AP Precalculus Unit 1 Rational Functions: Structure, Asymptotes, Holes, and Solving

Rational Functions and Vertical Asymptotes

What a rational function is (and why you should care)

A rational function is a function that can be written as a ratio of two polynomials:

f(x) = \frac{P(x)}{Q(x)}

Here P(x) and Q(x) are polynomials and Q(x) is not the zero polynomial. Rational functions matter because they create graphs with “breaks” and “barriers” that polynomial graphs don’t have—things like vertical asymptotes and holes. In many applied settings, rational functions model “rate per something” relationships (for example, average cost per item, or intensity per unit area), where division naturally appears.

A big idea you’ll use constantly: the denominator controls where the function is undefined, and those undefined inputs often drive the interesting graph features.

Domain: where the function is allowed to exist

The domain of f(x) = \frac{P(x)}{Q(x)} is all real numbers except where the denominator equals zero.

• If Q(a)=0, then x=a cannot be in the domain (you cannot divide by zero).
• Whether that “problem point” becomes a vertical asymptote or a hole depends on whether a factor cancels (you’ll learn that in the holes section).

Vertical asymptotes: the “barriers” in the graph

A vertical asymptote is a vertical line x=a that the graph approaches as x gets close to a (from the left or right), and the function values grow without bound (to positive or negative infinity).

Informally: as you get closer and closer to a, the function shoots up or down.

For rational functions in factored form, vertical asymptotes typically occur at zeros of the denominator that do not cancel with the numerator.

If

f(x) = \frac{(x-2)(x+1)}{(x-3)(x+1)}

then x=-1 is not a vertical asymptote because x+1 cancels (that creates a hole). But x=3 stays in the denominator, so x=3 is a vertical asymptote.

Why the graph behaves wildly near a vertical asymptote

Near x=a, the denominator Q(x) is very close to zero. Dividing by a very small number creates a very large magnitude result. The _sign_ (positive or negative) depends on the signs of numerator and denominator on each side of a.

A practical way to predict behavior is to test values just left and right of the asymptote (or do sign analysis using factors).

Example 1: Find vertical asymptotes and describe one-sided behavior

Consider

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

1) Find where the denominator is zero:

x-1=0
x=1

So x=1 is a candidate vertical asymptote.

2) Check if it cancels: There is no common factor between x+2 and x-1, so it does not cancel.

Therefore, x=1 is a vertical asymptote.

3) One-sided behavior:

• If x is slightly less than 1 (like 0.9), then x-1 is negative and small in magnitude, while x+2 is positive. Positive divided by negative small gives a large negative.
• If x is slightly greater than 1 (like 1.1), then x-1 is positive and small, and x+2 is positive. Positive divided by positive small gives a large positive.

So:

• as x approaches 1 from the left, f(x) goes to negative infinity
• as x approaches 1 from the right, f(x) goes to positive infinity

Example 2: Vertical asymptotes from factored form

Consider

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

The denominator is zero at x=4 and x=2. But x-4 cancels, so:

• x=2 is a vertical asymptote
• x=4 is not a vertical asymptote (it becomes a hole)

Exam Focus

• Typical question patterns:
  • “Find the vertical asymptote(s) and the domain of the rational function.”
  • “Given a graph, write a rational function with a specified vertical asymptote and intercepts.”
  • “Describe the behavior as x approaches a vertical asymptote from the left and right.”
• Common mistakes:
  • Treating every zero of the denominator as a vertical asymptote even when it cancels (that point is a hole).
  • Forgetting to state domain restrictions (points where the original denominator is zero are always excluded).
  • Assuming both sides of a vertical asymptote go to the same infinity without checking signs.

End Behavior and Slant Asymptotes of Rational Functions

End behavior: what happens “far away”

End behavior describes what the function does as x becomes very large positive or very large negative. For rational functions, end behavior is strongly controlled by the highest-degree terms of the numerator and denominator.

If

f(x) = \frac{P(x)}{Q(x)}

and P(x) and Q(x) are polynomials, then for large |x| the leading terms dominate. This is why degree comparisons are so powerful.

Horizontal and slant asymptotes (the “long-run” guide)

An asymptote is a line the graph approaches. For end behavior, you mainly care about:

• Horizontal asymptote: a constant line y=c that the graph approaches as x goes to positive or negative infinity.
• Slant (oblique) asymptote: a line y=mx+b (with m \neq 0) that the graph approaches as x goes to positive or negative infinity.

These matter because they help you sketch and interpret graphs quickly: you can see what the function “settles into” at extreme inputs.

Degree rules for end behavior (core AP Precalculus skill)

Let n be the degree of P(x) and d be the degree of Q(x).

In AP Precalculus, the most common non-horizontal case you’ll see is the slant asymptote when the numerator’s degree is exactly one more than the denominator’s.

How to find a slant asymptote (polynomial long division)

If \deg(P)=\deg(Q)+1, divide P(x) by Q(x):

\frac{P(x)}{Q(x)} = S(x) + \frac{R(x)}{Q(x)}

• S(x) is the quotient (a linear function), and it is the slant asymptote.
• R(x) is the remainder, with degree less than the degree of Q(x).
• As x goes to infinity, \frac{R(x)}{Q(x)} goes to 0, so the graph approaches y=S(x).

Example 1: Horizontal asymptote by degree comparison

Consider

f(x) = \frac{3x^2 - 5}{2x^2 + x + 1}

Both numerator and denominator have degree 2. The leading coefficients are 3 and 2, so the horizontal asymptote is:

y = \frac{3}{2}

This tells you that far left and far right, the function values get close to 1.5.

Example 2: Slant asymptote by long division

Consider

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

The numerator degree is one more than the denominator degree, so expect a slant asymptote. Divide:

1) Divide leading terms:

\frac{x^2}{x} = x

2) Multiply back and subtract:

(x^2 + 3x + 1) - (x^2 - 2x) = 5x + 1

3) Divide again:

\frac{5x}{x} = 5

4) Multiply back and subtract:

(5x + 1) - (5x - 10) = 11

So

\frac{x^2 + 3x + 1}{x - 2} = x + 5 + \frac{11}{x - 2}

As x becomes very large in magnitude, \frac{11}{x-2} approaches 0, so the slant asymptote is:

y = x + 5

A key interpretation: the graph of g(x) looks more and more like the line y=x+5 far away, except it still has a vertical asymptote at x=2.

Example 3: Using end behavior to compare left and right tails

Consider

h(x) = \frac{-2x^3 + 1}{x^3 + 4x}

Degrees match (both 3), so the horizontal asymptote is the ratio of leading coefficients:

y = \frac{-2}{1} = -2

Even without a full graph, you can say both tails approach y=-2 as x goes to positive infinity and negative infinity.

Exam Focus

• Typical question patterns:
  • “Determine the horizontal or slant asymptote using degrees or long division.”
  • “Given a rational function, describe end behavior as x approaches infinity and negative infinity.”
  • “Match a rational function to a graph using asymptotes and intercepts.”
• Common mistakes:
  • Using the ratio of leading coefficients when degrees are not equal.
  • Forgetting that slant asymptotes come from division (students sometimes try to guess the line).
  • Assuming a horizontal asymptote means the graph never crosses it (it can cross a horizontal asymptote).

Holes in Rational Functions

What a hole is (removable discontinuity)

A hole is a point where the function is undefined, but the graph does not “blow up” to infinity. Instead, it’s like a single missing point in an otherwise smooth curve. In calculus language, it’s a removable discontinuity.

Holes happen when a factor in the denominator also appears in the numerator and cancels. The original expression is undefined where that factor equals zero, but after cancellation the simplified formula gives the behavior everywhere else.

Why this matters: holes are easy to miss on a graphing window, but they are testable because they connect algebra (factor cancellation), domain, and graphical discontinuities.

How to identify a hole

Given

f(x) = \frac{P(x)}{Q(x)}

1) Factor P(x) and Q(x).
2) Cancel common factors.
3) Any cancelled factor of the form x-a indicates a hole at x=a.
4) To find the hole’s y-value, plug x=a into the simplified function (after cancellation). The result is the height of the open circle.

Important subtlety: even though the simplified expression produces a number at x=a, the original function is still undefined there. So you report a hole, not a filled-in point.

Example 1: Find the hole’s location

Consider

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

1) Factor the numerator:

x^2 - 9 = (x-3)(x+3)

So

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

2) Cancel x-3 (but keep in mind x \neq 3 in the original function):

f(x) = x+3

3) The cancelled factor is x-3, so there is a hole at x=3.

4) Find the corresponding y-value using the simplified function:

y = 3 + 3 = 6

So the graph is the line y=x+3 with an open circle at the point \left(3,6\right).

How holes differ from vertical asymptotes

Both holes and vertical asymptotes come from places where the denominator is zero. The difference is what the numerator does there:

• Hole: a factor cancels, so the function approaches a finite value.
• Vertical asymptote: the factor does not cancel, so the function grows without bound.

A useful mental picture: a cancelled factor “softens” the discontinuity from an infinite blow-up to a missing point.

Example 2: A function with both a hole and a vertical asymptote

Consider

g(x) = \frac{(x-1)(x+2)}{(x-1)(x-3)}

• x-1 cancels, so there is a hole at x=1.
• x-3 remains in the denominator, so there is a vertical asymptote at x=3.

Simplify:

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

Hole’s y-value comes from the simplified function at x=1:

g(1) = \frac{1+2}{1-3} = \frac{3}{-2} = -\frac{3}{2}

So there is an open circle at \left(1,-\frac{3}{2}\right) and a vertical asymptote at x=3.

Common misconception: “If it cancels, it’s gone completely”

Algebraically, the factor cancels in the simplified formula, but the original rational function still has the domain restriction. That’s why holes are still real features of the original function’s graph.

Exam Focus

• Typical question patterns:
  • “Determine whether a discontinuity at x=a is a hole or a vertical asymptote.”
  • “Find the coordinates of the hole for a simplified rational function.”
  • “Given a graph with an open circle, write a rational function that could produce it.”
• Common mistakes:
  • Cancelling factors incorrectly across addition (you can only cancel common factors, not common terms).
  • Finding the hole’s x-value but forgetting to compute the corresponding y-value.
  • Treating the simplified function’s value at x=a as included in the original graph (it must be an open circle).

Solving Rational Equations and Inequalities

Why rational solving is different

Rational equations and inequalities involve fractions with variables in denominators. The main complication is that some values make denominators zero, so you must track excluded values. Also, when you multiply both sides by an expression involving a variable, you might create extraneous solutions—answers that satisfy the multiplied equation but not the original.

A reliable process keeps you safe: identify restrictions, clear denominators carefully, solve, then check.

Solving rational equations

A rational equation is an equation that contains a rational expression, such as:

\frac{x+1}{x-2} = 3

Step-by-step method

1) State domain restrictions: find values that make any denominator zero and exclude them.
2) Multiply both sides by the LCD (least common denominator) to clear fractions.
3) Solve the resulting equation (often linear or quadratic).
4) Check solutions in the original equation (or at least verify they don’t violate restrictions and they make the original true).

Example 1: A one-fraction equation

Solve:

\frac{x+1}{x-2} = 3

1) Restriction:

x-2 \neq 0
x \neq 2

2) Multiply both sides by x-2:

x+1 = 3(x-2)

3) Solve:

x+1 = 3x - 6
7 = 2x
x = \frac{7}{2}

4) Check restriction: \frac{7}{2} \neq 2, so it’s allowed. Substitution confirms it works.

So the solution is:

x = \frac{7}{2}

Example 2: An equation that can produce extraneous solutions

Solve:

\frac{2}{x-1} + \frac{3}{x+1} = 1

1) Restrictions:

x \neq 1
x \neq -1

2) LCD is (x-1)(x+1). Multiply both sides:

2(x+1) + 3(x-1) = (x-1)(x+1)

3) Expand:

2x+2 + 3x-3 = x^2 - 1

Combine like terms:

5x - 1 = x^2 - 1

Bring all terms to one side:

0 = x^2 - 5x

Factor:

0 = x(x-5)

So:

x=0
x=5

4) Check restrictions: both 0 and 5 are allowed (neither is 1 nor -1). Substituting back verifies both satisfy the original.

So the solutions are:

x=0
x=5

Solving rational inequalities

A rational inequality looks like:

\frac{P(x)}Q(x) > 0

The key idea: a rational expression changes sign at

• zeros of the numerator (where the expression equals zero), and
• zeros of the denominator (where the expression is undefined).

So you solve rational inequalities by sign analysis across intervals.

Step-by-step method (sign chart / test intervals)

1) Rewrite the inequality so one side is 0.
2) Factor numerator and denominator.
3) Find critical numbers:

• numerator zeros (potentially included if the inequality allows equality)
• denominator zeros (never included)
  4) Put critical numbers on a number line to form intervals.
  5) Determine the sign of the expression on each interval (by test points or factor sign reasoning).
  6) Choose intervals that satisfy the inequality, respecting strict vs non-strict symbols.

Example 3: A rational inequality

Solve:

\frac{x-2}{x+1} \le 0

1) It is already compared to 0.

2) Critical numbers:

• Numerator zero at x=2 (this makes the expression 0, so it can be included because \le allows equality)
• Denominator zero at x=-1 (this is undefined, so it must be excluded)

3) Make intervals: (-\infty,-1), (-1,2), (2,\infty)

4) Test a point in each interval:

• For (-\infty,-1) pick x=-2:

\frac{-2-2}{-2+1} = \frac{-4}{-1} > 0

• For (-1,2) pick x=0:

\frac{0-2}{0+1} = -2 < 0

• For (2,\infty) pick x=3:

\frac{3-2}{3+1} = \frac{1}{4} > 0

5) We want \le 0, so we take the interval where it’s negative and include where it equals zero:

• Negative on (-1,2)
• Equals zero at x=2
• Exclude x=-1

Solution:

(-1,2]

Inequalities and holes: a subtle point

If a factor cancels, the simplified expression might suggest the function is defined at that point, but the original expression is not. In inequalities, that cancelled point is still excluded from the solution set because it’s not in the domain.

For example, consider

\frac{(x-1)(x+2)}{(x-1)(x-3)} > 0

Even though it simplifies to \frac{x+2}{x-3}, the original expression is still undefined at x=1 (a hole) and x=3 (vertical asymptote). Both points must be excluded when writing the final solution.

A good habit: write restrictions from the original denominator before canceling anything.

Exam Focus

• Typical question patterns:
  • “Solve a rational equation and state any extraneous solutions that must be rejected.”
  • “Solve a rational inequality and express the solution in interval notation.”
  • “Identify excluded values and explain how they affect the solution set.”
• Common mistakes:
  • Multiplying by an expression containing x and forgetting that it could be zero (leading to lost restrictions or extraneous solutions).
  • In inequalities, including denominator zeros as solutions (they must always be excluded).
  • Cancelling factors and then forgetting to exclude the hole’s x-value in the final answer.