The three characteristics of a rational graph — x-intercepts from the numerator's zeros, a vertical asymptote at each zero of the denominator, and a horizontal asymptote decided by comparing degrees — worked through all three degree cases, and an optimization problem solved with a rational model.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 8 — Rational Functions
Graph General Rational Functions
Objectives
Five outcomes. Three questions to ask of every rational function, and one comparison that settles the hardest of them.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.3 Graph General Rational Functions §8.3, pp. 565-569 — the lesson these objectives are drawn from
Warm-up
Lesson 8.2 handled a linear over a linear, where the horizontal asymptote was the ratio of leading coefficients.
Discussion prompt
For y equals 6 over x squared plus 1, try x equal to 10, then 100. What is happening, and does the old rule about leading coefficients still apply?
Hint: Compare the sizes of the numerator and the denominator.
Answer:
\[ x = 10: \; y = \tfrac{6}{101} \approx 0.059; \qquad x = 100: \; y = \tfrac{6}{10001} \approx 0.0006 \]
The outputs shrink toward zero, not toward some ratio of coefficients. The denominator grows quadratically while the numerator stays fixed, so it wins outright.
The rule that decides the horizontal asymptote is not one formula but a comparison of degrees, and Lesson 8.2's rule was only the case where the two happened to be equal.
Concept
The x-intercepts come from the numerator's zeros and the vertical asymptotes from the denominator's. The horizontal asymptote is settled by comparing the two degrees: lower means the horizontal axis, equal means the ratio of leading coefficients, higher means none at all.
end behavior — What a graph does as the inputs run far out in both directions. For a rational function with numerator degree m above denominator degree n, it matches the graph of the ratio of leading coefficients times x to the m minus n.
\[ m < n: \; y = 0; \quad m = n: \; y = \tfrac{a_m}{b_n}; \quad m > n: \; \text{none} \]
The three cases are not three rules to memorise but one idea: whichever polynomial grows faster far from the origin decides what the fraction does out there.
Figure (svg): The three characteristics of a rational graph, with the degree rule for the horizontal asymptote
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.3 Graph General Rational Functions §8.3, pp. 565-565
Section
Section 1
Concept
For a rational function with no common factors, the x-intercepts are the real zeros of the numerator, there is a vertical asymptote at each real zero of the denominator, and there is at most one horizontal asymptote, decided by the two degrees.
\[ f(x) = \frac{p(x)}{q(x)} \]
The requirement that the two polynomials share no common factor matters. A shared factor would cancel, and the cancelled zero leaves a hole rather than an asymptote.
Figure (svg): The three characteristics of a rational graph, with the degree rule for the horizontal asymptote
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.3 Graph General Rational Functions §8.3, pp. 565-565 — Graphs of Rational Functions
Picture it
The characteristics and the degree rule in one place.
Figure (svg): The three characteristics of a rational graph, with the degree rule for the horizontal asymptote
Only the third question involves both polynomials. The first two look at one of them alone, which is why they are the quickest to answer.
Worked example
Applying the key concept to Example 3's function.
\[ \text{For } y = \frac{x^2+3x-4}{x-2}, \text{ find the intercepts, vertical asymptotes and degrees.} \]
Factor the numerator
Why: X squared plus 3x minus 4 factors into two binomials.
\[ (x + 4) (x - 1) \]
Read the x-intercepts
Why: Set each factor to zero.
\[ x = -4\text{ and } x = 1 \]
Find the vertical asymptote
Why: Set the denominator to zero.
\[ x = 2 \]
Compare the degrees
Why: Two on top and 1 underneath.
\[ m > n \]
Figure (svg): The three characteristics of a rational graph, with the degree rule for the horizontal asymptote
\[ \text{intercepts } -4, 1; \; \text{asymptote } x = 2 \]
Verify: check that no factor cancels
Why: The denominator x minus 2 is not a factor of the numerator: substituting 2 into x squared plus 3x minus 4 gives 6, not 0. So there is a genuine vertical asymptote at 2 rather than a hole, and the key concept's no-common-factors condition is satisfied.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.3 Graph General Rational Functions §8.3, pp. 566-566
Matching
Each comes from one part of the fraction.
Match the pairs
Why: The last row is why the key concept opens with a condition rather than a formula. A common factor cancels, and what looked like an asymptote turns out to be a single missing point.
Worked example
Guided Practice 1 to 3.
\[ \text{For } \frac{4}{x^2+2}, \; \frac{3x^2}{x^2-1}, \; \frac{x^2-5}{x^2+1}, \text{ find the intercepts and vertical asymptotes.} \]
First: the numerator is a nonzero constant
Why: It has no zeros, and x squared plus 2 has no real zeros either.
Second: numerator zero at 0
Why: And x squared minus 1 is zero at 1 and negative 1.
\[ \text{intercept } 0;\text{ asymptotes at } 1\text{ and } -1 \]
Third: numerator zero where x squared is 5
Why: The square roots of 5, positive and negative.
\[ \text{intercepts about } 2.24\text{ and } -2.24 \]
Third: the denominator
Why: X squared plus 1 is never zero for real x.
Figure (svg): The solution to Worked example three more sets of characteristics shown as a ladder of expressions, one row per algebraic move
\[ \text{none}; \; 0, \; x = \pm 1; \; \pm\sqrt{5}, \text{ none} \]
Verify: check why two denominators have no real zeros
Why: Both x squared plus 2 and x squared plus 1 are sums of a square and a positive number, so both are at least 2 and at least 1 respectively — never zero. A denominator that is never zero means a domain of all real numbers and a graph with no break anywhere.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.3 Graph General Rational Functions §8.3, pp. 567-567
Trap
\[ y = \frac{x^2-4}{x-2} \]
Report a vertical asymptote at x equals 2
Why: The denominator vanishes there, so an asymptote is assumed.
\[ \text{asymptote } x = 2 \quad \text{(wrong)} \]
The numerator factors as x plus 2 times x minus 2, so the x minus 2 cancels. Near 2 the function behaves like x plus 2, which is perfectly tame.
\[ y = \frac{(x+2)(x-2)}{x-2} = x+2, \; x \neq 2 \]
Factor and cancel before applying the rules
Why: The key concept requires no common factors before any of its statements apply.
\[ \text{a hole at } (2, 4), \text{ not an asymptote} \]
The graph is the line y equals x plus 2 with a single point punched out. Factoring first is what distinguishes a hole from an asymptote.
Fill the middle
Example 3.
Fill in the blanks
x^2+3x-4 = (x+4)(x-1) \;\Longrightarrow\; \text___ -4 \text___ 1
Why: The factors 4 and negative 1 multiply to negative 4 and add to 3, so the numerator is x plus 4 times x minus 1. Factoring the numerator is how every x-intercept is found.
Sorting
A sum of a square and a positive number never vanishes.
Sort into buckets
Sort each denominator.
A denominator with no real zeros gives a domain of all real numbers and a graph in one unbroken piece — unusual for a rational function and worth noticing when it happens.
Prediction
Commit before reasoning.
Predict first
For y equals x squared minus 4, over x minus 2, what is at x equal to 2?
Correct: A hole, since the common factor cancels and leaves y = x + 2.
\[ \frac{(x+2)(x-2)}{x-2} = x+2 \text{ for } x \neq 2 \]
Why: Factoring gives x plus 2 times x minus 2, all over x minus 2, so the function equals x plus 2 everywhere except at x equal to 2, where it is undefined. The graph is a straight line with one point removed at the height 4. Near x equal to 2 the outputs approach 4 rather than running away, which is exactly what distinguishes a hole from an asymptote.
Section
Section 2
Concept
If the numerator's degree is less than the denominator's, the horizontal asymptote is the horizontal axis. Far from the origin the denominator grows faster, so the fraction shrinks toward zero.
\[ m < n \;\Longrightarrow\; y = 0 \]
Combined with a denominator that has no real zeros, this gives the tidiest of all rational graphs: a single unbroken hump with no break and no intercept.
Figure (svg): A rational graph whose numerator has lower degree than its denominator
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.3 Graph General Rational Functions §8.3, pp. 565-565 — Graph a rational function (m < n)
Picture it
Example 1: six over x squared plus 1.
Figure (svg): A rational graph whose numerator has lower degree than its denominator
The peak is at the origin's height 6, and the curve falls symmetrically toward the horizontal axis on both sides without ever reaching it.
Worked example
Example 1.
\[ \text{Graph } y = \frac{6}{x^2+1} \text{ and state the domain and range.} \]
Look for x-intercepts
Why: The numerator 6 is never zero.
Look for vertical asymptotes
Why: X squared plus 1 is at least 1, never zero.
Compare the degrees
Why: Zero on top, 2 underneath.
\[ y = 0 \]
Plot points and read the range
Why: The largest value is at x equal to 0.
\[ 0 < y \le 6 \]
Figure (svg): A rational graph whose numerator has lower degree than its denominator
\[ \text{domain: all reals}; \quad 0 < y \leq 6 \]
Verify: justify both ends of the range
Why: The denominator is smallest at x equal to 0, where it is 1, so the largest output is 6 and the range is closed at that end. And the denominator can be made as large as you like, so the outputs get arbitrarily small without reaching zero — which is why the range is open at the other end.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.3 Graph General Rational Functions §8.3, pp. 565-565
Fill the middle
Example 1.
Fill in the blanks
y = \frac6___ \text___ x = 0 \text___ y = ___
Why: At x equal to 0 the denominator is at its smallest, 1, so the output is at its largest, 6. That single evaluation gives the closed end of the range.
Worked example
Guided Practice 1.
\[ \text{Graph } y = \frac{4}{x^2+2} \text{ and state the domain and range.} \]
Check for intercepts and asymptotes
Why: The numerator never vanishes; the denominator is at least 2.
Compare the degrees
Why: Zero against 2 again.
\[ y = 0 \]
Find the peak
Why: The denominator is smallest at x equal to 0, where it is 2.
\[ \text{highest value } 2 \]
State the range
Why: The outputs run down toward zero from that peak.
\[ 0 < y \le 2 \]
Figure (svg): The solution to Worked example a second lower-degree case shown as a ladder of expressions, one row per algebraic move
\[ \text{domain: all reals}; \quad 0 < y \leq 2 \]
Verify: compare the two peaks
Why: Example 1's peak was 6 over 1, and this one is 4 over 2, or 2. Both peaks sit at x equal to 0 because that is where each denominator is smallest — and the peak is always the numerator divided by that smallest denominator value.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.3 Graph General Rational Functions §8.3, pp. 567-567
Error analysis
A student graphs a function whose denominator is a sum of squares.
Annotate
On: \( y = \frac{6}{x^2+1}: \; \text{vertical asymptote at } x = -1 \)
Solving the denominator for zero is right; accepting a non-real solution as an asymptote is not. Vertical asymptotes come from REAL zeros only.
Sorting
Compare the degrees.
Sort into buckets
Sort each function by the degree comparison.
The comparison takes a second and determines the whole far-out behaviour, which is why it is the first thing to check.
Comparison
Fill the blanks for six over x squared plus one.
Comparison matrix
| Feature | Where it comes from | Result here |
|---|---|---|
| x-intercepts | zeros of the numerator | none; 6 is never zero |
| Vertical asymptotes | real zeros of the denominator | none; x^2 + 1 is never zero |
| Horizontal asymptote | degree comparison, m < n | y = 0 |
| Range | the peak and the asymptote | 0 < y <= 6 |
Two of the four rows came back empty, which is what makes this case the simplest: no intercepts and no breaks leaves a single smooth curve.
Prediction
Commit before reasoning.
Predict first
For y equals 6 over x squared plus 1, is there an input giving output exactly 0?
Correct: No — setting the fraction to zero would need the numerator 6 to be zero.
\[ \frac{6}{x^2+1} = 0 \;\Longrightarrow\; 6 = 0, \text{ impossible} \]
Why: A fraction equals zero exactly when its numerator does, and 6 never does. So the horizontal asymptote is approached but never attained, which is why the range is written with a strict inequality at that end. This is the same reasoning that ruled out x-intercepts: both questions ask whether the numerator can vanish.
Section
Section 3
Concept
If the two degrees are equal, the horizontal asymptote is the ratio of the leading coefficients. Far from the origin the two polynomials grow at the same rate, so their quotient settles at that ratio.
\[ m = n \;\Longrightarrow\; y = \frac{a_m}{b_n} \]
This is Lesson 8.2's rule for a linear over a linear, generalised: it never depended on the degree being 1, only on the two degrees matching.
Figure (svg): A rational graph whose numerator and denominator have equal degree, with two vertical asymptotes
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.3 Graph General Rational Functions §8.3, pp. 566-566 — Graph a rational function (m = n)
Picture it
Example 2: two x squared over x squared minus 9.
Figure (svg): A rational graph whose numerator and denominator have equal degree, with two vertical asymptotes
The two vertical asymptotes cut the graph into three pieces. The outer two rise above the height 2 and the middle one dips below the horizontal axis.
Worked example
Example 2.
\[ \text{Graph } y = \frac{2x^2}{x^2-9}. \]
Find the x-intercept
Why: Two x squared is zero only at x equal to 0.
Find the vertical asymptotes
Why: X squared minus 9 is zero at 3 and negative 3.
\[ x = 3\text{ and } x = -3 \]
Find the horizontal asymptote
Why: Both degrees are 2, so divide 2 by 1.
\[ y = 2 \]
Plot points in all three regions
Why: Beyond both asymptotes and between them.
Figure (svg): A rational graph whose numerator and denominator have equal degree, with two vertical asymptotes
\[ x = \pm 3, \quad y = 2 \]
Verify: check one point in each region
Why: At x equal to 5 the value is 50 over 16, or about 3.1 — above the asymptote. At x equal to 2 it is 8 over negative 5, or negative 1.6 — below the horizontal axis. At x equal to negative 5 it is 3.1 again, by symmetry. Each region behaves differently, which is why points are needed in all three.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.3 Graph General Rational Functions §8.3, pp. 566-566
Fill the middle
Example 2.
Fill in the blanks
y = \frac1___: \; y = \frac______} = 2
Why: The denominator's leading coefficient is 1, since x squared is 1 times x squared. Dividing 2 by 1 gives the asymptote y equal to 2.
Worked example
Guided Practice 2 and 3.
\[ \text{Graph } y = \frac{3x^2}{x^2-1} \text{ and } f(x) = \frac{x^2-5}{x^2+1}. \]
First: intercept and asymptotes
Why: Numerator zero at 0; denominator zero at 1 and negative 1.
\[ \text{intercept } 0; x = 1, x = -1 \]
First: horizontal asymptote
Why: Three over 1.
\[ y = 3 \]
Second: intercepts
Why: X squared minus 5 is zero at the square roots of 5.
\[ \text{about } 2.24\text{ and } -2.24 \]
Second: asymptotes
Why: The denominator is never zero, and 1 over 1 is 1.
Figure (svg): The solution to Worked example two more equal-degree functions shown as a ladder of expressions, one row per algebraic move
\[ y = 3, \; x = \pm 1; \qquad y = 1, \text{ no vertical} \]
Verify: check the second's behaviour at the origin
Why: At x equal to 0 the value is negative 5 over 1, which is negative 5 — far below the horizontal asymptote at 1. So this graph rises from negative 5 up toward 1 on both sides, crossing the horizontal axis at the square roots of 5. A horizontal asymptote describes far-out behaviour only, and near the origin the graph is nowhere near it.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.3 Graph General Rational Functions §8.3, pp. 567-567
Trap
\[ y = \frac{2x^2}{x^2-9}, \; \text{asymptote } y = 2 \]
Conclude that every output is above 2
Why: The simple hyperbolas of Lesson 8.2 never crossed their horizontal asymptotes.
\[ y > 2 \text{ always} \quad \text{(wrong)} \]
At x equal to 2 the value is negative 1.6, well below 2. The middle piece lives entirely under the horizontal axis.
\[ \text{outer pieces: } y > 2; \quad \text{middle piece: } y \leq 0 \]
Check each region separately
Why: A horizontal asymptote constrains only the far-out behaviour, not the middle of the graph.
\[ \text{at } x = 2: \; y = \tfrac{8}{-5} = -1.6 \]
With two vertical asymptotes there are three regions, and each has to be sampled. This is the main reason Example 2 plots seven points rather than four.
Matching
Equal degrees means divide the leaders.
Match the pairs
Why: The first three have equal degrees and their asymptotes are the leading-coefficient ratios. The fourth has a lower-degree numerator, so a different rule applies and the answer is the horizontal axis.
Ranking
Graphing an equal-degree rational function.
Put in order
Why: Step two has to come before step four, because the vertical asymptotes are what divide the axis into regions. With two of them there are three regions, and sampling only one gives a badly wrong picture of the other two.
Prediction
Commit before reasoning.
Predict first
Why does two x squared over x squared minus 9 approach 2 far from the origin?
Correct: Because for huge x the -9 is negligible beside x squared, leaving 2x^2 over x^2.
\[ \frac{2x^2}{x^2-9} \approx \frac{2x^2}{x^2} = 2 \text{ for } |x| \text{ large} \]
Why: At x equal to 100 the denominator is 9991 against x squared's 10,000 — a difference of less than a tenth of a percent. So the fraction is within a hair of 2x squared over x squared, which is 2. The lower-degree terms matter enormously near the origin, where the graph is nothing like its asymptote, and not at all far away. That is the same argument that justified Lesson 8.2's rule.
Section
Section 4
Concept
If the numerator's degree exceeds the denominator's, there is no horizontal asymptote. The graph runs away at both ends, and its end behavior matches the ratio of leading coefficients times x raised to the difference of the degrees.
\[ m > n \;\Longrightarrow\; \text{end behavior like } \frac{a_m}{b_n}x^{m-n} \]
The exponent m minus n is what is left over after the denominator has cancelled as much of the numerator as it can. A difference of 1 gives end behavior like a line, a difference of 2 like a parabola.
Figure (svg): A rational graph whose numerator has higher degree, so there is no horizontal asymptote
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.3 Graph General Rational Functions §8.3, pp. 566-566 — Graph a rational function (m > n)
Picture it
Example 3: x squared plus 3x minus 4, over x minus 2.
Figure (svg): A rational graph whose numerator has higher degree, so there is no horizontal asymptote
There is no height the graph settles toward. Far to the right it rises like the line y equals x, and far to the left it falls the same way.
Worked example
Example 3.
\[ \text{Graph } y = \frac{x^2+3x-4}{x-2}. \]
Factor the numerator for intercepts
Why: X plus 4 times x minus 1.
\[ x = -4\text{ and } x = 1 \]
Find the vertical asymptote
Why: The denominator vanishes at 2.
\[ x = 2 \]
Compare the degrees
Why: Two against 1, so the numerator outranks.
Describe the end behavior
Why: The leading ratio is 1 over 1, and the degree difference is 1.
\[ \text{like } y = x \]
Figure (svg): A rational graph whose numerator has higher degree, so there is no horizontal asymptote
\[ x = 2; \quad \text{end behavior } y = x \]
Verify: test the end behavior far out
Why: At x equal to 100 the value is 10,296 over 98, or about 105 — close to 100, which is what end behavior like y equals x predicts. At x equal to negative 100 it is 9696 over negative 102, about negative 95, again close to the input. The graph tracks the line y equals x with a small and shrinking offset.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.3 Graph General Rational Functions §8.3, pp. 566-566
Fill the middle
Example 3.
Fill in the blanks
m - n = 2 - 1 = 1 \;\Longrightarrow\; \text1 y = x^___}
Why: The degree difference is 1, so the end behavior matches a first-degree function — a line. With a leading ratio of 1 over 1, that line is y equals x.
Worked example
Guided Practice 4.
\[ \text{Graph } y = \frac{x^2-2x-3}{x-4}. \]
Factor the numerator
Why: X minus 3 times x plus 1.
\[ \text{intercepts } 3\text{ and } -1 \]
Find the vertical asymptote
Why: X minus 4 vanishes at 4.
\[ x = 4 \]
Compare the degrees
Why: Two against 1 again.
Describe the end behavior
Why: Leading ratio 1 over 1, degree difference 1.
\[ \text{like } y = x \]
Figure (svg): The solution to Worked example one more higher-degree case shown as a ladder of expressions, one row per algebraic move
\[ x = 4; \quad \text{end behavior } y = x \]
Verify: check that no factor cancels
Why: Substituting 4 into the numerator gives 16 minus 8 minus 3, which is 5 — not zero, so x minus 4 is not a factor and the asymptote at 4 is genuine. Both intercepts lie to the left of it, which means the whole right branch stays above the horizontal axis.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.3 Graph General Rational Functions §8.3, pp. 567-567
Error analysis
A student finds the horizontal asymptote of a quadratic over a linear.
Annotate
On: \( y = \frac{x^2+3x-4}{x-2}: \; \text{horizontal asymptote } y = 1 \)
Compare the degrees before choosing a rule. When the numerator outranks, there is no horizontal asymptote to find.
Sorting
Compare the degrees first.
Sort into buckets
Sort each function.
Two of the five have no horizontal asymptote at all, and both are a quadratic over a linear. Reaching for the leading-coefficient rule there would give a confident wrong answer.
Comparison
Fill the blanks. One comparison decides everything.
Comparison matrix
| Case | Horizontal asymptote | Example |
|---|---|---|
| m < n | y = 0 | 6/(x^2 + 1) |
| m = n | y = the ratio of leading coefficients | 2x^2/(x^2 - 9) |
| m > n | none | (x^2 + 3x - 4)/(x - 2) |
| m > n, end behavior | like the leading ratio times x^(m - n) | like y = x |
The first three rows are the whole rule. The fourth is what replaces a horizontal asymptote when there is none: a description of the shape rather than a height.
Prediction
Commit before reasoning.
Predict first
What is the end behavior of a cubic over a linear, with leading coefficients 1 and 1?
Correct: Like a parabola, since the degree difference is 2.
\[ \frac{x^3}{x} = x^2 \text{ far from the origin} \]
Why: Three minus 1 is 2, so far from the origin the graph follows y equals x squared: rising steeply at both ends rather than in opposite directions. The degree difference is what is left after the denominator has cancelled as much of the numerator as it can, and that leftover degree is what shapes the ends. A difference of 1 gives opposite-direction ends; a difference of 2 gives matching ones.
Section
Section 5
Concept
A can of fixed volume can be tall and thin or short and wide. A tall thin can wastes material on its side; a short wide one wastes it on its two ends. Writing the surface area as a function of the radius alone gives a rational function whose minimum is the best compromise.
\[ S = 2\pi r^2 + \frac{684}{r} \]
The substitution is what makes this work: the height is eliminated using the fixed volume, leaving one variable and therefore a curve that can be minimised.
Figure (svg): Surface area of a fixed-volume can plotted against its radius, with the minimum marked
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.3 Graph General Rational Functions §8.3, pp. 567-567 — Multi-step problem
Picture it
Example 4: surface area against radius for a 342 cubic centimeter can.
Figure (svg): Surface area of a fixed-volume can plotted against its radius, with the minimum marked
The curve falls steeply, bottoms out near a radius of 3.79 centimeters, and climbs again. That bottom is the least material.
Worked example
Example 4.
\[ \text{Find the dimensions of a cylinder of volume } 342 \text{ cubic cm using the least material.} \]
Use the volume to eliminate the height
Why: Pi r squared h equals 342, so h is 342 over pi r squared.
\[ h = \frac{342}{\pi r ^{2}} \]
Write the surface area
Why: Two circular ends plus the curved side.
\[ S = 2 \pi r ^{2} + 2 \pi r h \]
Substitute and simplify
Why: The pi and one r cancel in the second term.
\[ S = 2 \pi r ^{2} + \frac{684}{r} \]
Find the minimum
Why: Graph and read the lowest point.
\[ r\text{ about } 3.79, S\text{ about } 271 \]
Figure (svg): Surface area of a fixed-volume can plotted against its radius, with the minimum marked
\[ r \approx 3.79 \text{ cm}, \quad h \approx 7.58 \text{ cm} \]
Verify: notice the shape of the answer
Why: The height 7.58 is twice the radius 3.79, so the height equals the diameter — the can is exactly as tall as it is wide. That relationship holds for every cylinder minimising surface area at fixed volume, whatever the volume, which is a much more useful fact than the two numbers themselves.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.3 Graph General Rational Functions §8.3, pp. 567-567
Fill the middle
Example 4, Step 1.
Fill in the blanks
342 = \pi r^2h \;\Longrightarrow\; h = \frac2___}}}
Why: The base area is pi r squared, so dividing the volume by it gives the height. This substitution is what turns a two-variable problem into a one-variable one.
Worked example
Guided Practice 5.
\[ \text{Repeat for a volume of } 544 \text{ cubic centimeters.} \]
Rewrite the height
Why: Five hundred forty-four over pi r squared.
\[ h = \frac{544}{\pi r ^{2}} \]
Rewrite the surface area
Why: Twice 544 is 1088 in the second term.
\[ S = 2 \pi r ^{2} + \frac{1088}{r} \]
Find the minimum
Why: The lowest point sits near a radius of 4.42.
\[ r\text{ about } 4.42 \]
Find the height
Why: Five hundred forty-four over pi times 19.57.
\[ h\text{ about } 8.85 \]
Figure (svg): The solution to Worked example a larger can shown as a ladder of expressions, one row per algebraic move
\[ r \approx 4.42 \text{ cm}, \quad h \approx 8.85 \text{ cm} \]
Verify: check the height against the diameter again
Why: Twice 4.42 is 8.84, and the height is 8.85 — equal to within rounding. The same relationship appeared for the smaller can, which is the point of doing a second one: it turns a computed answer into a general rule worth remembering.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.3 Graph General Rational Functions §8.3, pp. 567-567
Trap
\[ S = 2\pi r^2 + 2\pi rh \]
Graph this and look for a minimum
Why: The surface area formula is used as it stands.
\[ \text{cannot be graphed as a curve} \quad \text{(wrong)} \]
With both r and h free, the surface area can be made as small as you like — and the volume would not stay at 342.
\[ \pi r^2h = 342 \;\Longrightarrow\; h = \frac{342}{\pi r^2} \]
Use the constraint to eliminate one variable first
Why: The fixed volume is what ties the two dimensions together.
\[ S = 2\pi r^2 + \frac{684}{r}, \text{ a function of } r \text{ alone} \]
Every optimization problem has this shape: a quantity to minimise and a constraint. Substituting the constraint into the quantity is what reduces it to one variable.
Ranking
Finding the most efficient can.
Put in order
Why: Steps one and two set up the substitution and step four performs it. Skipping straight to step five with two variables still present is the single most common failure, and it produces a formula that cannot be graphed as a curve at all.
Sorting
Read the model against the can.
Sort into buckets
Sort each part of S equals 2 pi r squared plus 684 over r.
The two terms pull in opposite directions, which is exactly why a minimum exists rather than the surface area simply falling forever.
Prediction
Commit before reasoning.
Predict first
Both worked cans came out with height equal to diameter. Is that a coincidence?
Correct: No — it holds for every cylinder minimising surface area at fixed volume.
\[ h = 2r \text{ at the minimum, for every fixed volume} \]
Why: The volume 342 gave 3.79 and 7.58; the volume 544 gave 4.42 and 8.85. In both the height is twice the radius. Setting up the general problem gives the same conclusion for any fixed volume, so the most efficient cylinder is always as tall as it is wide. Real soup cans are taller and thinner than this, because material is not the only cost — shelves, labels and hands matter too, which is where the mathematical answer meets its limits.
Comparison
Fill the blanks. One comparison decides everything far from the origin.
Comparison matrix
| Comparison | Horizontal asymptote | What the graph does far out |
|---|---|---|
| m < n | y = 0 | flattens toward the horizontal axis |
| m = n | y = the ratio of leading coefficients | flattens toward that height |
| m > n by 1 | none | follows a line |
| m > n by 2 | none | follows a parabola |
The last two rows show what replaces a horizontal asymptote when there is none: a shape rather than a height, given by x raised to the leftover degree.
Pattern
Four questions, then a sketch.
With two vertical asymptotes there are three regions and each behaves differently. Sampling only one of them gives a badly wrong picture.
OpenStax Algebra and Trigonometry 2e, §5.6 Rational Functions §5.6
Check
Compare the degrees first.
Check your understanding
What is the horizontal asymptote of y = 6/(x^2 + 1)?
Answer: A
Why: The numerator's degree, 0, is less than the denominator's, 2.
Check
Equal degrees. Divide the leaders.
Check your understanding
What are the asymptotes of y = 2x^2/(x^2 - 9)?
Answer: A
Why: The denominator vanishes at plus and minus 3, and 2 over 1 is 2.
Check
Higher degree on top.
Check your understanding
What is the end behavior of y = (x^2 + 3x - 4)/(x - 2)?
Answer: A
Why: The leading ratio is 1 over 1 and the degree difference is 2 minus 1.
Real world
A drug's concentration in the blood t hours after a dose is C equals 5t over t squared plus 4, in milligrams per litre. A concentration below 0.5 is too low to be effective.
Discussion prompt
Find the horizontal asymptote and say what it means clinically, then estimate roughly how long the dose stays effective.
Hint: Compare the degrees, then solve the equation for 0.5.
Answer:
\[ m = 1 < n = 2 \;\Longrightarrow\; \text{horizontal asymptote } C = 0 \]
\[ 0.5 = \frac{5t}{t^2+4} \;\Longrightarrow\; 0.5t^2 - 5t + 2 = 0 \;\Longrightarrow\; t^2 - 10t + 4 = 0 \]
\[ t = \frac{10 \pm \sqrt{84}}{2} \approx 0.41 \text{ and } 9.59 \]
The asymptote at zero says the drug eventually clears completely — the concentration falls toward nothing without a lasting residue, which is exactly what the degree comparison predicted.
The dose rises above the effective level about 25 minutes after taking it and drops back below it after about 9 hours and 35 minutes, giving roughly 9 hours of effective coverage. That is how a dosing interval gets chosen: the next dose has to arrive before the curve falls back through the threshold. Both answers came from this lesson — one from comparing degrees, one from setting the rational function equal to a value and clearing the denominator, which is Lesson 8.6's method arriving early.
Commit first
Answer, then rate your confidence honestly.
Predict first
Can the graph of a rational function cross its horizontal asymptote?
Correct: Yes — a horizontal asymptote describes far-out behaviour only, so the middle of the graph may cross it.
\[ \frac{x^2-5}{x^2+1} = 1 \;\Longrightarrow\; -5 = 1, \text{ so that one does NOT cross} \]
Even so, many do — the middle piece of Example 2's graph sits far below its asymptote at 2 while the outer pieces sit above it.
Why: A VERTICAL asymptote is never crossed, because the function is undefined there. A horizontal one is a statement about what happens as the inputs run far out, and says nothing about the middle. The function x squared minus 5, over x squared plus 1, has horizontal asymptote y equal to 1 and passes through negative 5 at the origin, so it must cross the height 1 twice on its way up. The simple hyperbolas of Lesson 8.2 never crossed theirs, which makes the false rule easy to believe.
Explain it
They can find where a denominator is zero and think that is all there is to it.
Discussion prompt
In four sentences or fewer, explain how to decide whether a rational function has a horizontal asymptote.
Hint: Compare the highest powers.
Answer:
Look at the highest power on top and the highest power on the bottom, and ask which one grows faster for huge inputs.
If the bottom wins, the fraction shrinks to nothing and the asymptote is the horizontal axis. If they tie, divide the two leading coefficients and that is the height. If the top wins, the fraction grows without limit and there is no horizontal asymptote at all.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the degree case, write m and n above the fraction before doing anything else. For common factors, factor both polynomials first, every time. For plotting, count the vertical asymptotes and remember the regions number one more than that. For optimization, write the constraint first and use it to eliminate a variable before touching the quantity you are minimising.
Connect it up
Paper. Fifteen minutes.
Draw it
Build a rational-graph page. Top: write the three characteristics and, beside the third, the full degree rule with all three cases and the end-behavior formula. Left column: graph Example 1, marking the absence of intercepts and vertical asymptotes and writing the range with a justification for each end. Middle column: graph Example 2 in full, with both vertical asymptotes drawn and at least two points in each of the three regions, and note which regions sit above and below the horizontal asymptote. Right column: graph Example 3, marking both intercepts and the asymptote, and write out the end-behavior computation. Bottom: work the can problem from constraint to substitution to minimum, and write one sentence on why the answer always has height equal to diameter. In a margin, write one function with a common factor and show where its hole is.
If your Example 2 sketch has the middle piece above the horizontal asymptote, recheck it at x equal to 2: the value there is negative 1.6.
Recap
Five things, all resting on one comparison.
| If you see | Then |
|---|---|
| A zero of the numerator | An x-intercept |
| A zero of the denominator | A vertical asymptote |
| A factor in both | Cancel it; there is a hole, not an asymptote |
| m less than n | Horizontal asymptote y = 0 |
| m equal to n | Horizontal asymptote at the ratio of leading coefficients |
| m greater than n | No horizontal asymptote; end behavior like that ratio times x to the difference |
Lesson 8.4 leaves graphs behind and starts operating on rational expressions themselves, multiplying and dividing them by factoring and cancelling.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.3 Graph General Rational Functions §8.3, pp. 565-569 — everything on these slides traces back here
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