3: Rational Functions

A progression-based guide to simplifying, analyzing, graphing, and solving problems involving rational functions, equations, and inequalities.

Simplifying Rational Expressions

A is a quotient of polynomials:

f(x)=\frac{p(x)}{q(x)},\qquad q(x)\neq 0.

The denominator is central: it determines which inputs are allowed and often determines the graph's discontinuities. These functions are useful for representing rates, ratios, average cost, concentration, and other quantities involving division.

To simplify a rational expression:

Factor the numerator and denominator completely.
Find the values that make the original denominator zero.
Cancel common factors.
State the restrictions from the original denominator.

For example,

\frac{x^2-9}{x^2-3x} =\frac{(x-3)(x+3)}{x(x-3)} =\frac{x+3}{x},

but the original denominator is zero at $x=0$ and $x=3$ . Therefore, the simplified result must be written with $x\neq 0,3$ . The canceled factor changes the formula but does not restore the excluded input.

Takeaway: Always determine restrictions from the original denominator before canceling.

Domains, Restrictions, and Discontinuities

The domain of a contains every real number except the zeros of its original denominator. To find it, set the denominator equal to zero, solve, and exclude the resulting values.

For

f(x)=\frac{2x+1}{(x-4)(x+2)},

the restrictions are $x\neq 4$ and $x\neq -2$ , so the domain is

(-\infty,-2)\cup(-2,4)\cup(4,\infty).

Factoring also distinguishes the two main discontinuity types. In

f(x)=\frac{x^2-1}{(x-1)(x+2)} =\frac{(x-1)(x+1)}{(x-1)(x+2)} =\frac{x+1}{x+2}, \qquad x\neq 1,-2,

the factor $x-1$ cancels, so $x=1$ is a hole. The factor $x+2$ remains in the denominator, so $x=-2$ is a . The hole's output is found from the simplified function:

y=\frac{1+1}{1+2}=\frac{2}{3},

so the hole is at $\left(1,\frac{2}{3}\right)$ .

Takeaway: A canceled denominator factor creates a hole; a denominator factor that remains creates a .

Asymptotes and End Behavior

An asymptote describes long-term or unbounded behavior. A graph may approach an asymptote without being forbidden from crossing it.

For vertical behavior, simplify first. If $q(a)=0$ and the factor responsible for that zero remains in the denominator, then $x=a$ is a .

For end behavior, compare the degrees of the numerator and denominator in

f(x)=\frac{p(x)}{q(x)}.

If $\deg p<\deg q$ , the is $y=0$ .
If $\deg p=\deg q$ , the is the ratio of the leading coefficients.
If $\deg p>\deg q$ , there is no ; polynomial division may reveal a slant or other polynomial asymptote.

Examples:

\frac{3x^2-1}{x^3+4}\quad\Rightarrow\quad y=0,

because the numerator has lower degree, and

\frac{4x^2+1}{2x^2-7}\quad\Rightarrow\quad y=\frac{4}{2}=2,

because the degrees are equal. For

h(x)=\frac{x^2+1}{x-1},

division gives

h(x)=x+1+\frac{2}{x-1},

so the graph has the slant asymptote $y=x+1$ .

Takeaway: Use factors to identify vertical behavior and degree comparison or division to identify end behavior.

Graphing Rational Functions

A reliable graphing process combines algebraic information with a few test points:

Factor the numerator and denominator.
State all domain restrictions.
Identify holes and vertical asymptotes.
Find the horizontal or slant asymptote.
Find intercepts.
Test points in each interval separated by restrictions and intercepts.
Sketch each branch using the calculated behavior.

For

f(x)=\frac{x-2}{x+1},

the restriction $x\neq -1$ gives the $x=-1$ . Since the numerator and denominator have equal degree, the is $y=1$ . The numerator is zero at $x=2$ , giving the intercept $(2,0)$ , and

f(0)=\frac{-2}{1}=-2,

so the other intercept is $(0,-2)$ . Two useful test values are

f(-2)=4,\qquad f(3)=\frac{1}{4}.

These values help position the two branches on either side of $x=-1$ . As $x$ approaches the , the function becomes unbounded; as $x$ approaches positive or negative infinity, the graph approaches $y=1$ .

Takeaway: A complete sketch should show restrictions, holes, asymptotes, intercepts, and the behavior of each branch.

Solving Rational Equations

To solve a , preserve the original restrictions throughout the algebra:

Factor every denominator.
State values that make any denominator zero.
Find the least common denominator.
Multiply every term by the least common denominator.
Solve the resulting equation.
Check each candidate in the original equation.

Consider

\frac{2}{x-1}+\frac{1}{x+1}=1.

The restrictions are $x\neq 1,-1$ , and the least common denominator is $(x-1)(x+1)$ . Multiplying through gives

2(x+1)+(x-1)=(x-1)(x+1).

After simplifying,

3x+1=x^2-1,

x^2-3x-2=0.

The quadratic formula produces

x=\frac{3\pm\sqrt{17}}{2}.

Neither value is restricted, so both are solutions. A candidate produced during the algebra must be rejected if it makes an original denominator zero; such a candidate is extraneous.

Takeaway: Clearing denominators simplifies the equation, but only the original equation determines whether a candidate is valid.

Solving Rational Inequalities

A is easiest to solve with a sign chart. First move all terms to one side so the comparison is with zero. Then factor the numerator and denominator.

For

\frac{x-1}{x+3}\geq 0,

the critical numbers are $x=1$ , where the numerator is zero, and $x=-3$ , where the denominator is zero. They divide the number line into

(-\infty,-3),\qquad (-3,1),\qquad (1,\infty).

Testing one value from each interval gives the signs:

The expression is positive on $(-\infty,-3)$ .
The expression is negative on $(-3,1)$ .
The expression is positive on $(1,\infty)$ .

Because the inequality includes equality, include the numerator zero $x=1$ . Never include $x=-3$ , because the expression is undefined there. The solution is

(-\infty,-3)\cup[1,\infty).

Takeaway: Include zeros of the numerator when the inequality permits equality, but always exclude zeros of the denominator.

Checking Reasoning and Common Errors

Several errors recur across rational-function problems:

Canceling a factor without recording its original restriction removes a hole from the written work.
Calling every denominator zero a confuses a hole with an unbounded discontinuity.
Using the original numerator to find an intercept after cancellation can incorrectly label a hole as an intercept.
Including a denominator zero in an inequality solution includes a value where the expression is undefined.
Failing to check solutions to an equation can retain an extraneous value.
Assuming that a graph cannot cross an asymptote treats a limiting description as an absolute barrier.

A dependable final check asks:

Did I find restrictions from every original denominator?
Did I distinguish canceled factors from remaining factors?
Did I use the simplified expression for intercepts and asymptotes while retaining original restrictions?
Did I check equation solutions and exclude denominator zeros from inequality answers?

Together, these checks connect the algebraic form, domain, graph, and solution set of a rational expression.