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# Rational Function Holes – Explanation and Examples

Ever noticed those hollowed dots or points that functions sometimes have? These are called the holes of rational functions. Curious as to why these points remain unfilled?

*The holes in a rational function are the result of it sharing common factors shared by the numerator and denominator.*

*These are coordinates that the function passes through but are not part of the functionâ€™s domain and range.*

When a function contains holes, we actually need them as guide points when graphing the functionâ€™s curve. But to highlight that they are not part of the functionâ€™s solutions, we leave them as unfilled dots.

Donâ€™t worry. Weâ€™ll learn more about holes and how we can manipulate rational functions to find them in the next sections. Weâ€™ll also learn how to find the expressions of rational functions based on their holes, intercepts, and asymptotes.

## What is a hole in a rational function?

When a rational functionâ€™s numerator and denominator share a common factor, $x- a$, **a hole is found at **$\boldsymbol{(a, f(a))}$. This also means that $a$ will not be included in the domain of the function.

What do these holes represent? These are the values or the coordinates that a functionâ€™s graph may pass through but are not defined by the function, hence, the unfilled dots.

Letâ€™s take a look at the graph shown below to understand a rational functionâ€™s holes better.

The graph shown above contains three unfilled points at: $(-1, -2)$, $\left(0, -\dfrac{1}{2}\right)$, and $\left(3, \dfrac{2}{5}\right)$. This means that the function has three holes as well.

The graph shows that the graph has discontinuities at the hole and its vertical asymptote at $x = -2$. Hence, our particular example has the following domain and range:

Domain | $(-\infty, -2) \cup (-2, -1) \cup (-1, 0) \cup (0, 3) \cup (3, \infty)$ |

Range | $(-\infty, -2) \cup (-2, -0.5) \cup (-0.5, 0.4) \cup (0.4, 1) \cup (1, \infty)$ |

## How to find holes in a rational function?

Now that we understand the importance of understanding what the holes of rational functions represent, itâ€™s time that we learn how to determine holes that a function may have.

Here are some helpful steps to remember when finding the holes of a rational function:

- Express the rational functionâ€™s numerator and denominator in factored form.
- Look out for common factors shared by the numerator and denominator.
- Equate each common factor to $0$, then solve for $x$.
- Simplify the functionâ€™s expression.
- Substitute the values of $x$ from Step 3 into the function’s simplified expression to find the holeâ€™s $y$-coordinate.
- Write the hole as a coordinate, $(x, y)$, using the values from Step 3 and Step 5.

Yes, you might have guessed right. Finding holes in rational functions will require us to apply our knowledge on factoring, so check out these factoring techniques weâ€™ve written in the past if you need a refresher:

- Learn about factoring trinomials here.
- Review your knowledge of factoring by grouping.
- Whenever possible, donâ€™t forget to apply helpful properties such as the difference of two squares and the perfect square trinomial.

Why donâ€™t we apply the six steps mentioned to find the holes of $f(x) = {-3x^3 + 6x^2 + 3x â€“ 6}{4x^3 â€“ 4x}$?

First, letâ€™s express both the numerator and denominator of $f(x)$ in factored form.

Now that $f(x)$ is in factored form note the common factors shared between the numerator and denominator. For our case, we have $x â€“ 1$ and $x + 1$.

This means that there are holes at $x= -1$ and $x = 1$. To find their corresponding $y$-coordinates, substitute these values for $x$ into the simplified form of $f(x)$.

$\begin{aligned}f(x) &= \dfrac{-3\cancel{(x – 1)}\cancel{(x + 1)}(x – 2)}{ 4x\cancel{(x – 1)}\cancel{(x + 1)}}\\&=-\dfrac{3(x-2)}{4x}\end{aligned}$

$\boldsymbol{x}$ | $\boldsymbol{f(x)} =-\dfrac{3(x-2)}{4x}$ | $\boldsymbol{(x, y)}$ |

$-1$ | $-\dfrac{9}{4}$ | $\left( -1, \dfrac{9}{4}\right)$ |

$1$ | $\dfrac{3}{4}$ | $\left( 1, \dfrac{3}{4}\right)$ |

This means that $f(x)$ has holes at $\left( -1, \dfrac{9}{4}\right)$ and $\left( 1, \dfrac{3}{4}\right)$.

These will be presented as unfilled points that the graph of $f(x)$ will pass through.

Hereâ€™s the graph of $f(x)$ along its holes and asymptotes. This confirms that the function’s curve can pass through these holes, but these values will not be part of the functionâ€™s domain and range.

Why donâ€™t we write down $f(x)$â€™s range and domain?

Domain | $(\infty, -1) \cup (-1, 0) \cup (0, 1) \cup (1, \infty)$ or $ x\in \mathbb{R}: \{x \neq -1, x \neq 0, x \neq 1\}$ |

Range | $\left(\infty, -\dfrac{9}{4}\right) \cup \left(-\dfrac{9}{4}, -\dfrac{3}{4}\right) \cup \left(-\dfrac{3}{4}, \dfrac{3}{4}\right) \cup \left(\dfrac{3}{4}, \infty\right)$ or $y\in \mathbb{R}: \left \{y \neq – \dfrac{9}{4}, y \neq – \dfrac{3}{4}, y \neq \dfrac{3}{4} \right\}$ |

Ready to try out more problems? Donâ€™t worry, weâ€™ve prepared some for you! Make sure to review all the techniques and steps mentioned in this article.

*Example 1*

Find the holes found in the following rational functions.

a. $f(x) = \dfrac{-2(x â€“ 1)(x + 2)(x + 3)}{x^2(x – 1)(x + 4)(x – 3)}$

b. $g(x) = \dfrac{x^2 – 25}{x^2 â€“ 9x + 20}$

c. $h(x) = \dfrac{x^3 â€“ 7x + 6}{x^4 + 4x^3 +x^2 â€“ 6x}$

Solution

Since $f(x)$â€™s numerator and denominator are already in factored form, we can check them for common factors. For this function, we can see that $(x -1)$ is a factor shared by the numerator and denominator, so there is a hole at $x = 1$.

Simplify $f(x)$ then substitute $x = 1$ into the simplified form of $f(x)$.

$ \begin{aligned} f(x) &= \dfrac{-2\cancel{(x â€“ 1)}(x + 2)(x + 3)}{x^2\cancel{(x â€“ 1)}(x + 4)(x – 3)}\\ &=\dfrac{-2(x + 2)(x+3)}{(x+4)(x-3)}\\\\f(1)&= \dfrac{-2(1 +2)(1 +3)}{(1 + 4)(1 – 3)}\\&=\dfrac{-2(12)}{5(-2)}\\&=\dfrac{12}{5}\end{aligned}$

a. This means that $f(x)$ has a hole at $\left(1, \dfrac{12}{5}\right)$.

Factor the numerator and the denominator of $g(x)$ first then rewrite $g(x)$.

$\begin{aligned} g(x)&=\dfrac{x^2 – 25}{x^2 â€“ 9x + 20} \\&= \dfrac{(x – 5)(x +5)}{(x-4)(x-5)}\end{aligned}$

We can see that $x â€“ 5$ is a common factor shared by the numerator and denominator, so simplify $g(x)$ then substitute $x = 5$ into the simplified form of $g(x)$.

$\begin{aligned} g(x)&=\dfrac{\cancel{(x – 5)}(x +5)}{(x-4)\cancel{(x-5)}}\\&=\dfrac{x+5}{x-4}\\\\g(5)&= \dfrac{5 + 5}{5 – 4}\\&=10\end{aligned}$

b. From this, we can see that $g(x)$ has a hole at $(5, 10)$.The numerator and denominator of $h(x)$ have higher degrees than the previous examples, so letâ€™s break down the factoring process done on both.

Numerator | $ \begin{aligned} x^3 â€“ 7x + 6 &= x^3 – x – 6x + 6\\&=x(x^2-1) -6(x -1)\\&=x(x-1)(x + 1) -6(x -1)\\&=(x-1)[x(x+1)-6]\\&=(x-1)(x^2 + x -6)\\&=(x -1)(x – 2)(x +3)\end{aligned}$ |

Denominator | $ \begin{aligned} x^4 + 4x^3 +x^2 â€“ 6x &= x(x^3 + 4x^2 +x – 6)\\&=x(x^3 +3x^2 +x^2+x-6)\\&=x[x^2(x+3)+ (x-2)(x +3)]\\&=x[(x+3)(x^2 + x -2)]\\&=x[(x+3)(x -1)(x +2)]\end{aligned}$ |

$h(x) = \dfrac{( x -1)(x – 2)(x +3)}{ x(x+3)(x -1)(x+2)}$ |

We can see that $h(x)$â€™s numerator and denominator share common factors of $x â€“ 1$ and $x + 3$. Simplify $h(x)$ and substitute each of these values into the simplified form of $h(x)$.

$ \begin{aligned} h(x) &= \dfrac{\cancel{( x -1)}(x – 2)\cancel{( x + 3)}}{ x\cancel{( x + 3)}\cancel{( x -1)}(x+2)}\\&= \dfrac{x – 2}{x(x + 2)}\end{aligned}$

$\boldsymbol{x}$ | $\boldsymbol{h(x)} =\dfrac{x-2}{x( x + 2)}$ | $\boldsymbol{(x, y)}$ |

$-3$ | $-3 $ | $(-3, -15)$ |

$1$ | $1$ | $(1, -3)$ |

c. This means that $h(x)$ has holes at $(-3, -15)$ and $(1, -3)$.

*Example 2*

The graph of $f(x)$ passes through the $x$-intercepts, $(-2, 0)$ and $(-1, 0)$, and its graph is also shown below.

Use the information provided to find an expression that may represent $f(x)$.

__Solution__

Since we have $(-2, 0)$ and $(-1, 0)$ as the functionâ€™s $x$-intercept, the numerator of $f(x)$ has $(x + 2)$ and $(x + 1)$ as factors.

The graph shows a hole at $x = 1$, so $x â€“ 1$ is a common factor shared by the numerator and the denominator.

Why donâ€™t we write down what we have so far?

$f(x) = a\dfrac{(x + 2)(x + 1)(x â€“ 1)}{(x â€“ 1)}$

The variable $a$ represents the constant that we may need for $f(x)$.

Considering that there is a vertical asymptote at $x = 2$, the denominator has $x -2$ as a factor.

$f(x) = a\dfrac{(x + 2)(x + 1)(x â€“ 1)}{(x â€“ 1)(x â€“ 2)}$

When $f(x)$ is simplified and we substitute $x = 1$, we can equate the result to $-12$ to find $a$.

$ \begin{aligned} f(1)&=a\dfrac{(1 + 2)(1 + 1)}{1 â€“ 2}\\&=a \cdot \dfrac{6}{-1}\\&=-6a\\\\-12&=-6a\\a&=2\end{aligned}$

This means that the graph can be represented by $f(x) = \dfrac{2(x + 2)(x + 1)(x â€“ 1)}{(x â€“ 1)}$.

Check out these additional problems to test your knowledge of rational functions and their holes.

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### Practice Questions

*Images/mathematical drawings are created with GeoGebra.*