Rational Functions and Holes - AP PreCalculus Study Guide

The Core Idea: Rational Functions and Holes

A rational function is defined as a ratio of two polynomial functions, $f(x)=\frac{p(x)}{q(x)}$. The behavior of these functions is particularly interesting at $x$-values where the denominator, $q(x)$, equals zero, as this leads to a potential division by zero. This topic explores a specific type of discontinuity known as a "hole," or a removable discontinuity. A hole occurs at an $x$-value, $x=c$, where a factor in the denominator that would cause a division by zero is "canceled out" by an identical factor in the numerator.

Conceptually, a hole is a single point that is missing from the graph of the function. The function's graph will approach this point from both sides, but the point itself is undefined. The conditions for the existence of a hole depend on the zeros of both the numerator and the denominator, specifically on the multiplicity (the number of times a factor is repeated) of the common zero. By simplifying the rational expression, we can determine the precise coordinates of this missing point.

Key Definitions and Rules

Conditions for a Hole

A rational function $f(x)=\frac{p(x)}{q(x)}$ has a hole at $x=c$ if two conditions are met:

1. Common Zero: Both the numerator and the denominator are equal to zero when evaluated at $x=c$.

   $$p(c)=0\text{and}q(c)=0$$

2. Multiplicity Rule: The multiplicity of the zero $c$ in the numerator's polynomial, $p(x)$, must be greater than or equal to the multiplicity of the zero $c$ in the denominator's polynomial, $q(x)$.

• Multiplicity refers to the number of times a particular factor $(x-c)$ appears in the factored form of a polynomial. For example, in $p(x)=(x-2{)}^3(x+1)$, the zero $x=2$ has a multiplicity of 3.

Finding the Coordinates of a Hole

If the conditions for a hole at $x=c$ are met, its coordinates $(c,y)$ can be determined as follows:

1. $x$-coordinate: The $x$-coordinate of the hole is simply the value $c$.

2. $y$-coordinate: To find the $y$-coordinate, first simplify the rational expression $f(x)$ by canceling the common factor $(x-c)$. Let the simplified function be $g(x)$. The $y$-coordinate of the hole is the value of this simplified function evaluated at $x=c$.

   $$y=g(c)$$

Understanding Multiplicity in Holes

The concept of multiplicity is critical for distinguishing between a hole and other types of discontinuities like vertical asymptotes. The rule states that for a hole to exist at a common zero $x=c$, the multiplicity of the factor $(x-c)$ in the numerator must be greater than or equal to its multiplicity in the denominator.

Let $m$ be the multiplicity of the zero $c$ in the numerator $p(x)$, and let $n$ be the multiplicity of the zero $c$ in the denominator $q(x)$. The function can be written as:

$$f(x)=\frac{(x-c{)}^m\cdot (\text{other factors})}{(x-c{)}^n\cdot (\text{other factors})}$$

A hole exists at $x=c$ if and only if $p(c)=0$, $q(c)=0$, and $m\ge n$.

When $m\ge n$, we can cancel out all $n$ factors of $(x-c)$ from the denominator.

• If $m=n$, the factor $(x-c)$ is completely removed from the expression after simplification.

• If $m>n$, the factor $(x-c{)}^{m-n}$ remains in the numerator after simplification.

In both cases, the division by zero at $x=c$ is eliminated in the simplified function, which is why the discontinuity is "removable." The simplified function is what is used to find the $y$-coordinate of the hole.

Core Concepts & Rules

• A hole is a removable discontinuity in the graph of a rational function.

• For a rational function $f(x)=\frac{p(x)}{q(x)}$ to have a hole at $x=c$, the value $c$ must be a zero of both the numerator $p(x)$ and the denominator $q(x)$.

• The multiplicity of the zero $c$ in the numerator must be greater than or equal to its multiplicity in the denominator.

• The $x$-coordinate of the hole is $c$.

• The $y$-coordinate of the hole is found by first algebraically simplifying the expression for $f(x)$ (by canceling the common factors of $(x-c)$) and then evaluating the resulting simplified function at $x=c$.

Step-by-Step Example 1: Identifying a Hole

Problem: Find the coordinates of the hole in the graph of the function $f(x)=\frac{x^2+x-6}{x-2}$.

Step 1: Factor the numerator and denominator.

The denominator is already factored. We factor the numerator:

$$p(x)=x^2+x-6=(x-2)(x+3)$$

So, the function in factored form is:

$$f(x)=\frac{(x-2)(x+3)}{x-2}$$

Step 2: Identify common zeros and check conditions.

The denominator is zero when $x-2=0$, so at $x=2$. Let's check if the numerator is also zero at $x=2$.

$$p(2)=(2-2)(2+3)=(0)(5)=0$$

Since both numerator and denominator are zero at $x=2$, a discontinuity exists at $x=2$.

Step 3: Check the multiplicity rule.

The factor $(x-2)$ appears with a multiplicity of 1 in the numerator ($m=1$) and a multiplicity of 1 in the denominator ($n=1$).

Since $m\ge n$ ($1\ge 1$), the discontinuity at $x=2$ is a hole.

Step 4: Find the coordinates of the hole.

The $x$-coordinate is $c=2$.

To find the $y$-coordinate, we simplify the function $f(x)$:

$$g(x)=\frac{\cancel{(x-2)}(x+3)}{\cancel{x-2}}=x+3$$

Now, evaluate the simplified function $g(x)$ at $x=2$:

$$y=g(2)=2+3=5$$

Conclusion: The function $f(x)$ has a hole at the coordinates $(2,5)$.

Step-by-Step Example 2: A Case with Higher Multiplicity

Problem: A function is defined by $h(x)=\frac{(x-4{)}^3(x+1)}{(x-4{)}^2(x-1)}$. Determine the location of any holes.

Step 1: Identify common zeros.

The function is already in factored form. We can see that the factor $(x-4)$ appears in both the numerator and the denominator. This corresponds to a common zero at $x=4$.

Step 2: Check the multiplicity rule for the common zero.

The zero at $x=4$ has a multiplicity of $m=3$ in the numerator and $n=2$ in the denominator.

We check if $m\ge n$.

$$3\ge 2$$

The condition is met, so there is a hole at $x=4$.

Step 3: Find the coordinates of the hole.

The $x$-coordinate of the hole is $c=4$.

To find the $y$-coordinate, we simplify the expression for $h(x)$ by canceling the common factors.

$$g(x)=\frac{(x-4{)}^{\cancel{3}}(x+1)}{(x-4{)}^{\cancel{2}}(x-1)}=\frac{(x-4)(x+1)}{x-1}$$

Now, we evaluate this simplified function $g(x)$ at $x=4$:

$$y=g(4)=\frac{(4-4)(4+1)}{(4-1)}=\frac{(0)(5)}{3}=\frac{0}{3}=0$$

Conclusion: The function $h(x)$ has a hole at the coordinates $(4,0)$. Note that the simplified function still has a zero at $x=4$, which is why the $y$"-coordinate of the hole is 0.

Using Your Calculator

The identification of holes is an analytical process. A graphing calculator is not used to find the hole directly but is an excellent tool for verifying your analytical result.

To verify a hole at $x=c$:

1. Graph the Function: Enter the original rational function into your calculator's graphing utility (e.g., in Y1=). For $f(x) = \frac{x^2 + x - 6}{x-2}, you would enter (x^2+x-6)/(x-2)`.

2. Inspect the Graph: Graph the function. The graph will likely look like a continuous line (in this case, the line $y=x+3$). The calculator screen does not have high enough resolution to show a single missing point. You may see a tiny pixel gap if you zoom in very closely, but this is not reliable.

3. Use the Table Feature:

   • Go to TBLSET (Table Setup). Set TblStart to a value near your suspected hole, and set ΔTbl (delta table) to a small increment like 0.1 or even smaller.

   • Go to TABLE. Scroll to the $x$-value where you believe the hole exists. For our example, scroll to $x=2$.

   • You will see an ERROR or $undefined$ message in the $y$-column corresponding to $x=2$. This confirms the function is undefined at that point. You will see defined $y$-values for $x$-values immediately surrounding it (e.g., at $x=1.9$ and $x=2.1$), which is characteristic of a hole.

4. Use the Trace or Value Feature:

   • While viewing the graph, press the TRACE button. If you try to trace exactly to $x=2$, the cursor will disappear, and the $y$-value will be blank.

   • Alternatively, use the CALC menu and select $1:value$. When prompted for $X$, enter $2$. The calculator will return a blank or undefined $y$-value, confirming the discontinuity.

These calculator methods confirm the existence and $x$-coordinate of the hole. They do not find the $y$-coordinate; that must be done analytically using the simplified function.

AP Exam Quick Hit

Common Question Types

• Finding Coordinates: Given a rational function in either expanded or factored form, you will be asked to find the coordinates of any removable discontinuities (holes).

  • Example: "Find the coordinates of the hole in the graph of $f(x)=\frac{2x^2-18}{x+3}$."

• Creating a Function: You will be given properties of a rational function, such as the location of its holes, zeros, or other discontinuities, and asked to construct a possible equation for the function.

  • Example: "Write the equation of a rational function $g(x)$ that has a hole at $(5,10)$." (A possible answer would involve a numerator and denominator with a common factor of $(x-5)$, where the simplified function equals 10 when $x=5$. For instance, $g(x)=\frac{2(x-5)(x)}{x-5}$ simplifies to $g(x)=2x$, and $2(5)=10$).

Common Mistakes

• Confusing Holes and Vertical Asymptotes: Students see that a denominator is zero at $x=c$ and immediately conclude there is a vertical asymptote. It is critical to always check if the numerator is also zero at $x=c$. If both are zero, you must investigate the multiplicities to determine if it is a hole.

• Plugging into the Original Function: A common error is to find the correct $x$-coordinate of the hole, $c$, and then try to find the $y$-coordinate by calculating $f(c)$. This will always result in the indeterminate form $\frac{0}{0}$. The $y$-coordinate must be found using the simplified function.

• Factoring Errors: The entire analytical process depends on correctly factoring the polynomials in the numerator and denominator. A simple factoring mistake will lead to missing a common factor or identifying an incorrect one.

• Ignoring Multiplicity: Seeing a common factor $(x-c)$ and immediately concluding there is a hole without comparing the multiplicities. If the multiplicity in the denominator is greater than in the numerator, the discontinuity is a vertical asymptote, not a hole.