Rational Functions: One Six-Step Routine That Covers AP Precalculus 1.7 to 1.10

There is a particular kind of student who understands rational functions perfectly while the teacher is talking, and then stares at a blank page that night. Nothing is missing conceptually. What's missing is a first move — a reliable thing to write down before understanding has to show up.

That's worth naming, because it changes what practice should look like. If the problem were comprehension, the fix would be another explanation. It isn't. The fix is a routine short enough to recall under pressure.

Four Sections Are Really One Procedure

Here's the good news about AP Precalculus sections 1.7 through 1.10. They look like four separate topics — end behavior, zeros, vertical asymptotes, holes — and they're taught across four lessons, usually a week apart. But they are one procedure with four different questions asked of it. Learn the procedure once and all four sections collapse into something you can do half-asleep.

Every rational function question is some subset of these six moves, in this order:

1. Set the original denominator equal to zero → domain restrictions
2. Factor the top and the bottom completely
3. Cancel → gone from the bottom means hole; still on the bottom means vertical asymptote
4. The hole's y-value comes from the simplified function
5. Compare degrees → horizontal or slant asymptote
6. Check signs just left and right of each vertical asymptote

Notice that steps 1 through 4 are sections 1.9 and 1.10, step 5 is 1.7, and the zeros that fall out along the way are 1.8. One list, four sections. If factoring is the part that slows you down, the algebra formula reference has the difference-of-squares and cubic patterns that cover most of what shows up here.

1. Step One Is the One People Skip

Set the original denominator equal to zero before anything cancels.

This is the single highest-value habit in the whole unit, and it's the one students drop first, because it feels like it isn't doing anything yet.

Take f(x) = (x² − x − 6)/(x² − 9). The bottom is zero at x = 3 and x = −3. Write those down now. If you factor and cancel first, the (x − 3) disappears from your paper — and x = 3 quietly stops being part of your answer, even though the function is still undefined there.

Those two values are also the only places anything interesting can happen. Every hole and every vertical asymptote this function has is somewhere on that list. Step 1 isn't preliminary work; it's the entire list of candidates.

2. Hole or Asymptote? Count the Powers

Now factor and cancel:

f(x) = (x − 3)(x + 2)        (x + 2)
       ───────────────  =    ───────  ,  x ≠ 3
       (x − 3)(x + 3)        (x + 3)

The rule is short enough to say out loud:

A factor that leaves the bottom gives a hole. A factor still on the bottom gives a vertical asymptote.

So x = 3 is a hole and x = −3 is a vertical asymptote. Same function, both features. That answers the question students ask constantly — can a function have both? Yes, routinely. What it cannot have is both at the same x-value. A given restriction is one or the other, and the deciding question is always whether any power of that factor survives downstairs.

Three versions worth memorizing as a set:

That third row catches almost everyone. A squared factor on top is not a warning sign — it's the opposite. Direction matters more than repetition.

3. The Hole's Height

A hole is a point, so it needs two coordinates, and the y-value comes from the simplified function:

y = (3 + 2)/(3 + 3) = 5/6   →   hole at (3, 5/6)

Substituting into the original gives 0/0, which isn't an answer — it's the reason there's a hole in the first place. The simplified function agrees with f everywhere except that one point, so it tells you the height the graph is reaching for and never attains.

This is the formal idea behind the term removable discontinuity: redefine the function at that one point and the break disappears. Nothing about the curve elsewhere changes. Compare that with a vertical asymptote, which is non-removable — no single redefinition can patch it, because the function isn't approaching a finite height there at all.

4. End Behavior: Section 1.7 in One Idea

For large |x|, only the leading terms matter:

f(x) ≈ (leading term of top) / (leading term of bottom)

Every rule about horizontal asymptotes is a consequence of that single line — which is why it's worth understanding rather than memorizing four separate cases.

(x + 1)/(x² − 9)    → y = 0
(2x² + 1)/(x² − 4)  → y = 2
(x² + 2)/(x − 1)    → y = x + 1     since it equals x + 1 + 3/(x − 1)
(x⁴)/(x² − 1)       → none; behaves like y = x²

Use the degrees of the original function. Cancelling strips the same factors from top and bottom, so the comparison never changes.

Two things students get backwards here. First, when the degrees match, use the leading coefficients — not the constants. Second: a graph may cross its horizontal asymptote. A horizontal asymptote describes the far left and far right only; in the middle the curve can do as it pleases. Vertical asymptotes are the untouchable ones.

If a question asks whether a graph crosses its HA, don't guess — set them equal. For (x² − 1)/(x² + 1) = 1 you get −1 = 1, which is impossible, so that one never crosses. For a function with HA y = 0 and an x-intercept at x = −3, it obviously does.

The AP exam also wants this stated in limit notation, so get comfortable writing it: lim x→∞ f(x) = 1 for a horizontal asymptote, and lim x→−3⁺ f(x) = −∞ for the behavior just to the right of a vertical one. If the limit notation itself is new, the calculus formula reference is a compact place to see how the one-sided arrows are written.

5. Zeros and Multiplicity: Section 1.8

Zeros come from the numerator — but from the simplified numerator.

• Odd multiplicity → the graph crosses the x-axis
• Even multiplicity → it touches and turns around

And here's the trap that hides inside 1.8 and 1.10 together: a hole that lands at height zero is not an x-intercept. Consider

k(x) = (x − 8)(x + 2)²
       ─────────────────
       x(x − 1)(x + 2)

The (x + 2) cancels once and leaves a copy on top, so x = −2 is a hole — and its y-value works out to 0. The simplified numerator is zero there, so it looks exactly like an x-intercept. It isn't. The point is missing. The only x-intercept is (8, 0).

Worked end to end, that function gives:

Domain: x ≠ −2, 0, 1
Hole: (−2, 0)
VAs: x = 0  (left → −∞, right → +∞)
     x = 1  (left → +∞, right → −∞)
HA: y = 1          degrees 3 and 3, leading coefficients 1 and 1
Intercepts: (8, 0) only — no y-intercept, since x = 0 is an asymptote

For the direction on each side of a vertical asymptote, test a value a tenth of a unit out and keep only the sign. You never need the value — and if you'd rather not do six sign checks by hand, the scientific calculator will burn through them in a few seconds. Shortcut worth knowing: an odd surviving power sends the graph opposite ways on the two sides; an even one sends it the same way on both.

6. Two Real Curves That Make the Rules Concrete

The rules feel arbitrary until you meet named curves that obey them.

The Witch of Agnesi is y = 8a³/(x² + 4a²) — described by Maria Gaetana Agnesi in her 1748 calculus textbook Instituzioni Analitiche, and one of the first curves a beginner meets that is rational all the way down. Run the routine on it. The denominator x² + 4a² is a sum of squares, so it is never zero for any real x: no domain restrictions, no holes, no vertical asymptotes at all. Degree 0 on top against degree 2 on the bottom gives a horizontal asymptote at y = 0, which is exactly the bell shape flattening out on both sides. One function, four sections answered, and three of the answers are "none."

Now contrast the Cissoid of Diocles, studied around 180 BC as an attempt to double the cube. It creeps toward the line x = 2a forever without reaching it — a vertical asymptote in the wild, more than two millennia before anyone wrote it as a limit. Or the Folium of Descartes, whose parametric form 3at/(1 + t³) has a denominator that vanishes at t = −1: step 1 of the routine, applied to a curve Descartes posed as a challenge to Fermat in 1638.

The Traps, Collected

• Cancelling before writing the domain — the hole's x stays excluded forever
• Forgetting the GCF: x³ − 4x = x(x − 2)(x + 2). Miss the x and you lose the asymptote at 0. If pulling out common factors is shaky, the GCF calculator is good for checking your instinct on the numeric part
• Plugging a hole's x into the original function, getting 0/0, and stopping
• Calling a hole at height 0 an x-intercept
• Assuming any cancellation kills the asymptote — check what power is left
• Claiming a graph can't cross a horizontal asymptote
• Keeping the remainder in a slant asymptote answer
• Trying to factor a sum of squares — x² + 1, x² + 4 and 4x² + 25 are all prime over the reals, which is precisely why the Witch of Agnesi has no asymptotes to find

And one that costs points even when the math is right: answer in the correct form. Holes are points — (3, 5/6). Asymptotes are lines — x = −3, y = 1. A bare number is not an asymptote.

What to Copy Onto Scratch Paper

1. bottom = 0  (BEFORE cancelling)   → domain
2. factor both (GCF first!)
3. cancel → gone from bottom = hole | left on bottom = VA
4. hole y from simplified            → point
5. degrees: <  → y = 0
            =  → ratio of leading coefficients
            +1 → divide, keep the quotient
            +2 → neither
6. signs left/right of each VA

Six lines. If a student can reproduce those from memory, 1.7 through 1.10 are done — and the same six lines handle the rational-function questions that show up on the SAT and PSAT, where the wording changes but the procedure doesn't.

Where to Go Next

Official material. The AP Precalculus Course and Exam Description is the authoritative source for what each section is responsible for — it's where the 1.7/1.8/1.9/1.10 split comes from, and the Unit 1 learning objectives are worth skimming directly. For what the questions actually look like, College Board posts past free-response questions with scoring guidelines. Reading a scoring guideline is the fastest way to understand why answer format matters so much.

Lessons by section. The Algebros' free Flipped Math course maps one-to-one onto this post: 1.7 on end behavior, 1.8 on zeros, 1.9 on vertical asymptotes and 1.10 on holes — each with a video, a packet and worked solutions.

Extra reps. Khan Academy's rational functions unit has targeted exercise sets; end behavior and points of discontinuity are the two that hit this material hardest.

See it. Type a function into a grapher after you've worked it by hand — never before. Predict the hole and the asymptotes on paper, then look. Being wrong and seeing it is worth more than three correct problems.

Try Them Yourself

• Algebra Formulas — the factoring patterns that make step 2 fast
• Calculus Formulas — limit notation, written the way the exam wants it
• Analytic Geometry Formulas — asymptotes are lines, and this is where lines live
• GCF Calculator — check the common factor before you factor the rest
• Scientific Calculator — run the sign checks either side of each vertical asymptote
• Witch of Agnesi — a rational function with no holes and no vertical asymptotes, and the reason why
• Cissoid of Diocles — a vertical asymptote drawn by the Greeks around 180 BC
• Folium of Descartes — a denominator that vanishes at a single value of the parameter

The One Line to Remember

If a student takes exactly one thing from all four sections, make it this:

Set the bottom to zero first. Then ask, for each restriction separately: after cancelling, is that factor still down there?

Everything else in 1.7 through 1.10 is bookkeeping around that question.