2
$\begingroup$

Create a formula for a rational function which has all of the following characteristics:

  1. A domain of all real numbers except 3
  2. A hole in the graph at $(3,4)$
  3. A horizontal asymptote of $y=0$ on both sides of the graph

In the above problem, I know how to create a formula that satisfies the first requirement and part of the second requirement. For the first one, since the domain is all real numbers except 3, the only factor of the denominator which can equal zero must be $x-3$.For the second one, since a hole in the graph at $(3,4)$, $x-3$ must be a factor of the numerator.

Can anyone help me on requirements 2 and 3? Thank you!

  • 0
    You can reverse engineer [this article](http://lps.lexingtonma.org/cms/lib2/MA01001631/Centricity/Domain/833/Rational%20Functions%20with%20holes%20and%20asymptotes.pdf). Answer coming up. – 2017-02-22

1 Answers 1

2

You have already established that $x-3$ appears in numerator and denominator. So we can set $\frac{x-3}{x-3}$ aside as a subexpression that only causes a hole at $x=3$.

While it is true that the first requirement means $x-3$ is the only linear factor that can appear in the denominator, it does not prevent irreducible quadratic factors that can never reach zero such as $1+x^2$ from appearing. Thus we can set the remaining part of our rational function as $\frac N{1+x^2}$, where $N$ is a constant that ensures the third ($y=0$ asymptote) requirement from degree considerations. At $x=3$, $N$ has to be 40 to yield the required removable singularity at $(3,4)$. So the final rational function satisfying all three conditions is $$\frac{40(x-3)}{(1+x^2)(x-3)}$$

  • 0
    In $\frac{40(x-3)}{(1+x^2)(x-3)}$, why $(x-3)$ do not cancel out? – 2017-02-22
  • 0
    @math They do whenever $x\ne3$. The purpose of them is to create the hole at $x=3$. – 2017-02-23