Single Variable Calculus, Chapter 4, 4.4, Section 4.4, Problem 42

Determine the formula for a function that has vertical asymptotes $x = 1$ and $x = 3$ and horizontal asymptotes $y = 1$.

The vertical asymptotes must make the denominator of the function 0. So we have,


$
\begin{equation}
\begin{aligned}

(x - 1)(x - 3) =& 0
\\
\\
x^2 - 3x - x + 3 =& 0
\\
\\
x^2 - 4x + 3 =& 0

\end{aligned}
\end{equation}
$


Since we know that the degree of the denominator is 2, we must introduce a variable $x$ in the numerator with degree of 2 to make the horizontal asymptote equal to 1, because to find the horizontal asymptote we need to find the highest degree of $x$ in numerator and denominator. So we have,

$y = x^2$

Hence,

$\displaystyle y = \frac{x^2}{x^2} = 1$

Therefore,

The equation of the function is $\displaystyle f(x) = \frac{x^2}{x^2 + 4x + 3}$ or $\displaystyle f(x) = \frac{x^2}{(x - 1)(x - 3)}$