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4.3 — Rational Thinking
Human review completed .
Use polynomial degrees and leading coefficients to predict rational-function asymptotes, and calculate intercepts from the rule.
1. In class: A Solidify Understanding Task
A ratio of polynomials
A rational function can be written \(f(x)=\frac{P(x)}{Q(x)}\), where P and Q are polynomials and Q is not the zero polynomial. Real inputs that make Q(x)=0 are excluded. The word “rational” describes the rule as a ratio of polynomials; it does not require every output to be a rational number.
The degree of a nonzero polynomial is its largest exponent after like terms have been combined. A nonzero constant such as 1 has degree 0. In a product, add factor degrees and multiply leading coefficients: \((2x+5)(2x-5)=4x^2-25\) has degree 2 and leading coefficient 4.
Review 3.9: polynomial end behavior (draft pending human review) for why leading terms control large-input behavior.
Why the degrees predict the horizontal asymptote
As \(x\) becomes very large in either direction, the highest-power terms—the leading terms—dominate the numerator and denominator. The lower-power terms have less and less effect on the ratio, so we can use the leading terms to see what value the function approaches.
For example, in \(\frac{2x^2+3}{4x^2-25}\), the constants 3 and −25 become less significant compared with the \(x^2\) terms. The fraction gets closer to \(\frac{2x^2}{4x^2}=\frac12\), so the horizontal asymptote is \(y=\frac12\). We are describing what the fraction approaches, not saying the lower-power terms actually disappear.
If the numerator’s degree is smaller, the denominator eventually grows faster in magnitude, so \(f(x)\to0\) at both ends. The horizontal asymptote is \(y=0\).
If the degrees are equal, the same power of x appears in both leading terms and cancels in their ratio. The horizontal asymptote is the ratio of leading coefficients:
Another way to see why this works
Multiply the numerator and denominator by \(1/x^a\), where \(a\) is the degree of the denominator. This is the same as dividing every term in both by \(x^a\), and it keeps the fraction’s value the same for \(x\ne0\).
The key idea: As \(x\) gets very large in either direction, \(1/x\), \(1/x^2\), and every \(1/x^n\) with a positive whole-number exponent get closer and closer to 0. For example, when \(x=10,100,1000\), the values of \(1/x\) are \(0.1,0.01,0.001\). Multiplying these small fractions by a fixed number still gives a quantity approaching 0.
Main 2 example: \(\frac{x}{(x+2)(x-3)}=\frac{x}{x^2-x-6}\). The denominator has degree 2, so multiply the numerator and denominator by \(1/x^2\):
Here \(1/x\) and \(6/x^2\) approach 0. The numerator therefore approaches 0 and the denominator approaches 1, giving the horizontal asymptote \(y=0\).
Main 6 example: multiply the numerator and denominator by \(1/x^2\) again:
Here \(3/x^2\) and \(25/x^2\) approach 0. The numerator approaches 2 and the denominator approaches 4, so the fraction approaches \(2/4=1/2\). This makes the leading-coefficient rule visible in the algebra.
The horizontal asymptote is determined by end behavior. It can be crossed at finite inputs, as Main 2 does at (0,0).
Find vertical asymptotes and intercepts
- Domain: exclude real zeros of the original denominator.
- Vertical asymptotes in these examples: set each denominator factor equal to zero. The numerator is nonzero at those inputs, so outputs grow without bound near them. An excluded input alone does not always establish a vertical asymptote; if both numerator and denominator are zero, examine the expression further before applying this shortcut.
- x-intercepts: solve P(x)=0 and keep only inputs in the domain. Write each as (x,0).
- y-intercept: substitute x=0, if allowed, and write (0,f(0)).
Original problem: For \(y=\frac{3x-1}{x+4}\), find degrees, asymptotes and intercepts, then use technology to graph it.
Worked solution: both polynomials have degree 1, so the horizontal asymptote is y=3. The denominator is zero at −4, where the numerator equals −13; the vertical asymptote is x=−4. Set the numerator to zero: \(3x-1=0\Rightarrow x=\frac13\). Substitute zero for x to get \(-\frac14\).
Answers: degrees (1,1); asymptotes x=−4, y=3; x-intercept \((\frac13,0)\), y-intercept \((0,-\frac14)\). Check the graph against these algebraic features.
Main 1–6 — Complete checking results
Degree pairs below mean (numerator degree, denominator degree). Separate all branches at the vertical asymptotes when checking your technology graphs.
| Main | Rule | Degrees | Horizontal | Vertical | Intercepts: x; y |
|---|---|---|---|---|---|
| 1 | \(\frac{x+3}{x-2}\) | (1,1) | \(y=1\) | \(x=2\) | \((-3,0);\ (0,-\frac32)\) |
| 2 | \(\frac{x}{(x+2)(x-3)}\) | (1,2) | \(y=0\) | \(x=-2,3\) | \((0,0)\) is both intercepts |
| 3 | \(\frac1{(x+4)^2}\) | (0,2) | \(y=0\) | \(x=-4\) | No x-intercept; \((0,\frac1{16})\) |
| 4 | \(\frac{3x-1}{x+4}\) | (1,1) | \(y=3\) | \(x=-4\) | \((\frac13,0);\ (0,-\frac14)\) |
| 5 | \(\frac{x^2-3x}{(x-2)(x+1)(x+3)}\) | (2,3) | \(y=0\) | \(x=2,-1,-3\) | \((0,0),(3,0)\); y-intercept (0,0) |
| 6 | \(\frac{2x^2+3}{(2x+5)(2x-5)}\) | (2,2) | \(y=\frac12\) | \(x=\pm\frac52\) | No real x-intercept; \((0,-\frac3{25})\) |
Main 2 graph: the graph crosses its horizontal asymptote at (0,0). A horizontal asymptote is not a barrier.
Plots not shown — Main 1, 3, 5, 6: use your graphing tool as requested. Along with the features above, check these points:
| Main | Sample points |
|---|---|
| 1 | \((1,-4),(3,6),(7,2)\) |
| 3 | \((-5,1),(-3,1)\) and \((-2,\frac14)\) |
| 5 | \((-2,\frac52),(1,\frac14),(4,\frac2{35})\) |
| 6 | \((-2,-\frac{11}9),(2,-\frac{11}9),(3,\frac{21}{11})\) |
Main 9 — How to find the asymptotes and intercepts
The organizer on printed p. 15 asks for general methods, using the patterns from Problems 1–8. It does not give one function with one set of asymptotes. An asymptote is a line, so write its equation: \(x=\text{a number}\) for a vertical line, or \(y=\text{a number}\) for a horizontal line.
| Degree relationship | How to find vertical asymptotes | How to find the horizontal asymptote | How to find intercepts |
|---|---|---|---|
| Numerator degree is less than denominator degree | Factor and cancel common factors first, keeping the original excluded inputs. Set the remaining denominator equal to 0 and solve. Each real solution \(a\) gives a vertical asymptote \(x=a\). A canceled factor may give a hole instead. | The horizontal asymptote is \(y=0\). | For x-intercepts, solve numerator = 0 and keep only inputs in the original domain; write each as \((x,0)\). For the y-intercept, substitute \(x=0\), if allowed, and write \((0,f(0))\). |
| Numerator and denominator have equal degrees | Use the same factoring and denominator-solving steps above. | Divide the numerator’s leading coefficient by the denominator’s leading coefficient. The horizontal asymptote is \(y=\frac{\text{numerator’s leading coefficient}}{\text{denominator’s leading coefficient}}\). | Use the same intercept steps above. |
For example: For \(y=\frac{2x+1}{x-2}\), nothing cancels. Solving \(x-2=0\) gives the vertical asymptote \(x=2\). The degrees are equal, so the horizontal asymptote is \(y=\frac21=2\). These are two different lines, even though both equations contain the number 2.
2. Ready — Add and multiply numerical fractions
For addition, use a common denominator so the parts have the same size, then add numerators. For multiplication, multiply the numerators and multiply the denominators; common factors can be divided out before or after. Keep fractions in lowest terms. Try the exercises before checking the steps.
Check your work — Ready 1–9
| Problem | Checked calculation |
|---|---|
| 1 | \(\frac5{17}+\frac8{17}=\frac{13}{17}\) |
| 2 | \(\frac3{10}+\frac{16}{25}=\frac{15}{50}+\frac{32}{50}=\frac{47}{50}\) |
| 3 | \(\frac45+\frac7{11}=\frac{44}{55}+\frac{35}{55}=\frac{79}{55}\) |
| 4 | \(\frac23\cdot\frac45=\frac8{15}\) |
| 5 | \(\frac23\cdot\frac9{16}=\frac{18}{48}=\frac38\) |
| 6 | \(\frac{10}{33}\cdot\frac{11}{15}=\frac{110}{495}=\frac29\) |
7a: add the numerators and keep the common denominator; simplify if possible.
7b: find a common denominator, rewrite both fractions equivalently by multiplying top and bottom by the same factor, then add as in 7a.
8: multiply numerator by numerator and denominator by denominator; divide out common nonzero factors to simplify.
9: both orders give the same result. Simplifying first often keeps numbers smaller and reduces arithmetic. For 6, \(\frac{10\cdot11}{33\cdot15}\) becomes \(\frac{2\cdot1}{3\cdot3}=\frac29\), dividing out factors 5 and 11.
3. Set — Find rational graph features
Use the degree rules and intercept procedure, then mark the requested features on the supplied graphs. The task asks you to add asymptotes and intercepts to existing curves.
Check your work — Set 10–13
| Problem | Rule | Degrees | Horizontal asymptote | Vertical asymptotes | y-intercept | x-intercepts |
|---|---|---|---|---|---|---|
| 10 | \(\frac{x^2}{(x+6)(x-4)}\) | (2,2) | \(y=1\) | \(x=-6,4\) | \((0,0)\) | \((0,0)\) |
| 11 | \(\frac{x-6}{x+3}\) | (1,1) | \(y=1\) | \(x=-3\) | \((0,-2)\) | \((6,0)\) |
| 12 | \(\frac{10x}{(x+3)^2}\) | (1,2) | \(y=0\) | \(x=-3\) | \((0,0)\) | \((0,0)\) |
| 13 | \(\frac{x+1}{(x+2)(x-5)}\) | (1,2) | \(y=0\) | \(x=-2,5\) | \((0,-\frac1{10})\) | \((-1,0)\) |
Annotated plots not shown: on each packet graph, draw the stated asymptote lines dashed and mark each intercept at its coordinate. In 10, the double numerator zero makes the middle branch touch the x-axis at the origin; it is still an x-intercept.
4. Go — Simplify numerical fractions and explain why
Dividing numerator and denominator by the same nonzero factor leaves the value unchanged: \(\frac{ca}{cb}=\frac ab\). Find the greatest common factor and divide both parts by it. Keep improper fractions as fractions. The packet skips number 16; no exercise is missing from this key.
Check your work — Go 14–15, 17–20
| Problem | Lowest form | Explanation |
|---|---|---|
| 14 | \(\frac45\) | Divide 12 and 15 by 3. |
| 15 | \(\frac{26}{11}\) | 26 and 11 have no common integer factor greater than 1, so no reduction. |
| 17 | 3 | Divide 51 and 17 by 17; \(\frac31=3\). |
| 18 | \(\frac6{13}\) | 6 and 13 have no common integer factor greater than 1. |
| 19 | \(\frac{38}9\) | Divide 114 and 27 by 3. The new numerator and denominator are relatively prime. |
| 20 | −1 | Divide −14,529 and 14,529 by 14,529; the quotient is negative. |
Additional learning resources
The Khan Academy videos focus on individual skills from this lesson. Both Professor Leonard videos are very comprehensive and good choices if you want a full explanation. They also include topics beyond this lesson, as noted below.
| Resource | What it helps with and scope |
|---|---|
| Khan Academy — Graphing rational functions according to asymptotes | Close match: identifies horizontal and vertical asymptotes and distinguishes a hole from a vertical asymptote. The cancellation example extends the noncanceling examples in this lesson. |
| Khan Academy — Graphs of rational functions: y-intercept | Close match: uses the value at x=0 to identify which graph can represent a rational function. Supports the organizer’s y-intercept method. |
| Khan Academy — Graphs of rational functions: horizontal asymptote | Close match: uses the horizontal asymptote to select a possible graph, connecting the leading coefficients to end behavior. |
| Khan Academy — Analyzing horizontal asymptotes of rational functions | Explains horizontal asymptotes using leading terms and end behavior. Relevant to the degree comparisons in this lesson; the broader discussion can also serve as an extension. |
| Professor Leonard — Finding Vertical Asymptotes of Rational Functions | A very comprehensive video, recommended for a full explanation of vertical asymptotes, holes, and how factor multiplicity affects the graph. Worked examples start at 14:30. Broader than this lesson: the multiplicity discussion goes beyond the organizer’s basic locating rules. |
| Professor Leonard — Finding Asymptotes and Holes of Rational Functions | A very comprehensive video, recommended for a full explanation of finding asymptotes and holes before graphing rational functions. Broader than this lesson: covers more than the two degree cases emphasized in 4.3. Optional viewing for students who want the complete explanation. |
Teacher/parent review of the source packet
No substantial mathematical error was found in the numbered student exercises. Go, p. 20, skips number 16; the printed numbering is retained.
Teacher notes and Main 9 organizer: denominator zeros identify vertical asymptotes for these examples because the numerator is nonzero there. Avoid teaching “every excluded input is a vertical asymptote” as an unconditional rule. Horizontal asymptotes describe end behavior and can be crossed; Main 2 supplies a concrete example.
Coverage
| Packet component | Support in this lesson |
|---|---|
| Main introduction / 1–6 · pp. 15–17 | Rational definition, degree/leading coefficients, both horizontal cases with reasoning, complete feature results. Graphs 2/4 shown; 1/3/5/6 plots not shown, sample points supplied. |
| Main 7–9 · p. 17 | Two labeled sample functions, features/points and completed organizer with reasoning; own-function plots not shown. |
| Ready 1–9 · p. 18 | All six arithmetic calculations and explanations 7a/7b/8/9. |
| Set 10–13 · p. 19 | All degree pairs, asymptotes and intercept coordinates; annotations not shown. |
| Go 14–15, 17–20 · p. 20 | All lowest forms and explanations; no printed 16. |
Data and brief prompts adapted from Mathematics Vision Project, Secondary Mathematics III, Module 4 student packet, Section 4.3, printed pp. 15–20, under CC BY 4.0. Explanations, solutions, and diagrams were prepared for this supplement.
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