### Holes and Rational Functions

A **hole** on a graph looks like a hollow circle. It represents the fact that the function approaches the point, but is not actually defined on that precise \begin{align*}x\end{align*}

Take a look at the graph of the following equation:

\begin{align*}f(x)=(2x+2) \cdot \frac{\left(x+\frac{1}{2}\right)}{\left(x+\frac{1}{2}\right)}\end{align*}

The reason why this function is not defined at \begin{align*}-\frac{1}{2}\end{align*}

This is the essence of dealing with holes in rational functions. You should cancel what you can and graph the function like normal making sure to note what \begin{align*}x\end{align*}

Watch the first part of this video and focus on holes in rational equations.

### Examples

#### Example 1

Earlier, you were asked what happens to the equation \begin{align*}f(x)=\frac{(3x+1)(x-1)}{(x-1)}\end{align*}

#### Example 2

Graph the following rational function and identify any removable discontinuities.

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

This function requires some algebra to change it so that the removable factors become obvious. You should suspect that \begin{align*}(x-1)\end{align*}

\begin{align*}f(x)=\frac{(-x^2+2x+4)(x-1)}{(x-1)}\end{align*}

The removable discontinuity occurs at (1, 5).

#### Example 3

Graph the following rational function and identify any removable discontinuities.

\begin{align*}f(x)=\frac{x^6-6x^5+5x^4+27x^3-48x^2-9x+54}{x^3-7x-6}\end{align*}

This is probably one of the most challenging rational expressions with only holes that people ever try to graph by hand. There are multiple ways to start, but a good habit to get into is to factor everything you possibly can initially. The denominator seems less complicated with possible factors \begin{align*}(x \pm 1), (x \pm 2), (x \pm 3), (x \pm 6)\end{align*}

\begin{align*}f(x)=\frac{x^6-6x^5+5x^4+27x^3-48x^2-9x+54}{(x+1)(x+2)(x-3)}\end{align*}

The factors of the denominator are strong hints as to the factors of the numerator so use polynomial division and try each. When you fully factor the numerator you will have:

\begin{align*}f(x)=\frac{(x^3-6x^2+12x-9)(x+1)(x+2)(x-3)}{(x+1)(x+2)(x-3)}\end{align*}

Note the factors that cancel \begin{align*}(x=-1,-2,3)\end{align*}

\begin{align*}f(x)=x^3-6x^2+12x-9\end{align*}

At this point it is probably reasonable to make a table and plot points to get a sense of where this cubic function lives. You also could notice that the coefficients are almost of the pattern 1 3 3 1 which is the binomial expansion. By separating the -9 into -8 -1 you can factor the first four terms.

\begin{align*}f(x)=x^3-6x^2+12x-8-1=(x-2)^3-1\end{align*}

This is a cubic function that has been shifted right by two units and down one unit.

Note that there are two holes that do not fit in the graph window. When this happens you still need to note where they would appear given a properly sized window. To do this, substitute the invalid \begin{align*}x\end{align*}

\begin{align*}f(x)=(x-2)^3-1\end{align*}

Holes: (3, 0); (-1, -28); (-2, -65)

#### Example 4

Without graphing, identify the location of the holes of the following function.

\begin{align*}f(x)=\frac{x^3+4x^2+x-6}{x^2+5x+6}\end{align*}

First factor everything. Then, identify the \begin{align*}x\end{align*}

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

Holes: (-3, -4); (-2, -3)

#### Example 5

What is a possible equation for the following rational function?

The function seems to be a line with a removable discontinuity at (1, -1). The line is has slope 1 and \begin{align*}y\end{align*}

\begin{align*}f(x)=x-2\end{align*}

The removable discontinuity must not allow the \begin{align*}x\end{align*}

\begin{align*}f(x)=\frac{(x-2)(x-1)}{x-1}\end{align*}

### Review

1. How do you find the holes of a rational function?

2. What’s the difference between a hole and a removable discontinuity?

3. If you see a hollow circle on a graph, what does that mean?

Without graphing, identify the location of the holes of the following functions.

4. \begin{align*}f(x)=\frac{x^2+3x-4}{x-1}\end{align*}

5. \begin{align*}g(x)=\frac{x^2+8x+15}{x+3}\end{align*}

6. \begin{align*}h(x)=\frac{x^3+6x^2+2x-8}{x^2+x-2}\end{align*}

7. \begin{align*}k(x)=\frac{x^3+6x^2+2x-8}{x^2-3x+2}\end{align*}

8. \begin{align*}j(x)=\frac{x^3+4x^2-17x-60}{x^2-9}\end{align*}

9. \begin{align*}f(x)=\frac{x^3+4x^2-17x-60}{x^2-5x+4}\end{align*}

10. \begin{align*}g(x)=\frac{x^3-4x^2-19x-14}{x^2-8x+7}\end{align*}

11. What is a possible equation for the following rational function?

12. What is a possible equation for the following rational function?

Sketch the following rational functions.

13. \begin{align*}f(x)=\frac{x^3+4x^2-17x-60}{x^2-x-12}\end{align*}

14. \begin{align*}g(x)=\frac{x^3+4x^2-17x-60}{x^2+8x+15}\end{align*}

15. \begin{align*}h(x)=\frac{x^3-4x^2-19x-14}{x^2-6x-7}\end{align*}

### Review (Answers)

To see the Review answers, open this PDF file and look for section 2.7.