# Writing a rational function with given characteristics

I've got two homework question that have me stumped. Both lack an x-intercept, and the second one throws an oblique asymptote into the mix.

Problem one provides the following characteristics:

Vertical asymptotes at $x=-2$, and $x=5$, Hole in graph at $x=0$, Horizontal asymptote at $y=3$

What has me stumped is what am I supposed to do with the numerator? All the previous question had an x-intercept.

Here's what I have so far: $\dfrac{x}{x} \cdot \dfrac{3(???)}{(x+2)(x-5)}$

Problem two also does not provide an x-intercept. This problem also has an oblique asymptote that I don't know how to handle.

Here are the characteristics: Vertical asymptotes at $x=2$ and $x=-4$, Oblique asymptote at $y=2x$

Any help would be greatly appreciated!

-

1. The function has to have $\lim_{x\rightarrow\pm\infty}=3$ . To do this, the numerator must be a polynomial of the same degree as the denominator (so neither overpowers the other), with a $3$ as the coefficient of the largest term. Basically a number of functions will work, such as:

$$\frac{3x(x^2+1)}{x(x+2)(x+5)}$$

Both cubics, with a $3x^3$ on top and an $x^3$ on the bottom. So as $|x|$ increases the smaller terms ($x^2$,etc.) will drop away to leave $3$. There are no $x$ intercepts, since $x^2+1\neq 0$ for any $x$.

For the oblique asymptote the idea is the same, but now the numerator should be larger than the denominator, so that the two largest terms divide to give $2x$. Try it yourself, and I'll edit this answer if you're still stuck.

-
I checked the graph on my TI-84 and it appears that the graph crosses the horizontal asymptote of 3. –  user35623 Jul 11 '12 at 21:11
Here's what I put into the TI-84: (3x(X^2+1)) / (x(x+2)(x-5)) –  user35623 Jul 11 '12 at 21:12
@user35623: It’s perfectly acceptable for a graph to cross one of its horizontal asymptotes. For example, the graph of $$y=\frac{x}{x^2+1}$$ has $y=0$ as asymptote in both directions and crosses that line at $x=0$. –  Brian M. Scott Jul 11 '12 at 21:37
Thank you for the explanation and example! I'll give problem 2 a shot now. –  user35623 Jul 11 '12 at 21:50
Would the second answer be: $\dfrac{4x(x^2+1)}{2x(x-2)(x+4)}$ –  user35623 Jul 12 '12 at 0:26