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5. Rational Functions
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Algebra 2

5.03 Other rational functions

Lesson
Practice

Removable points of discontinuity

As we saw with rational parent functions f\left(x\right)=\dfrac{1}{x} and f\left(x\right)=\dfrac{1}{x^2}, their characteristics include:

  • domain

  • range

  • zeros

  • x- and y-intercepts

  • increasing and decreasing intervals

  • end behavior

  • asymptotes

These characteristics also apply to rational functions in other forms, where the numerator is an algebraic expression.

A rational function is any function that can be expressed as a quotient of two polynomials, with a non-zero denominator.

f\left(x\right)=\dfrac{P\left(x\right)}{Q\left(x\right)},\quad Q\left(x\right) \neq 0Where P\left(x\right) and Q\left(x\right) are polynomials.

Rational functions can sometimes be simplified by factoring the numerator and denominator and then simplifying any common factors between them.

A characteristic of rational functions that have been simplified is a removable point of discontinuity.

Removable point of discontinuity

A point at which a function is not defined, but the graph of the function approaches that point from both sides. Sometimes called a "hole"

A removable point of discontinuity will occur at x=a if x=a is a zero of both the numerator and the denominator.

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Notice that the denominator of the graphed function can be factored and simplified further:\begin{aligned}y&=\dfrac{x-1}{x^2-1}\\&=\dfrac{x-1}{\left(x+1\right)\left(x-1\right)}\end{aligned}From here, we can see that x=-1 is only a zero of the denominator. This is why there is a vertical asymptote at x=-1.

Notice that x = 1 is a zero of both the numerator and denominator, so there is a removable point of discontinuity at x = 1.

To find the y-value of a hole, substitute the x-value into the simplified form of the function.

Points of discontinuity, like asymptotes, impact the domain and range of the function because the function does not exist at that point.

Examples

Example 1

Consider the function f \left( x \right) = \dfrac{x+3}{x^2 + 7x + 12}.

a

What is the domain of the function?

Worked Solution
Create a strategy

As a rational function, the domain will be all real values of x except for those which make the denominator equal to 0. So we will set the denominator to be equal to 0 and solve for x.

Apply the idea
\displaystyle x^2 + 7x + 12\displaystyle =\displaystyle 0Set denominator equal to 0
\displaystyle \left(x + 3\right)\left(x + 4\right)\displaystyle =\displaystyle 0Factor the quadratic
\displaystyle x + 3 = 0,\quad x + 4\displaystyle =\displaystyle 0Set each factor equal to 0
\displaystyle x = -3,\quad x\displaystyle =\displaystyle -4Solve each equation for x

These are the values that make the denominator equal to 0, which would result in an undefined expression. As such, these values must be removed from the domain.

The domain of the function is \left(-\infty, -4\right) \cup \left(-4, -3\right) \cup \left(-3, \infty\right).

b

For each value of x not in the domain, determine whether there is a vertical asymptote or a removable point of discontinuity at that value.

Worked Solution
Create a strategy

We need to compare the factors of the numerator with the factors of the denominator.

If a factor in the denominator is not in the numerator, or is in the numerator but with a lower multiplicity, then there will be a vertical asymptote at the corresponding value of x.

If a factor is in the denominator and the numerator with the same or higher multiplicity, there will be a removable point of discontinuity instead.

Apply the idea
\displaystyle f\left(x\right)\displaystyle =\displaystyle \dfrac{x+3}{x^2 + 7x + 12}State the function
\displaystyle =\displaystyle \dfrac{x+3}{\left(x + 3\right)\left(x + 4\right)}Previous factoring of the denominator

We can see that the factor \left(x + 3\right) appears in both the numerator and denominator, while the factor \left(x + 4\right) only appears in the denominator.

As such, x = -3 will correspond to a removable point of discontinuity, while x = -4 will be the equation of a vertical asymptote.

Reflect and check

We can see why there is a removable point of discontinuity by analyzing the graph of the simplified form of f\left(x\right)= \dfrac{x+3}{x^2+7x+12}.

We know that \dfrac{\left (x+3 \right)}{\left (x+3 \right)}=1 when x\neq -3, so the function can be written as:

\displaystyle f\left(x\right)\displaystyle =\displaystyle \dfrac{x+3}{x^2+7x+12}Original function
\displaystyle =\displaystyle \dfrac{\left (x+3 \right)}{\left (x+3 \right) \left(x+4 \right)}Factored form
\displaystyle =\displaystyle \dfrac{\cancel{\left (x+3 \right)}}{\cancel{\left (x+3 \right)} \left(x+4 \right)}For x\neq -3
\displaystyle =\displaystyle \dfrac{1}{x+4}

Consider the graph of f\left(x\right) = \dfrac{1}{x+4}, x\neq -3:

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The vertical asymptote at x=-4 can still be seen in the simplified form of the function, but x=-3 would lead to an undefined function as well.

We have to exclude both x=-3 and x=-4 from the domain of the original function, leaving a hole at x=-3.

c

Describe the end behavior of the function as x \to -\infty and as x \to \infty.

Worked Solution
Create a strategy

The end behavior of a graph is how the graph behaves as x approaches infinity or negative infinity. Consider what happens to f\left(x\right) for very small values of x and very large values of x.

If the degree of the denominator is larger than the degree of the numerator, there is a horizontal asymptote of y=0, which will determine the end behavior of the function.

Apply the idea
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Since the degree of the denominator is greater than the numerator, there is a horizontal asymptote at y=0.

As x gets increasingly small, f\left(x\right) approaches 0 from below the x-axis.

As x gets increasingly large, f\left(x\right) approaches 0 from above the x-axis.

Using proper notation, the function's end behavior is as follows:

  • As x\to -\infty, f\left(x\right)\to 0

  • As x\to \infty, f\left(x\right)\to 0

Example 2

Abdul used a calculator to graph the function f\left(x\right)=\dfrac{x^2-2x-15}{x+3}. He was surprised to see the graph of a line instead of a rational function.

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a

Explain why the graph appears to be linear.

Worked Solution
Create a strategy

A straight line implies that the given rational function can be simplified. Recall that common factors in the numerator and denominator can lead to removable points of discontinuity, which affect the graph of the function.

Apply the idea

First, we will factor the rational expression.

f\left(x\right)=\dfrac{x^2-2x-15}{x+3} = \dfrac{(x-5)(x+3)}{x+3}

If we restrict the domain so that it does not include x=-3, we can divide out the common factor of (x+3). This simplifies to f\left(x\right)=x-5 for all x\neq -3, which is the equation of the line graphed.

Reflect and check

Notice there is one important distinction between f\left(x\right) = x-5 and our rational function. Rational functions exclude any values of x that result in a 0 in the denominator. Therefore, x = -3 should be excluded from the domain of the linear graph since it is a removable discontinuity.

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You should always be careful when graphing with technology. Not all graphing calculators will show points of discontinuity. It is important to consider the equation and context for possible domain restrictions. You can generally use the tracing tool of the graphing calculator to see that the function is, in fact, undefined at x=-3

b

Identify the domain and range.

Worked Solution
Create a strategy

The domain is all the real values of x, except for the value where the removable point of discontinuity lies.

The range is all the real values of y, except for the value where the removable point of discontinuity lies.

Apply the idea

In part (a), we found the only value excluded from the domain is x=-3.

Therefore, the domain is \left(-\infty,-3\right)\cup\left(-3,\infty\right).

To find the range algebraically, we would need to substitute the x-value of the removable point of discontinuity into the simplified function.

From part (a), we know the simplified function is:

f\left(x\right) = x-5

To find the y-value of the hole, the value that should be excluded from the range, we substitute x=-3 into f\left(x\right)=x-5:

\displaystyle f(3)\displaystyle =\displaystyle \left(-3\right)-5
\displaystyle =\displaystyle -8

This means the range is \left(-\infty,-8\right)\cup\left(-8,\infty\right).

c

Find the intercepts.

Worked Solution
Create a strategy

Look at the graph to identify the points where the line crosses the x- and y-axis.

Apply the idea

The x-intercept is \left(5,0\right).

The y-intercept is \left(0,-5\right).

Reflect and check

We can find the intercepts algebraically using the simplified form of the function.

  • x-intercept:\begin{aligned}y&=x-5\\0&=x-5\\5&=x\end{aligned}

  • y-intercept:\begin{aligned}y&=x-5\\y&=0-5\\y&=-5\end{aligned}

Example 3

Consider the function f\left(x\right) = \dfrac{x}{3x - 3} and the graph of g\left(x\right).

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a

Compare the increasing intervals of f\left(x\right) and g\left(x\right).

Worked Solution
Create a strategy

Consider how each function changes as you move across the domain.

Apply the idea

From the graph, we can see that g\left(x\right) increases over its entire domain. So, the increasing interval of g\left(x\right) is \left( -\infty, \infty \right).

We can use technology to graph f\left(x\right).

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The function f\left(x\right) has an asymptote at x=1, and it is decreasing on either side of the asymptote. So, the increasing intervals of f\left(x\right) are \left( -\infty, 1 \right) and \left( 1, \infty \right).

b

Compare the zeros of f\left(x\right) and g \left(x\right).

Worked Solution
Create a strategy

Look at the graph to identify the zeros of g \left(x\right). Determine when the numerator is 0 to identify possible zeros of f\left(x\right).

Apply the idea

The function g \left(x\right) intercepts the x-axis at x = 0. So x=0 is the zero of g \left(x\right).

The function f\left(x\right) has a 0 in the numerator when x = 0. Since this value is not excluded from the domain, x = 0 is the zero of f\left(x\right).

Both functions have the same zero at the point \left(0,0\right).

c

State a value in the domain of g\left(x\right) that is not in the domain of f\left(x\right).

Worked Solution
Create a strategy

Look for asymptotes and removable discontinuities to find discrepancies between the domains.

Apply the idea

f\left(x\right) has an asymptote at x=3, but the domain of g\left(x\right) includes x=3.

So x=3 is in the domain of g\left(x\right), but it is excluded from the domain of f\left(x\right).

Idea summary

The characteristics of rational functions include:

  • domain

  • range

  • zeros

  • x- and y-intercepts

  • increasing and decreasing intervals

  • end behavior

  • asymptotes

  • removable points of discontinuity (holes)

Rational functions can sometimes be simplified by factoring the numerator and denominator and then simplifying any common factors between them.

A removable point of discontinuity will occur at x = a if x=a is a zero of both the numerator and the denominator.

To find the y-value of a hole, substitute the x-value into the simplified form of the function.

Asymptotes

Recall that vertical asymptotes occur for values of x that make the denominator 0. We can find the asymptotes by factoring the denominator and setting each factor equal to 0.

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The graph shows this function: f\left(x\right) = \dfrac{1}{x^2+2x-15}

Simplified, we have: f\left(x\right) = \dfrac{1}{(x-3)(x+5)}

Setting each factor in the denominator to 0 and solving, we get the asymptotes x = 3 and x=-5.

Whether or not the function will have a horizontal asymptote depends on the degree (highest exponent) of the polynomials in the numerator and denominator of the function.

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If the degree of the numerator is less than the degree of the denominator, then y = 0 (the x-axis) is the horizontal asymptote.

Consider the function f\left(x\right) = \dfrac{4}{x-2}:

  • Degree of numerator is 0

  • Degree of denominator is 1

Since 0<1, the horizontal asymptote is at y=0.

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If the degree of the numerator is equal to the degree of the denominator, then the horizontal asymptote is the ratio of the leading coefficients.

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

  • Degree of numerator is 1

  • Degree of denominator is 1

Since 1=1, the horizontal asymptote is at y=\dfrac{3}{2}.

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If the degree of the numerator is greater than the degree of the denominator, then there are no horizontal asymptotes.

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

  • Degree of numerator is 2

  • Degree of denominator is 1

Since 2>1, this function has no horizontal asymptote.

Examples

Example 4

Consider the function y = \dfrac{x^2 - 5x + 4}{x + 1}

a

Determine the coordinates of the intercepts of the function.

Worked Solution
Create a strategy

The y-intercept of a function occurs where x = 0. We can substitute this into the function to find the y-coordinate.

Similarly, the x-intercept(s) of a function occur where y = 0. In particular, for a rational function, this occurs at values of x which make the numerator equal to 0 (and not the denominator).

Apply the idea

For the y-intercept, we have:

\displaystyle y\displaystyle =\displaystyle \dfrac{0^2 - 5\left(0\right) + 4}{0 + 1}Substitute x = 0 into the function
\displaystyle =\displaystyle \dfrac{4}{1}Simplify numerator and denominator
\displaystyle =\displaystyle 4Evaluate the division

So the y-intercept occurs at \left(0, 4\right).

For the y-intercepts, we have:

\displaystyle \dfrac{x^2 - 5x + 4}{x + 1}\displaystyle =\displaystyle 0Set y = 0
\displaystyle \dfrac{\left(x - 4\right)\left(x - 1\right)}{x + 1}\displaystyle =\displaystyle 0Factor the quadratic in the numerator
\displaystyle \left(x - 4\right)\left(x - 1\right)\displaystyle =\displaystyle 0Multiply both sides by x + 1
\displaystyle x - 4 = 0,\quad x - 1\displaystyle =\displaystyle 0Set each factor equal to 0
\displaystyle x = 4,\quad x\displaystyle =\displaystyle 1Solve each equation for x

The function is undefined at x = -1 (the value which makes the denominator 0). This is neither of the values we just solved for, and so we have x-intercepts at both \left(1, 0\right) and \left(4, 0\right).

Reflect and check

It is important to consider values of x which make the denominator equal to 0, as the function cannot have an intercept if it is undefined at that value of x.

Note that the step of work "multiply both sides by x + 1" fails if x = -1. This is why we factored the numerator first, so that we could see that there was no corresponding factor of x + 1 in the numerator.

b

Identify any asymptotes or points of discontinuity.

Worked Solution
Create a strategy

A rational function has a horizontal asymptote if the degree of the numerator is less than or equal to the degree of the denominator. Otherwise, it does not have a horizontal asymptote.

A rational function has a vertical asymptote when the denominator of the simplified form is equal to zero.

Lastly, a rational function has a removable discontinuity where there are any common algebraic factors between the numerator and denominator.

Apply the idea

In this case, the function is y = \dfrac{x^2 - 5x + 4}{x + 1}, which has a numerator of degree 2 and a denominator of degree 1.

The degree of the numerator is larger than the degree of the denominator, so the function has no horizontal asymptote.

We can factor the function y = \dfrac{x^2 - 5x + 4}{x + 1} = \dfrac{(x-1)(x-4)}{x + 1}.

Since the numerator and denominator have no common factor, there are no removable discontinuities. The vertical asymptote occurs when the denominator is 0 which is at x = -1. So, x = -1 is the vertical asymptote.

c

State the domain of f\left(x\right).

Worked Solution
Create a strategy

Use the asymptotes and points of discontinuity to determine the domain.

Apply the idea

Since we have a vertical asymptote at x = -1, the domain does not include x=-1. In set builder notation, we say the domain is \{ x | x \neq -1 \}.

Example 5

Consider the function f\left(x\right)=-\dfrac{4x+1}{x-5}.

a

Determine any asymptotes of f\left(x\right).

Worked Solution
Create a strategy

To find the horizontal asymptotes, consider the degree of the numerator and denominator. To find the vertical asymptotes, consider what values make the denominator equal to zero.

Apply the idea

The degree of the numerator is 1, and the degree of the denominator is 1. Since the degrees are equal, the horizontal asymptote is the ratio of leading coefficients of the numerator and denominator.

  • Leading coefficient of numerator: -4

  • Leading coefficient of the denominator: 1

The horizontal asymptote is y=\dfrac{-4}{1}=-4.

The denominator of f\left(x\right) is 0 when x=5. Since f\left(x\right) does not simplify, there is a vertical asymptote at x = 5.

b

Describe the end behavior.

Worked Solution
Create a strategy

To find the end behavior, we need to first identify the horizontal asymptote. Then, we can use our knowledge of parent rational functions to find the end behavior as x \to -\infty and x\to \infty.

Apply the idea

Using technology to graph our function:

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The graph confirms that there is a horizontal asymptote at y=-4, and the y-values approach it as x tends toward negative infinity and positive infinity.

This means that as x \to -\infty, f\left(x\right) \to -4, and as x\to\infty, f\left(x\right) \to -4.

Reflect and check

We can also substitute negative values of x into the function to see what the y-values approach:

For x=-100:

\displaystyle -\dfrac{4x+1}{x-5}\displaystyle =\displaystyle -\dfrac{4\left(-100\right)+1}{\left(-100\right)-5}Substitute x=-100
\displaystyle =\displaystyle -\dfrac{-399}{-105}
\displaystyle =\displaystyle -3.8

For x=-500:

\displaystyle -\dfrac{4x+1}{x-5}\displaystyle =\displaystyle -\dfrac{4\left(-500\right)+1}{\left(-500\right)-5}Substitute x=-500
\displaystyle =\displaystyle -\dfrac{-1999}{-505}
\displaystyle \approx\displaystyle -3.9584

So as the x-values get larger in the negative direction, the y-values get closer to y = -4. The same could have been done to check the end behavior as x \to \infty.

c

Describe the increasing and decreasing intervals of f\left(x\right).

Worked Solution
Create a strategy

To find the increasing and decreasing intervals of f\left(x\right), we can observe the graph from the previous part. We also know that our horizontal asymptote is at y=-4, and the vertical asymptote is at x = 5.

Apply the idea

Using technology to graph our function:

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The function is increasing on either side of its vertical asymptote, x = 5. Therefore, f\left(x\right) is increasing on the intervals \left( - \infty, 5 \right) and \left( 5, \infty \right).

The function has no interval where it is decreasing.

d

Determine the range of f\left(x\right).

Worked Solution
Create a strategy

Use the fact that there is a horizontal asymptote at y = -4 to determine the range.

Apply the idea

Since y = -4 is excluded from the range, the range of f\left(x\right) can be written in interval notation as \left( -\infty, -4 \right) \cup \left( -4, \infty \right ).

Reflect and check

The range can also be written in set notation: \{y|y\neq -4\} or \{y|y<-4\text{ or }y>-4\}

Idea summary

For a rational function f\left(x\right)=\dfrac{P\left(x\right)}{Q\left(x\right)}:

  • If the degree of P\left(x\right)< the degree of Q\left(x\right), then y = 0 (the x-axis) is the horizontal asymptote.

  • If the degree of P\left(x\right)= the degree of Q\left(x\right), then the horizontal asymptote is the ratio of the leading coefficients of P\left(x\right) and Q\left(x\right).

  • If the degree of P\left(x\right)> the degree of Q\left(x\right), then there are no horizontal asymptotes.

To find any vertical asymptotes, factor the denominator and determine the values that make it equal to zero. If any factors are in both the numerator and denominator, it is a point of discontinuity. Otherwise, it represents a vertical asymptote.

Outcomes

A2.F.1

The student will investigate, analyze, and compare square root, cube root, rational, exponential, and logarithmic function families, algebraically and graphically, using transformations.

A2.F.1e

Compare and contrast the graphs, tables, and equations of square root, cube root, rational, exponential, and logarithmic functions.

A2.F.2

The student will investigate and analyze characteristics of square root, cube root, rational, polynomial, exponential, logarithmic, and piecewise-defined functions algebraically and graphically.

A2.F.2a

Determine and identify the domain, range, zeros, and intercepts of a function presented algebraically or graphically, including graphs with discontinuities.

A2.F.2b

Compare and contrast the characteristics of square root, cube root, rational, polynomial, exponential, logarithmic, and piecewise-defined functions.

A2.F.2c

Determine the intervals on which the graph of a function is increasing, decreasing, or constant.

A2.F.2f

For any value, x, in the domain of f, determine f(x) using a graph or equation. Explain the meaning of x and f(x) in context, where applicable.

A2.F.2g

Describe the end behavior of a function.

A2.F.2h

Determine the equations of any vertical and horizontal asymptotes of a function using a graph or equation (rational, exponential, and logarithmic).

A2.F.2j

Graph the inverse of a function as a reflection over the line y = x.

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