The equivalence of zeros, factors, solutions and x-intercepts, graphing a polynomial from its intercepts and end behaviour, turning points and the rules that count them, local maxima and minima, and maximising a volume model.
Subject: Algebra 2 · 65 slides · symbolic lesson
Open the interactive version of this deck
Title
Algebra 2 · Chapter 5 — Polynomials and Polynomial Functions
Analyze Graphs of Polynomial Functions
Objectives
Five outcomes. The first is a summary of the chapter; the rest are what it lets you draw.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.8 Analyze Graphs of Polynomial Functions §5.8, pp. 387-391 — the lesson these objectives are drawn from
Warm-up
This chapter has produced four different names for the same numbers, one per lesson.
Discussion prompt
For the function f of x equals the quantity x plus 3, times the quantity x minus 2, squared, write down everything you can say about the number 2 using the vocabulary of Lessons 5.4 to 5.7.
Hint: Zero, factor, solution, intercept.
Answer:
Two is a zero of f, since f of 2 is 0. So x minus 2 is a factor, by the factor theorem. So 2 is a solution of f of x equals zero. And since 2 is real, the graph passes through the x-intercept (2, 0).
Four statements, one number, and each was proved in a different lesson. This lesson uses them all at once to draw the graph.
Concept
A polynomial in factored form displays its real zeros directly, and those are the x-intercepts. Combining them with the end behaviour of Lesson 5.2 and a handful of extra points gives a confident sketch without any calculus.
turning point — A point where a graph changes from rising to falling or from falling to rising. Its y-coordinate is a local maximum if the point is higher than all nearby points, and a local minimum if it is lower.
\[ f(x) = \tfrac{1}{6}(x+3)(x-2)^2 \]
The number of turning points is not free either: a polynomial of degree n turns at most n minus 1 times, and if it has n distinct real zeros it turns exactly that many.
Figure (svg): Four equivalent statements about a number k, listed with the language each belongs to
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.8 Analyze Graphs of Polynomial Functions §5.8, pp. 387-388
Section
Section 1
Concept
For a polynomial function f, saying that k is a zero, that x minus k is a factor, and that k is a solution of the equation are three ways of saying one thing. If k is real, a fourth follows: the graph passes through the point with coordinates k and zero.
\[ f(k) = 0 \;\Longleftrightarrow\; (x-k) \mid f(x) \;\Longleftrightarrow\; k \text{ solves } f(x)=0 \]
The fourth statement is the only one with a condition attached. An imaginary zero satisfies the first three and has no point on the graph at all.
Figure (svg): Four equivalent statements about a number k, listed with the language each belongs to
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.8 Analyze Graphs of Polynomial Functions §5.8, pp. 387-387 — Zeros, Factors, Solutions, and Intercepts
Picture it
Each row came from a different lesson.
Figure (svg): Four equivalent statements about a number k, listed with the language each belongs to
The condition on the last row is why Lesson 5.7's quintic had five zeros but only two points on its graph — three of the five were repeated or imaginary.
Worked example
One number, four statements.
\[ \text{Given } f(x) = (x+3)(x-2)^2, \text{ say everything about } k = -3 \text{ and about } k = 2. \]
Start from the factorisation
Why: The factor x plus 3 appears once and x minus 2 appears twice.
Translate for negative 3
Why: It is a zero, x plus 3 is a factor, it solves the equation, and the graph crosses at (-3, 0).
Translate for 2
Why: The same four hold, but the factor appears twice, so 2 is a repeated zero.
\[ \text{multiplicity } 2 \]
Say what the multiplicity changes
Why: An even power makes the graph tangent rather than crossing.
\[ \text{tangent at } (2, 0) \]
Figure (svg): The solution to Worked example translate between the four shown as a ladder of expressions, one row per algebraic move
\[ (-3, 0) \text{ crossing}; \quad (2, 0) \text{ tangency} \]
Verify: check the values just either side of each
Why: Just left and right of negative 3 the function changes sign, since only one factor changes sign. At 2 the squared factor is positive on both sides, so no sign change occurs and the curve turns back. That is exactly the difference between crossing and touching.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.8 Analyze Graphs of Polynomial Functions §5.8, pp. 387-387
Sorting
Only one of the four has a condition.
Sort into buckets
For a zero k of a polynomial f, sort each statement.
The split is exactly the algebra-against-geometry line. Everything algebraic survives the move to complex numbers; nothing geometric does.
Worked example
The condition on the fourth row, tested.
\[ \text{For } f(x) = (x-2)(x^2+1), \text{ say what is true of } k = i. \]
Check whether i is a zero
Why: Substituting gives the quantity i minus 2 times i squared plus 1, which is i minus 2 times 0, so it is zero.
Check the factor statement
Why: X squared plus 1 is x minus i times x plus i, so x minus i is a factor.
Check the solution statement
Why: It satisfies the equation f of x equals zero.
Check the intercept statement
Why: The graph plots real x against real y, and i is not a real number, so there is no such point.
Figure (svg): The solution to Worked example which statements survive for an imaginary zero shown as a ladder of expressions, one row per algebraic move
\[ \text{zero, factor, solution — but no intercept} \]
Verify: count the intercepts against the zeros
Why: The function has degree 3 and so three zeros — 2, i and negative i — but its graph crosses the axis exactly once. Every discrepancy between the number of zeros and the number of intercepts comes from either an imaginary zero or a repeated one, and this is the first kind.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.8 Analyze Graphs of Polynomial Functions §5.8, pp. 387-387
Trap
\[ f(x) = (x+3)(x-2)^2 \]
Count the points where the graph meets the axis
Why: There are two such points, at negative 3 and 2.
\[ \text{f has two zeros} \quad \text{(wrong count)} \]
The function is cubic, so by Lesson 5.7's corollary it has exactly three zeros: negative 3, 2 and 2.
\[ \text{degree } 3 \;\Longrightarrow\; \text{three zeros: } -3, 2, 2 \]
Count from the degree, and read the graph only for their positions
Why: A graph shows where the real zeros are, not how many zeros there are.
\[ \text{two distinct zeros, three counted with multiplicity} \]
A graph undercounts twice over: a repeated zero shows as one point, and an imaginary zero shows as none. It is a locator, not a census.
Fill the middle
The warm-up function.
Fill in the blanks
(x - 2)^2 \text2 \;\Longrightarrow\; \text___ (___, 0)
Why: The factor vanishes at 2, so that is where the graph meets the axis, and the even power makes it a tangency rather than a crossing. Reading the intercept as the value that makes the factor zero is the same habit as everywhere else in this course.
Prediction
Commit before reasoning.
Predict first
Why can an imaginary zero not be an x-intercept?
Correct: The graph plots real inputs against real outputs.
\[ (2+i, 0) \text{ is not a point of the real plane} \]
Why: Every point on the graph has a real first coordinate, so a number like 2 plus i simply cannot be located there. The zero is perfectly genuine — it satisfies the equation and gives a factor — but the picture is drawn in a plane that has no room for it. Lesson 4.6's complex plane would show it, but that is a different picture entirely.
Matching
Each was proved somewhere in this chapter.
Match the pairs
Why: The chapter builds one idea across four lessons: what a zero is, how to recognise one, how many there are, and where to look for them. This lesson adds the fifth question — what they look like on a graph.
Section
Section 2
Concept
For a polynomial in factored form, plot the x-intercepts first, then evaluate at a few values between and beyond them to fix the heights, then use the end behaviour to draw the two ends and join everything smoothly.
\[ f(x) = \tfrac{1}{6}(x+3)(x-2)^2 \]
The degree and leading coefficient can be read off the factored form without expanding: three linear factors and a positive constant means a cubic with a positive leading coefficient.
Figure (svg): A cubic graphed from its intercepts, with a repeated zero producing a tangency
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.8 Analyze Graphs of Polynomial Functions §5.8, pp. 387-387 — Use x-intercepts to graph a polynomial function
Picture it
Example 1, with the tangency at the repeated zero.
Figure (svg): A cubic graphed from its intercepts, with a repeated zero producing a tangency
The curve crosses at negative 3 and turns back at 2. Four plotted points between and beyond the intercepts were enough to fix the shape.
Worked example
Example 1, in the book's four steps.
\[ \text{Graph } f(x) = \tfrac{1}{6}(x+3)(x-2)^2. \]
Plot the intercepts
Why: The factors vanish at negative 3 and at 2.
\[ (-3, 0)\text{ and } (2, 0) \]
Evaluate between and beyond
Why: At negative 2, negative 1, 0, 1 and 3 the values are eight thirds, 3, 2, two thirds and 1.
Read the end behaviour
Why: Three linear factors and a positive constant give a cubic with a positive leading coefficient.
Join smoothly
Why: The curve crosses at negative 3 and is tangent at 2, because that factor is squared.
Figure (svg): The solution to Worked example graph from factored form shown as a ladder of expressions, one row per algebraic move
\[ f(x) = \tfrac{1}{6}(x+3)(x-2)^2 \]
Verify: check the leading coefficient without expanding
Why: Multiplying the three leading terms gives a sixth times x times x times x, so the leading coefficient is a sixth — positive, and the degree is 3. Expanding would confirm it but is not needed; the factored form contains the information already.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.8 Analyze Graphs of Polynomial Functions §5.8, pp. 387-387
Ranking
Sketching from factored form.
Put in order
Why: Noting the even powers early matters because it changes what the extra points are telling you: near a tangency the values on both sides have the same sign, and a sketch drawn without that in mind will cross where it should turn.
Worked example
Guided Practice 1 and 2.
\[ \text{Graph } 0.25(x+2)(x-1)(x-3) \text{ and } 2(x-1)^2(x-4). \]
First: read the intercepts and shape
Why: Three distinct zeros at negative 2, 1 and 3, with a positive leading coefficient.
First: locate the turns
Why: Three distinct real zeros force exactly two turning points, one between each pair.
Second: read the intercepts
Why: A repeated zero at 1 and a simple zero at 4.
\[ \text{tangent at } 1,\text{ crosses at } 4 \]
Second: locate the turns
Why: The tangency at 1 is itself a turning point, and there is a minimum between 1 and 4.
\[ \text{local } \max(1, 0),\text{ local } \min(3, -8) \]
Figure (svg): The solution to Worked example two more from factored form shown as a ladder of expressions, one row per algebraic move
\[ \text{two cubics, three turns between them} \]
Verify: check the second's turning points
Why: The local maximum is at (1, 0), which is the tangency itself, and substituting 3 gives 2 times 4 times negative 1, or negative 8, so the local minimum is at (3, negative 8). A repeated zero is always a turning point, which is another way of saying the graph is tangent there.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.8 Analyze Graphs of Polynomial Functions §5.8, pp. 389-389
Error analysis
A student sketches a cubic with a squared factor.
Annotate
On: \( f(x) = \tfrac{1}{6}(x+3)(x-2)^2: \text{ the curve crosses the axis at both } -3 \text{ and } 2 \)
Even power touches, odd power crosses. Checking the sign just either side of the intercept settles it in two substitutions.
Fill the middle
Example 1, without expanding.
Fill in the blanks
\tfrac3___(x+3)(x-2)^2: \; \text___ \tfrac______ \cdot x \cdot x^2 = \tfrac______x^___}
Why: Multiplying the leading terms of the factors gives a sixth times x cubed, so the degree is 3 and the leading coefficient is one sixth. The exponents add, exactly as in Lesson 5.1, which is why the degree of a product is the sum of the degrees.
Sorting
Look at the power on each factor.
Sort into buckets
For f(x) = (x + 2)(x - 1)^2 (x - 4)^3, sort each intercept.
Only the parity matters, not the size of the power. A factor to the fifth crosses just as a linear one does, though it flattens noticeably as it goes through.
Comparison
Fill the blanks. Two ways to draw the same curve.
Comparison matrix
| Question | From a table | From the intercepts |
|---|---|---|
| What you need first | nothing but the function | the function in factored form |
| How the crossings are found | by spotting sign changes in the table | read from the factors |
| Evaluations needed | seven or so | four or five |
| Repeated zeros | easy to miss | visible as an even power |
Factored form is worth the work of getting it, and Lessons 5.4 to 5.6 exist largely to produce it.
Section
Section 3
Concept
The graph of a polynomial of degree n has at most n minus 1 turning points. If it has n distinct real zeros, it has exactly n minus 1, because the curve must turn between every pair of consecutive crossings.
\[ \deg f = n \;\Longrightarrow\; \text{at most } n-1 \text{ turns} \]
The second rule explains the first. Each crossing forces the curve to come back to the axis, so n crossings need n minus 1 turns, and the degree caps how many crossings there can be.
Figure (svg): Two rules relating degree and distinct zeros to the number of turning points
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.8 Analyze Graphs of Polynomial Functions §5.8, pp. 388-388 — Turning Points of Polynomial Functions
Picture it
What the degree allows and what distinct zeros force.
Figure (svg): Two rules relating degree and distinct zeros to the number of turning points
A cubic may turn twice, once in a degenerate sense, or not at all: y equals x cubed has no turning point despite being a cubic. The bound is real but it is only a bound.
Worked example
Four functions from earlier in the chapter.
\[ \text{How many turning points has } x^3, \; (x+2)(x-1)(x-3), \; \tfrac{1}{6}(x+3)(x-2)^2, \; x^4 - 6x^3 + 3x^2 + 10x - 3? \]
First: apply the bound
Why: Degree 3 allows at most 2 turns, but this function is always increasing.
Second: apply the exact rule
Why: Three distinct real zeros force exactly 2 turns.
Third: count the distinct zeros
Why: Only two distinct zeros, so the exact rule does not apply; the bound allows 2.
Fourth: count the distinct zeros
Why: Four distinct real zeros force exactly 3 turns.
Figure (svg): The solution to Worked example count the turns shown as a ladder of expressions, one row per algebraic move
\[ 0, \; 2, \; 2, \; 3 \]
Verify: check the first against the bound
Why: A cubic may turn twice, and x cubed turns not at all — so the bound is genuinely an upper limit rather than a prediction. The exact rule did not apply there because x cubed has only one distinct real zero, not three.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.8 Analyze Graphs of Polynomial Functions §5.8, pp. 388-388
Matching
Count the distinct real zeros first.
Match the pairs
Why: Only the second and fourth have the full number of distinct real zeros, so only they are covered by the exact rule. The other two fall under the bound, and they land at opposite ends of it — one with no turns and one with the maximum.
Worked example
Guided Practice 2 examined more closely.
\[ \text{How many turning points has } g(x) = 2(x-1)^2(x-4), \text{ and where are they?} \]
Count the distinct zeros
Why: Two: 1 and 4. The exact rule needs three for a cubic, so it does not apply.
\[ \text{the bound allows } 2 \]
Note that a tangency is a turn
Why: At a repeated zero the curve touches and turns back, so x equal to 1 is a turning point.
\[ \text{one turn at } (1, 0) \]
Find the other
Why: Between 1 and 4 the curve must come back up to cross at 4, so there is a minimum in between, at x equal to 3.
\[ (3, -8) \]
Classify them
Why: The first is a local maximum and the second a local minimum.
Figure (svg): The solution to Worked example turns and repeated zeros shown as a ladder of expressions, one row per algebraic move
\[ (1, 0) \text{ max}; \quad (3, -8) \text{ min} \]
Verify: check the value at the minimum
Why: Substituting 3 gives 2 times 2 squared times negative 1, which is negative 8. And the local maximum is at height zero, on the axis itself — a reminder that a local maximum need not be the largest value the function ever takes, since the curve rises without bound to the right.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.8 Analyze Graphs of Polynomial Functions §5.8, pp. 389-389
Error analysis
A student sketches a cubic with one real zero.
Annotate
On: \( \text{degree } 3 \;\Longrightarrow\; \text{two turning points, so the curve must wiggle} \)
Read which rule applies before drawing. The bound tells you what cannot happen; only the exact rule tells you what must.
Fill the middle
A polynomial of degree 6.
Fill in the blanks
\deg f = 6 \;\Longrightarrow\; \text5 ___ \text___
Why: Six minus 1 is 5. If the sextic also has six distinct real zeros then it has exactly five turns, one between each consecutive pair. Fewer distinct zeros means the count could be anything from 5 down to 1 — a polynomial of even degree always turns at least once.
Prediction
Commit before reasoning.
Predict first
A cubic crosses the axis at negative 2, 1 and 3. Why must it turn twice?
Correct: Between two crossings it must come back to the axis, which requires a turn.
\[ n \text{ distinct real zeros} \;\Longrightarrow\; n-1 \text{ gaps} \;\Longrightarrow\; n-1 \text{ turns} \]
Why: After crossing at negative 2 the curve moves away from the axis, and to cross again at 1 it must come back — which means changing direction somewhere in between. Three crossings give two such gaps and therefore two turns. This is why the second rule is exact where the first is only a bound: crossings force turns, whereas degree merely permits them.
Two truths and a lie
All three are about turning points.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. Three is the maximum, reached only when there are four distinct real zeros. The quartic x to the fourth has just one turning point, at the origin, and x to the fourth plus 1 also has one. Always is what breaks the claim; at most would make it true.
Section
Section 4
Concept
The y-coordinate of a turning point is a local maximum if the point is higher than all nearby points, and a local minimum if it is lower. Local is the essential word: a local maximum can be far below values the function takes elsewhere.
local maximum — The y-coordinate of a turning point that is higher than all nearby points. A local minimum is defined the same way with lower.
\[ g(x) = x^4 - 6x^3 + 3x^2 + 10x - 3 \]
For a polynomial of odd degree there is no highest or lowest value at all, because the ends run off in opposite directions. Only local extremes exist.
Figure (svg): A quartic with its four x-intercepts and three turning points labelled as local maxima and minima
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.8 Analyze Graphs of Polynomial Functions §5.8, pp. 388-388 — Find turning points
Picture it
Example 2b: a quartic with four intercepts.
Figure (svg): A quartic with its four x-intercepts and three turning points labelled as local maxima and minima
The local maximum at height 5.11 is nowhere near the function's largest value — the curve climbs past 5.11 again on both ends. Local means only in a neighbourhood.
Worked example
Example 2, both parts.
\[ \text{For } x^3 - 3x^2 + 6 \text{ and } x^4 - 6x^3 + 3x^2 + 10x - 3, \text{ find the intercepts and turning points.} \]
First: graph and count
Why: The cubic has one x-intercept and two turning points.
First: read the coordinates
Why: The intercept is about negative 1.20; the local maximum is at (0, 6) and the local minimum at (2, 2).
\[ (0, 6)\text{ and } (2, 2) \]
Second: graph and count
Why: The quartic has four x-intercepts and three turning points.
Second: read the coordinates
Why: The intercepts are about negative 1.14, 0.29, 1.82 and 5.03.
Second: classify the turns
Why: A local maximum at (1.11, 5.11) and local minima at (-0.57, -6.51) and (3.96, -43.04).
Figure (svg): The solution to Worked example find the turning points shown as a ladder of expressions, one row per algebraic move
\[ (0,6), (2,2); \quad (1.11, 5.11), (-0.57,-6.51), (3.96,-43.04) \]
Verify: check the counts against the rules
Why: The cubic has one distinct real zero and turns twice, which the bound allows. The quartic has four distinct real zeros and turns exactly three times, as the exact rule requires. Both agree with the rules, and a graph disagreeing with them would signal a reading error.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.8 Analyze Graphs of Polynomial Functions §5.8, pp. 388-388
Sorting
Is the point higher or lower than its neighbours?
Sort into buckets
For the quartic in Example 2b, sort each turning point.
A quartic with three turns always has two of one kind and one of the other, alternating along the curve — which is a useful check on a sketch.
Worked example
Guided Practice 3.
\[ \text{For } h(x) = 0.5x^3 + x^2 - x + 2, \text{ find the intercept and the turning points.} \]
Read the end behaviour
Why: Degree 3 with a positive leading coefficient: down-left, up-right.
Find the intercept
Why: The curve crosses at about negative 3.07.
\[ x\text{ about } -3.07 \]
Locate the local maximum
Why: Between the crossing and the origin the curve peaks at about (-1.72, 4.13).
Locate the local minimum
Why: Just right of the origin it bottoms out at about (0.39, 1.79).
Figure (svg): The solution to Worked example a cubic with one intercept shown as a ladder of expressions, one row per algebraic move
\[ x \approx -3.07; \quad (-1.72, 4.13); \quad (0.39, 1.79) \]
Verify: check the local minimum is above the axis
Why: The local minimum has height about 1.79, which is positive, so the curve never returns to the axis after that turn — which is exactly why there is only one x-intercept despite two turning points. Reading the sign of a turning point's height predicts the number of crossings.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.8 Analyze Graphs of Polynomial Functions §5.8, pp. 389-389
Trap
\[ g \text{ has a local maximum at } (1.11, 5.11) \]
Conclude that g never exceeds 5.11
Why: The maximum is taken as a ceiling on the whole function.
\[ g(5.5) = 915.0625 - 998.25 + 90.75 + 55 - 3 \approx 59.6 \quad \text{(far above 5.11)} \]
The quartic has even degree with a positive leading coefficient, so it rises without bound at both ends and has no largest value at all.
\[ \text{local maximum: higher than all NEARBY points} \]
Read local as a statement about a neighbourhood
Why: It compares the point with its immediate surroundings, not with the whole graph.
\[ \text{even degree, } a > 0: \; \text{a lowest value exists, but no highest} \]
For an odd-degree polynomial there is neither a highest nor a lowest value, since the two ends run in opposite directions. Local extremes are the only ones such a function has.
Fill the middle
Guided Practice 2.
Fill in the blanks
g(x) = 2(x-1)^2(x-4): \; g(3) = 2(4)(-1) = -8
Why: Substituting 3 gives 2 times the quantity 3 minus 1, squared, which is 4, times the quantity 3 minus 4, which is negative 1 — so the value is negative 8. The local minimum is at (3, negative 8), below the axis, which is why the curve has to come back up to cross at 4.
Prediction
Commit before reasoning.
Predict first
Which polynomials have a genuine largest value, not merely a local maximum?
Correct: Only those of even degree with a negative leading coefficient.
\[ \text{even degree}, a < 0 \;\Longrightarrow\; \text{a global maximum exists} \]
Why: An even degree makes both ends behave alike, and a negative leading coefficient sends both downward, so the curve has a highest point somewhere in the middle. An even degree with a positive leading coefficient has a lowest value instead, and any odd degree has neither, since one end runs to positive infinity and the other to negative. This is Lesson 5.2's end behaviour answering a question about extremes.
Comparison
Fill the blanks. The word local is doing real work.
Comparison matrix
| Question | Local maximum | Global maximum |
|---|---|---|
| Compared with | nearby points only | every point on the graph |
| How many can there be? | several | at most one value |
| Does a quartic with a > 0 have one? | yes | no |
| Does a cubic have one? | possibly | never |
The third row is the case to remember: an upward quartic has local maxima but no global one, because it climbs without bound at both ends.
Section
Section 5
Concept
A volume model built from a folded sheet is a cubic. Only part of its domain is physically possible, and the maximum is found by graphing on that interval and reading the highest point.
\[ V = (20-2x)(16-2x)x = 4x^3 - 72x^2 + 320x \]
The restriction matters as much as the model. The cubic rises without bound to the right, but a cut wider than 8 inches would remove more than the sheet has.
Figure (svg): A cardboard sheet with corner cuts and the volume function peaking near a cut of three inches
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.8 Analyze Graphs of Polynomial Functions §5.8, pp. 389-389 — Maximize a polynomial model
Picture it
Example 3: a 16 by 20 inch sheet with square corners removed.
Figure (svg): A cardboard sheet with corner cuts and the volume function peaking near a cut of three inches
The volume peaks at about 420 cubic inches when the cut is about 2.94 inches, giving a box roughly 3 by 10 by 14 inches.
Worked example
Example 3. Verbal model, then a graph.
\[ \text{From a } 16 \text{ by } 20 \text{ inch sheet, square corners of side } x \text{ are cut and the sides folded. Maximise the volume.} \]
Write the verbal model
Why: Volume is length times width times height.
\[ V = (20 - 2 x) (16 - 2 x) x \]
Expand into standard form
Why: The product of the two binomials is 320 minus 72x plus 4x squared, times x.
\[ V = 4 x ^{3} - 72 x ^{2} + 320 x \]
Restrict the domain
Why: The cut must be positive and less than half the shorter side.
\[ 0 < x < 8 \]
Graph on that interval and find the peak
Why: The maximum is about 420 at x about 2.94.
\[ \text{about } 420\text{ cubic inches} \]
State the dimensions
Why: The cut is about 3 inches, giving a box about 3 by 10 by 14.
\[ 3\text{ by } 10\text{ by } 14\text{ inches} \]
Figure (svg): The solution to Worked example maximise the box shown as a ladder of expressions, one row per algebraic move
\[ x \approx 2.94, \quad V \approx 420 \text{ in}^3 \]
Verify: evaluate the model at the peak
Why: At x equal to 2.94: 4 times 25.4 is about 102; minus 72 times 8.64 is about negative 622; plus 320 times 2.94 is about 941. The total is about 420 cubic inches. And the dimensions 20 minus 5.9, 16 minus 5.9 and 2.94 are about 14.1, 10.1 and 2.94, whose product is about 420 — the two routes agree.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.8 Analyze Graphs of Polynomial Functions §5.8, pp. 389-389
Fill the middle
Example 3, at the restriction.
Fill in the blanks
16 - 2x > 0 \;\Longrightarrow\; x < 8
Why: The width 16 minus 2x must be positive, so x is less than 8. The length gives the weaker condition x less than 10, so the shorter side is what binds. Checking both dimensions and taking the tighter restriction is the step that is easy to skip.
Worked example
Guided Practice 5. Everything scales, but not proportionally.
\[ \text{Repeat for a } 10 \text{ by } 15 \text{ inch sheet.} \]
Write the new model
Why: The dimensions become 15 minus 2x, 10 minus 2x and x.
\[ V = 4 x ^{3} - 50 x ^{2} + 150 x \]
Restrict the domain
Why: The cut must be less than half the shorter side, which is 5.
\[ 0 < x < 5 \]
Graph and read the peak
Why: The maximum is about 132 at x about 1.96.
\[ \text{about } 132\text{ cubic inches} \]
State the dimensions
Why: About 2 by 6.1 by 11.1 inches.
\[ \text{roughly } 2\text{ by } 6\text{ by } 11 \]
Figure (svg): The solution to Worked example a smaller sheet shown as a ladder of expressions, one row per algebraic move
\[ x \approx 1.96, \quad V \approx 132 \text{ in}^3 \]
Verify: compare the two sheets
Why: The smaller sheet has about half the area, and its best box has about a third of the volume — 132 against 420. Volume falls faster than area because the height shrinks along with the base, which is the same cubic scaling seen in Lesson 5.1's star comparison.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.8 Analyze Graphs of Polynomial Functions §5.8, pp. 389-389
Error analysis
A student maximises the box volume.
Annotate
On: \( V = 4x^3 - 72x^2 + 320x \text{ grows without bound, so make } x \text{ as large as possible} \)
The model extends beyond the situation, as every model does. Write the domain down before graphing and read the maximum only inside it.
Ranking
Maximising a polynomial model.
Put in order
Why: Step three comes before step four because a graph drawn over the wrong interval will show the wrong maximum, or none at all. Step five has two parts, and reporting only the value of x without converting it into dimensions leaves the question half answered.
Prediction
Commit before reasoning.
Predict first
The volume is zero at both ends of the allowed interval. What does that tell you?
Correct: The maximum must occur strictly inside the interval.
\[ V(0) = 0, \; V(8) = 0, \; V(x) > 0 \text{ between} \]
Why: At x equal to zero there is no height and at x equal to 8 there is no width, so the volume is zero at both ends and positive in between. A continuous curve that starts and finishes at zero and is positive in the middle must peak somewhere inside. That guarantees the maximum exists without finding it, which is a useful thing to establish first.
Two truths and a lie
All three are about the box model.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. Volume scales with the cube of a length, so doubling both sheet dimensions multiplies the best volume by 8, not by 2. The comparison of the two sheets in this idea shows the same effect: about half the area gave about a third of the volume rather than half.
Comparison
Fill the blanks. A graph is built from four independent readings.
Comparison matrix
| Feature | Read from | Lesson |
|---|---|---|
| x-intercepts | the real zeros, from the factors | 5.4 to 5.6 |
| Crossing or tangency | the parity of each factor's power | 5.7 |
| End behaviour | the degree and leading coefficient | 5.2 |
| Turning points | the degree, and the distinct zero count | 5.8 |
None of the four requires evaluating the function anywhere. A few points are still worth plotting, but only to fix heights.
Pattern
One routine for sketching any polynomial in factored form.
For a model, restrict the domain to what the situation allows before reading any maximum or minimum from the graph.
OpenStax Algebra and Trigonometry 2e, §5.3 Graphs of Polynomial Functions §5.3
Check
Reading a factored form.
Check your understanding
Where does the graph of f(x) = (1/6)(x + 3)(x - 2)^2 meet the x-axis, and how?
Answer: A
Why: The factor x + 3 is to the first power so the curve crosses; (x - 2)^2 is squared so it is tangent.
Check
Turning points. Which rule applies?
Check your understanding
A quartic has four distinct real zeros. How many turning points does its graph have?
Answer: A
Why: With n distinct real zeros a polynomial of degree n has exactly n - 1 turning points.
Check
A model. Mind the domain.
Check your understanding
For V = 4x^3 - 72x^2 + 320x from a 16 by 20 inch sheet, what is the largest sensible x?
Answer: A
Why: The width 16 - 2x must stay positive, so x < 8; the length gives the weaker condition x < 10.
Real world
An open-topped trough is made from a sheet of metal 12 inches wide by folding up a strip of width x along each long edge. The trough is 60 inches long.
Discussion prompt
Write the capacity as a function of x, state the sensible domain, and find the fold that maximises it. Then say what the graph looks like near its zeros.
Hint: The cross-section is a rectangle x deep and 12 minus 2x wide.
Answer:
\[ V(x) = 60x(12 - 2x) = -120x^2 + 720x, \quad 0 < x < 6 \]
This one is a quadratic, not a cubic, because only one dimension of the cross-section varies with x while the length is fixed. Its vertex is at x equal to 3, giving a capacity of 1080 cubic inches with a cross-section 3 inches deep and 6 inches wide.
Near its zeros, at x equal to 0 and x equal to 6, both factors are to the first power, so the parabola crosses the axis at both — no tangency. And because the leading coefficient is negative the curve has a genuine global maximum, not merely a local one, which is why the answer is unambiguous. Compare that with the box in Example 3, where the cubic's rise to the right had to be excluded by hand.
Commit first
Answer, then rate your confidence honestly.
Predict first
A polynomial's graph is tangent to the x-axis at x equal to 4. What does that tell you about its factorisation?
Correct: The factor x minus 4 appears an even number of times.
\[ (x-4)^2 \text{ or } (x-4)^4: \; \text{no sign change at } 4 \]
Why: A tangency means the function reaches zero without changing sign, and a factor changes sign exactly when it appears an odd number of times. So the multiplicity of 4 must be even — twice, four times, and so on. The degree is unconstrained: a degree-7 polynomial can perfectly well be tangent at 4 if that factor is squared and the rest supplies the other five.
Explain it
They can plot points and have never sketched from a factored form.
Discussion prompt
In four sentences or fewer, explain how to sketch a polynomial without making a long table.
Hint: Three readings and a few points.
Answer:
Read the zeros straight from the factors and mark them on the axis — those are the only places the curve can meet it. Then look at the power on each factor: an even power means the curve touches and turns back, an odd power means it goes through.
Multiply the leading terms of the factors to get the degree and the leading coefficient, which tells you what the two ends do. After that you only need a few points to fix the heights, and the shape is already decided.
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 crossings, look only at whether each power is odd or even. For the degree, add the exponents of the factors' leading terms. For turning points, ask whether the number of distinct real zeros equals the degree — if it does the count is exact, otherwise it is a bound. For domains, write down every quantity that must stay positive and take the tightest condition. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Choose one polynomial of your own in factored form, of degree at least 4, with one repeated factor and at least three distinct real zeros. Top left: list its real zeros with their multiplicities and say for each whether the graph crosses or is tangent. Top right: find the degree and leading coefficient without expanding, and state the end behaviour. Middle: state the maximum number of turning points and, if the exact rule applies, the actual number, then sketch the curve using only what you have written so far. Bottom: evaluate at four values to check the heights of your sketch, correcting it if anything disagrees. In a margin, write down what a graph cannot tell you about the zeros, in one sentence.
If your sketch has more turning points than the degree minus one, something has gone wrong: the bound is a hard limit, not a guideline.
Recap
Five things, and together they turn factored form into a picture.
| If you see | Then |
|---|---|
| A linear factor | The graph crosses there |
| A factor to an even power | The graph is tangent there |
| Factors multiplied | Add exponents for the degree |
| n distinct real zeros, degree n | Exactly n - 1 turning points |
| An odd degree | No global maximum or minimum |
| A model with a folded sheet | Restrict the domain first |
Lesson 5.9 closes the chapter by running everything backwards once more: given points or data, find the polynomial function itself.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.8 Analyze Graphs of Polynomial Functions §5.8, pp. 387-391 — everything on these slides traces back here
Want this taught 1-on-1? Alexander tutors Algebra 2 — $55/session, free consultation.