$. Somehow, on it's graph, the slant looks totally off. What is the problem with my math? The factored form that was given is $\displaystyle \frac{(x-5)^2(x+2)^2}{(x-3)^4}$. Thanks!

2. Originally Posted by nivek516
I have found the slant asymptote of $\displaystyle \frac{x^4-6x^3-11x^2+60x+100}{x^3-9x^2+27x-27}$ to be $\displaystyle y=x+3$. Somehow, on it's graph, the slant looks totally off. What is the problem with my math? The factored form that was given is $\displaystyle \frac{(x-5)^2(x+2)^2}{(x-3)^{\color{red}4}}$. Thanks!
Do you mean $\displaystyle (x-3)^3 = x^3 - 9x^2 + 27x-27$? If so then your expansions are correct and your slant asymptote is correct. The graph you're looking at might be wrong .... Is it a graph you've drawn using technology? Check how you entered the equation.

3. Hello nivek:

Your math looks fine; the equation of the slant asymptote is y = x + 3.

(There is a typographical error in your factored form; the exponent in the denominator should be 3.)

What do you mean by saying that the slant on your graph is off?

The slope is one; if the scales on both axes are the same, then the graph of y = x + 3 should make a 45-degree angle with the horizontal axis.

Cheers,

~ Mark

4. Yes, I typed in the wrong equation in my post. When I say the asymptote is off, I mean that the graph in quadrant 1 begins at around x=3 and avoids the asymptote. As $\displaystyle x\rightarrow \infty$, the graph starts to acknowlege the aymptote. Take a look. Could you explain why this happens?