Lesson: January 30, 2025. Quiz: February 13, 2025.

  • Graph rational functions by identifying asymptotes and end behavior.
  • Rewrite simple rational expressions in different forms using long division.

Assignment

Rational Expressions

Take a reciprocal function and open up the numerator and denominator to any polynomial and you have a rational function, for example $f(x)=\frac{4x}{x-3}$. Using division you can rewrite some of these so that they look like the reciprocal function, making analyzing the graph a bit easier. So, $\frac{4x}{x-3}= 4+\frac{12}{x-3}$ showing us a shift of 4 units up and 3 units to the right. There’s also a vertical stretch of 12 happening, but we’re not going to worry about that too much.

Do enough of these and you’ll start to notice some patterns that will save you from doing the division every time. It all starts with factoring, so take a look at both polynomials and factor everything as much as possible. Do not eliminate common factors.

Identify Vertical Asymptotes

Look at your denominator to find these. Any place where your denominator equals 0 is where you will run have a vertical asymptotes with one catch. If the zero factor generating the zero can be factored out, it will instead produce a hole in the graph.

\[\begin{align} f(x) &= \frac{x-2}{x^2+7x+12} \\ &= \frac{x-2}{(x+3)(x+4)} \end{align}\]

The function above will have asymptotes at $x=-3$ and $x=-4$, which you can see here. Now let’s modify that function slightly.

\[\begin{align} f(x) &= \frac{x+3}{x^2+7x+12} \\ &= \frac{x+3}{(x+3)(x+4)} \end{align}\]

Here’s that graphed in Desmos. Notice how the asymptote at $x=-3$ is gone and if you try and highlight value of the graph at $x=-3$ you’ll get $(-3,\text{undefined})$. This happens because the factor $x+3$ appears in both the numerator and denominator, meaning it can be eliminated.

You would think that would allow $x=-3$ to exist now since we don’t have to worry about dividing by 0, but the original restriction on the domain still exists. This is why I mentioned earlier to not eliminate common factors. They are needed to dictate your domain and range.

Identifying Horizontal Asymptotes

Here, it’s actually preferred to have the unfactored polynomials since we need to compare the leading terms. Specifically, the degrees of each leading term.

  • If the numerator’s degree is higher (meaning more powerful), there is no horizontal asymptote. Instead the function will behave like the polynomial in the numerator at very large values of $x$.
  • If the degree is the denominator is higher, then the horizontal asymptote will be at $y=0$. This is because at large values of $x$, the denominator will be much much larger than the numerator, so the function evaluates to a number that will get infinitely small and close to zero.
  • When the degrees are equal, we just look at the leading the coefficients since the degrees basically cancel each other out at large values of $x$. Take the two leading coefficients and divide them and you’ll have your horizontal asymptote. The function $f(x)=\frac{2x^2+x+1}{x^2-1}$ has a horizontal asymptote at $y=2/1=2$.

Graph a Rational Function

Let’s graph one using everything from above.

\[\begin{align} f(x) = \frac{2x+1}{3x-4} \end{align}\]

Nothing to factor and no common factors to worry about so we can jump right in. This one will have a vertical asymptote at $x=\frac{4}{3}$, and since the degrees are equal, a horizontal asymptote at $y=\frac{2}{3}$.

While we’re here, we can do the intercepts. See where the numerator evaluates to zero to get the $x$-intercepts, so only at $x=-\frac{1}{2}$ in this case. The $y$-intercept comes from the quotient of the constants, so $y=-\frac{1}{4}$.

Here is the graph on Desmos.