A hole, or removable point of discontinuity, in a rational function's graph occurs when there is a real zero in the numerator and denominator that divides out. Specifically, if the multiplicity of a real zero in the numerator is greater than or equal to its multiplicity in the denominator, then the graph of the rational function has a hole at the corresponding input value.

If the graph of a rational function r\left(x\right) has a hole at x=c, then the location of the hole can be determined by examining the input-output pairs close to c. If input values just to the left and right of c correspond to output values very close to a specific number, L then the hole is located at the point with coordinates \left(c,L\right).

-4

-3

-2

-1

-4

-3

-2

-1

r(x)

The function r\left(x\right)=\dfrac{\left(x+1\right)\left(x+4\right)}{2 \left(x+1\right)\left(x-2\right)} shown on the graph has a hole at x=-1 because the factor \left(x+1\right) appears in both the numerator and denominator.

By examining the limits as x approaches -1 from the left and right x=1 we can identify the coordinates of the hole.

From the left:

r\left(-1.25\right)=-0.423, \text{ } r\left(-1.1\right)=-0.468, \\ \text{ } r\left(-1.01\right)=-0.497

From the right:

r\left(-0.75\right)=-0.591, \text{ } r\left(-0.9\right)=-0.534, \text{ } r\left(-0.99\right)=-0.503

We can see from each direction that as the x-value gets extremely close to -1 the y-value approaches -0.5.

Or in limit notation:

\lim_{x \to c} r\left(x\right) = -0.5

Because:

\lim_{x \to c+} r\left(x\right) = -0.5 \\ \text{and} \\ \lim_{x \to c-} r\left(x\right) = -0.5

Examples

Example 1

Consider the rational function: f\left(x\right)=\dfrac{x^3-2x^2-x+2}{x^2-1}

Identify the x-values for which f\left(x\right) has holes.

Worked Solution

Create a strategy

Start by factoring the numeratior and denominator completely and then look for any common factors.

Apply the idea

Factor the numerator by grouping:

\displaystyle \text{numerator}	\displaystyle =	\displaystyle x^3-2x^2-x+2
	\displaystyle =	\displaystyle x^2\left(x-2\right)-1\left(x-2\right)
	\displaystyle =	\displaystyle \left(x^2-1\right)\left(x-2\right)
	\displaystyle =	\displaystyle \left(x-1\right)\left(x+1\right)\left(x-2\right)

Factor the denominator (difference of two squares):

\displaystyle \text{denominator}	\displaystyle =	\displaystyle x^2-1
\displaystyle \text{denominator}	\displaystyle =	\displaystyle \left(x-1\right)\left(x+1\right)

Rewriting f\left(x\right) in factored form we get: f\left(x\right)=\dfrac{\left(x-1\right)\left(x+1\right)\left(x-2\right)}{\left(x-1\right)\left(x+1\right)}.

There are two common factors between the numerator and denominator, they are \left(x-1\right) and \left(x+1\right).

This means there is a hole at x=1 and x=-1.

Identify the coordinates of any holes.

Worked Solution

Create a strategy

By examining the outputs very close to x=1 and x=-1, we can determine the y-values of the holes.

Apply the idea

One hole is located at x=1 so we can evaluate f\left(x\right) at x=0.999 and x=1.001.

f\left(0.999\right)=\dfrac{\left(0.999-1\right)\left(0.999+1\right)\left(0.999-2\right)}{\left(0.999-1\right)\left(0.999+1\right)}=-1.001

f\left(1.001\right)=\dfrac{\left(1.001-1\right)\left(1.001+1\right)\left(1.001-2\right)}{\left(1.001-1\right)\left(1.001+1\right)}=-0.999

These output values are getting extremely close to -1 as we approach the hole from each side so the coordinates of one hole are \left(1,-1\right).

The other hole is located at x=-1 so we can evaluate f\left(x\right) at x=-1.001 and x=-0.999.

f\left(-1.001\right)=\dfrac{\left(-1.001-1\right)\left(-1.001+1\right)\left(-1.001-2\right)}{\left(-1.001-1\right)\left(-1.001+1\right)}=-3.001

f\left(-0.999\right)=\dfrac{\left(-0.999-1\right)\left(-0.999+1\right)\left(-0.999-2\right)}{\left(-0.999-1\right)\left(-0.999+1\right)}=-2.999

These output values are getting extremely close to -3 as we approach the hole from each side so the coordinates of one hole are \left(-1,-3\right).

Example 2

Consider the limit notation for a rational function:

\lim_{x \to 3} f\left(x\right)=4

Interpret the location of a hole on the function's graph based on this information.

Worked Solution

Create a strategy

This notation means that as the input values approach 3 the output values approach 4.

Apply the idea

The hole is located at \left(3,4\right).

Idea summary

Holes in the graph of a rational function occur when the same zero exists in the numerator and denominator (as long as the multiplicity of the zero in the numerator is greater than or equal to its multiplicity in the denominator).

The location of a hole in a rational function can be determined by examining the limit of the function as it approaches the input value of the hole.

1.10 Rational functions and holes

Introduction

Ideas

Rational functions and holes

Examples

Example 1

Example 2

Outcomes

1.10.A

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