What counts as a polynomial function and how degree names it, evaluating by direct and by synthetic substitution, reading end behaviour from the degree and the leading coefficient, and graphing polynomial functions including a wave-energy model.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 5 — Polynomials and Polynomial Functions
Evaluate and Graph Polynomial Functions
Objectives
Five outcomes. The fourth is the one that makes a graph sketchable without a calculator.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.2 Evaluate and Graph Polynomial Functions §5.2, pp. 337-345 — the lesson these objectives are drawn from
Warm-up
Chapters 2 and 4 handled degree 1 and degree 2. This chapter removes the ceiling.
Discussion prompt
A linear function has degree 1 and a quadratic has degree 2. Write down what you know about the ends of each graph — what happens far to the left and far to the right — and see whether the two cases have anything in common.
Hint: Think about a line with positive slope, and about a parabola opening upward.
Answer:
A line with positive slope falls to the left and rises to the right: the two ends disagree. An upward parabola rises at both ends: the two ends agree.
That difference is not about being a line or a parabola. It is about the degree being odd or even, and it holds for every degree — which is what makes the ends of any polynomial graph predictable before a single point is plotted.
Concept
A polynomial function is a sum of terms with whole-number exponents. Far from the origin, the term with the highest exponent dominates every other, so the two ends of the graph are settled entirely by the degree and the sign of the leading coefficient.
polynomial function — A function of the form a-sub-n times x to the n, plus lower terms, where the leading coefficient is non-zero, every exponent is a whole number and every coefficient is real. Its degree is n and its constant term is the term with no x.
\[ f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 \]
The middle of the graph still has to be plotted, because that is where the lower-degree terms matter. But knowing the ends first turns a table of five points into a confident sketch.
Figure (svg): Four small graphs showing the end behaviour of polynomial functions by degree parity and sign of the leading coefficient
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.2 Evaluate and Graph Polynomial Functions §5.2, pp. 337-339
Section
Section 1
Concept
A polynomial is a monomial or a sum of monomials, and a polynomial function is a function built from one. Every exponent must be a whole number and every base must be the variable; the coefficients themselves may be any real numbers, rational or not.
\[ f(x) = a_nx^n + \cdots + a_0, \quad a_n \neq 0 \]
Standard form writes the terms in descending order of exponent, which puts the leading coefficient first and makes the degree readable at a glance.
Figure (svg): A table of polynomial function types by degree, from constant through quartic
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.2 Evaluate and Graph Polynomial Functions §5.2, pp. 337-337 — Common Polynomial Functions
Picture it
Example 1: two that qualify and two that do not.
Figure (svg): Four candidate functions sorted into polynomial and not, each with the reason
Pi and the square root of 3 appear as coefficients in a perfectly good polynomial. What disqualifies the other two is an exponent of negative 1 in one and a constant base in the other.
Worked example
Example 1, all four parts.
\[ \text{Which of } x^4 - \tfrac{1}{4}x^2 + 3, \; 7x - \sqrt{3} + \pi x^2, \; 5x^2 + 3x^{-1} - x, \; x + 2^x - 0.6x^5 \text{ qualify?} \]
First: check every exponent
Why: Four and 2 are whole numbers, and the function is already in descending order.
Second: reorder into standard form
Why: Descending order puts the x squared term first, so the leading coefficient is pi.
Third: look for a bad exponent
Why: The term 3x to the negative 1 has an exponent that is not a whole number.
Fourth: look for a bad base
Why: The term 2 to the x has a constant base with a variable exponent, which is not a monomial at all.
Figure (svg): The solution to Worked example identify polynomial functions shown as a ladder of expressions, one row per algebraic move
\[ \text{quartic}, \; \text{quadratic}; \quad \text{neither} \]
Verify: check what is and is not restricted
Why: The second function has an irrational coefficient, pi, and an irrational constant term, negative root 3, and is still a polynomial — coefficients are unrestricted. The third has a perfectly ordinary coefficient of 3 and fails anyway, because the restriction is on the exponent. Knowing which part of a term the rules govern is the whole of this classification.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.2 Evaluate and Graph Polynomial Functions §5.2, pp. 337-337
Sorting
Check every exponent and every base.
Sort into buckets
Sort each function.
The restriction is on exponents and bases only. A coefficient can be as ugly as you like.
Worked example
Guided Practice 1 to 3.
\[ \text{Classify } 13 - 2x, \; 9x^4 - 5x^{-2} + 4, \; 6x^2 + \pi - 3x. \]
First: reorder and describe
Why: Descending order gives negative 2x plus 13.
Second: check the exponents
Why: The term negative 5x to the negative 2 has a negative exponent.
Third: reorder
Why: Descending order gives 6x squared minus 3x plus pi.
Name the constant terms
Why: The first has constant 13 and the third has constant pi.
\[ 13\text{ and } \pi \]
Figure (svg): The solution to Worked example three more, with full descriptions shown as a ladder of expressions, one row per algebraic move
\[ -2x + 13; \quad \text{not a polynomial}; \quad 6x^2 - 3x + \pi \]
Verify: check the leading coefficient after reordering
Why: In the first, reading left to right before reordering would suggest a leading coefficient of 13, which is the constant term. The leading coefficient is always the one on the highest power, so reordering into standard form is not cosmetic — it is what makes the description correct.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.2 Evaluate and Graph Polynomial Functions §5.2, pp. 338-338
Trap
\[ f(x) = 6 + 2x^2 - 5x^4 \]
Read the first term as the leading term
Why: The expression is taken as written.
\[ \text{degree } 0, \text{ leading coefficient } 6 \quad \text{(wrong)} \]
The highest power present is x to the fourth, so the degree is 4 and the leading coefficient is negative 5.
\[ f(x) = -5x^4 + 2x^2 + 6 \]
Write the terms in descending order of exponent first
Why: Standard form is what makes the leading term the first one.
\[ \text{degree } 4 \text{ (quartic)}, \; a = -5, \; \text{constant term } 6 \]
The description matters beyond bookkeeping: the degree and the sign of the leading coefficient are exactly what the end-behaviour rule needs, and reading them off an unordered expression gets both wrong.
Matching
Reorder first, then read the highest power.
Match the pairs
Why: Two of these arrive out of standard form and must be reordered before the degree is read. Note that a constant function has degree 0 rather than no degree, because a constant is x to the zero times that constant.
Fill the middle
Exercise 1.
Fill in the blanks
f(x) = 6 + 2x^2 - 5x^4 = -5x^4 + 2x^2 + 6
Why: Descending order puts x to the fourth first, x squared next and the constant last, so the function is negative 5x to the fourth plus 2x squared plus 6. The degree is 4, the leading coefficient negative 5 and the constant term 6 — all three now readable in one glance.
Prediction
Commit before reasoning.
Predict first
What goes wrong with a term like 3x to the negative 1 that does not go wrong with 3x cubed?
Correct: It is undefined at x equal to zero, so the function has a gap.
\[ 3x^{-1} = \tfrac{3}{x}: \; \text{undefined at } x = 0 \]
Why: Three x to the negative 1 is 3 over x, which has no value at x equal to zero, so the graph breaks there. Every polynomial function, by contrast, is defined for every real number and its graph is one unbroken smooth curve — which is precisely the property that end behaviour and the later theorems of this chapter rely on. The restriction is what buys the good behaviour.
Section
Section 2
Concept
To evaluate a polynomial function at a value, substitute the value for the variable everywhere and follow the order of operations: evaluate the powers, then multiply by the coefficients, then add.
\[ f(3) = 2(3)^4 - 5(3)^3 - 4(3) + 8 = 23 \]
The powers are where the work is. A quartic needs x squared, x cubed and x to the fourth computed separately, which is why the alternative in the next idea is worth learning.
Figure (svg): Two columns comparing direct substitution with synthetic substitution
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.2 Evaluate and Graph Polynomial Functions §5.2, pp. 338-338 — Evaluate by direct substitution
Picture it
What each method costs.
Figure (svg): Two columns comparing direct substitution with synthetic substitution
Direct substitution needs no setup but many operations; synthetic substitution needs a row of coefficients but very few. For a quartic the difference is already noticeable.
Worked example
Example 2. Brackets around the value, every time.
\[ \text{Evaluate } f(x) = 2x^4 - 5x^3 - 4x + 8 \text{ at } x = 3. \]
Substitute 3 for every x
Why: Brackets keep the substituted value together with its exponent.
\[ 2(3) ^{4} - 5(3) ^{3} - 4(3) + 8 \]
Evaluate the powers
Why: Three to the fourth is 81 and 3 cubed is 27.
\[ 2(81) - 5(27) - 12 + 8 \]
Multiply by the coefficients
Why: Two times 81 is 162 and 5 times 27 is 135.
\[ 162 - 135 - 12 + 8 \]
Add and subtract left to right
Why: One hundred sixty-two minus 135 is 27; minus 12 is 15; plus 8 is 23.
\[ 23 \]
Figure (svg): The solution to Worked example evaluate by direct substitution shown as a ladder of expressions, one row per algebraic move
\[ f(3) = 23 \]
Verify: count the operations
Why: Two powers, two multiplications by coefficients, one further multiplication and three additions — about eight operations for a quartic, and it grows with the degree. The next idea does the same job in four multiplications and four additions, and the gap widens for higher degrees.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.2 Evaluate and Graph Polynomial Functions §5.2, pp. 338-338
Fill the middle
Example 2, at the second step.
Fill in the blanks
f(3) = 2(3)^4 - 5(3)^3 - 4(3) + 8 = 2(81) - 5(27) - 12 + 8
Why: Three to the fourth is 81, since 9 squared is 81. Evaluating the powers before multiplying by the coefficients is the order of operations, and doing it in the other order — multiplying 2 by 3 first — would give 6 to the fourth, which is 1296 rather than 162.
Worked example
Guided Practice 4 and 5. Watch the negative value.
\[ \text{Evaluate } x^4 + 2x^3 + 3x^2 - 7 \text{ at } x = -2, \text{ and } x^3 - 5x^2 + 6x + 1 \text{ at } x = 4. \]
First: evaluate the powers of negative 2
Why: Negative 2 to the fourth is 16, negative 2 cubed is negative 8, and negative 2 squared is 4.
\[ 16, -8, 4 \]
First: combine
Why: Sixteen plus 2 times negative 8 is 16 minus 16, plus 3 times 4 is 12, minus 7.
\[ 5 \]
Second: evaluate the powers of 4
Why: Four cubed is 64 and 4 squared is 16.
\[ 64, 16 \]
Second: combine
Why: Sixty-four minus 5 times 16, or 80, plus 6 times 4, or 24, plus 1.
\[ 9 \]
Figure (svg): The solution to Worked example two more evaluations shown as a ladder of expressions, one row per algebraic move
\[ f(-2) = 5, \qquad g(4) = 9 \]
Verify: check the signs of the powers
Why: For the first, the even powers of negative 2 came out positive and the odd power negative, which is the pattern from Lesson 5.1. If both had come out positive the answer would have been 37 instead of 5, so keeping the brackets on the negative value is not optional.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.2 Evaluate and Graph Polynomial Functions §5.2, pp. 338-338
Error analysis
A student evaluates a quartic at negative 2.
Annotate
On: \( f(-2) = -2^4 + 2(-2)^3 + 3(-2)^2 - 7 = -16 - 16 + 12 - 7 = -27 \)
Substitute with brackets every time, even when it looks unnecessary. The habit costs nothing and removes the error entirely.
Sorting
Look at the coefficient's sign and the parity of the exponent.
Sort into buckets
For x equal to negative 2, sort each term by its value's sign.
Two signs multiply, so the term is positive when they agree. Working this out before computing catches sign errors before they happen.
Ranking
Evaluating by direct substitution.
Put in order
Why: This is the order of operations applied to a formula, exactly as in Lesson 1.2. Multiplying before evaluating the power is the classic slip and it changes the answer by an enormous amount, since it effectively raises the coefficient to the power as well.
Prediction
Commit before reasoning.
Predict first
Direct substitution on a quartic needs about eight operations. Roughly how many for a polynomial of degree 10?
Correct: Many more — the powers alone take about ten multiplications.
\[ \text{direct: about } 2n \text{ operations} \qquad \text{synthetic: about } n \]
Why: Computing x squared through x to the tenth takes nine multiplications even when each is built from the last, and then each is multiplied by its coefficient and added. The count grows roughly with twice the degree. Synthetic substitution needs one multiplication and one addition per coefficient, which is about the degree itself — half the work, and simpler work, which is why it is worth the setup.
Section
Section 3
Concept
Write the coefficients in descending order with the input value to the left. Bring down the leading coefficient, then repeatedly multiply the last number written by the input and add it to the next coefficient. The final sum is the value of the function.
synthetic substitution — A method of evaluating a polynomial function using only its coefficients, with one multiplication and one addition per coefficient. Every missing power must be represented by a coefficient of zero.
\[ f(x) = 2x^4 - 5x^3 + 0x^2 - 4x + 8 \]
The zero for a missing term is the step people forget. Without it the coefficients line up against the wrong powers and every number after that point is wrong.
Figure (svg): The synthetic substitution tableau for a quartic evaluated at three, with the missing coefficient shown
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.2 Evaluate and Graph Polynomial Functions §5.2, pp. 338-338 — Evaluate by synthetic substitution
Picture it
Example 3: the same quartic at x equal to 3.
Figure (svg): The synthetic substitution tableau for a quartic evaluated at three, with the missing coefficient shown
The bottom row ends at 23, matching direct substitution. The other numbers in that row are not waste — Lesson 5.5 shows they are the coefficients of a quotient.
Worked example
Example 3, in the book's three steps.
\[ \text{Use synthetic substitution to find } f(3) \text{ for } f(x) = 2x^4 - 5x^3 - 4x + 8. \]
Write the coefficients, including zeros
Why: The x squared term is missing, so a 0 goes in its place: 2, negative 5, 0, negative 4, 8.
\[ 2, -5, 0, -4, 8 \]
Bring down the leading coefficient and multiply
Why: Bring down 2; 2 times 3 is 6, written under negative 5; adding gives 1.
\[ 2,\text{ then } 1 \]
Repeat across the row
Why: One times 3 is 3, under 0, giving 3; 3 times 3 is 9, under negative 4, giving 5.
\[ 3,\text{ then } 5 \]
Finish
Why: Five times 3 is 15, under 8, giving 23.
\[ 23 \]
Figure (svg): The solution to Worked example evaluate synthetically shown as a ladder of expressions, one row per algebraic move
\[ f(3) = 23 \]
Verify: compare with Example 2
Why: Direct substitution gave 23 as well, so the two methods agree — as they must, since synthetic substitution is just a compact way of organising the same arithmetic. The check is worth doing the first few times, until the tableau stops feeling like magic.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.2 Evaluate and Graph Polynomial Functions §5.2, pp. 338-338
Fill the middle
Example 3, at the setup.
Fill in the blanks
2x^4 - 5x^3 - 4x + 8 \;\Longrightarrow\; 2, \; -5, \; 0, \; -4, \; 8
Why: The x squared term is absent, so its coefficient is zero and a 0 must be written to hold its place. A degree-4 polynomial has five coefficients; counting them before starting the tableau catches this instantly.
Worked example
Guided Practice 6 and 7. The second has a negative input and a missing term.
\[ \text{Find } f(2) \text{ for } 5x^3 + 3x^2 - x + 7, \text{ and } g(-1) \text{ for } -2x^4 - x^3 + 4x - 5. \]
First: write the coefficients
Why: Nothing is missing: 5, 3, negative 1, 7.
\[ 5, 3, -1, 7 \]
First: run the row
Why: Bring down 5; 10 under 3 gives 13; 26 under negative 1 gives 25; 50 under 7 gives 57.
\[ f(2) = 57 \]
Second: insert the zero
Why: The x squared term is missing: negative 2, negative 1, 0, 4, negative 5.
\[ -2, -1, 0, 4, -5 \]
Second: run the row with negative 1
Why: Bring down negative 2; 2 under negative 1 gives 1; negative 1 under 0 gives negative 1; 1 under 4 gives 5; negative 5 under negative 5 gives negative 10.
\[ g(-1) = -10 \]
Figure (svg): The solution to Worked example two more tableaux shown as a ladder of expressions, one row per algebraic move
\[ f(2) = 57, \qquad g(-1) = -10 \]
Verify: check the second by direct substitution
Why: Negative 2 times 1, minus negative 1, plus 4 times negative 1, minus 5 is negative 2 plus 1 minus 4 minus 5, which is negative 10. The tableau agrees. Note that the multiplier was negative throughout, so the signs alternated down the middle row — normal, and not a signal of an error.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.2 Evaluate and Graph Polynomial Functions §5.2, pp. 339-339
Trap
\[ f(x) = 2x^4 - 5x^3 - 4x + 8 \]
Write only the coefficients that appear
Why: The row becomes 2, negative 5, negative 4, 8.
\[ \text{tableau on } 2, -5, -4, 8 \;\Longrightarrow\; f(3) = 5 \quad \text{(wrong)} \]
Four coefficients describe a cubic, not a quartic. The negative 4 has been treated as the coefficient of x squared.
\[ f(x) = 2x^4 - 5x^3 + 0x^2 - 4x + 8 \]
Write a zero for every power that does not appear
Why: The row must have one entry per power from the degree down to the constant.
\[ \text{five coefficients for degree } 4 \;\Longrightarrow\; f(3) = 23 \]
A quick count settles it: a polynomial of degree n has exactly n plus 1 coefficients. The book flags this in an Avoid Errors note, and counting the entries before starting takes two seconds.
Ranking
One pass across the row.
Put in order
Why: Every step after the third is the same step repeated, which is what makes the method mechanical enough to do quickly and reliably. The setup in steps one and two is where all the thinking is, and where the missing-zero error lives.
Comparison
Fill the blanks. Same answer, different cost.
Comparison matrix
| Question | Direct | Synthetic |
|---|---|---|
| What you write down | the whole function with 3 substituted | just the coefficients, plus zeros |
| Powers computed | 3^4 and 3^3 separately | none |
| Result at x = 3 | 23 | 23 |
| Extra output | none | a row of numbers used in Lesson 5.5 |
The last row is why synthetic substitution is taught here rather than presented as a curiosity: the same tableau divides polynomials three lessons from now.
Prediction
Commit before reasoning.
Predict first
The bottom row reads 2, 1, 3, 5, 23. The last is f of 3. What might the first four be?
Correct: The coefficients of a quotient, with 23 as a remainder.
\[ 2x^4 - 5x^3 - 4x + 8 = (x-3)(2x^3 + x^2 + 3x + 5) + 23 \]
Why: Dividing the quartic by x minus 3 gives 2x cubed plus x squared plus 3x plus 5, remainder 23 — exactly the bottom row. That is the remainder theorem, and it is the subject of Lesson 5.5. Noticing it now makes that lesson feel like a consequence rather than a new technique, which is why the tableau is introduced here.
Section
Section 4
Concept
As x grows large in either direction, the leading term overwhelms every other, so only the degree and the sign of the leading coefficient matter. An odd degree makes the two ends disagree; an even degree makes them agree; and a negative leading coefficient flips both.
end behaviour — What a function's graph does as x approaches positive infinity and as x approaches negative infinity. For a polynomial function it is determined entirely by the degree and the sign of the leading coefficient.
\[ f(x) \to +\infty \text{ or } -\infty \text{ as } x \to \pm\infty \]
The rule reduces to two facts about the leading term because the ratio of any lower term to the leading term shrinks to zero as x grows, so far enough out only the leading term is visible.
Figure (svg): Four small graphs showing the end behaviour of polynomial functions by degree parity and sign of the leading coefficient
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.2 Evaluate and Graph Polynomial Functions §5.2, pp. 339-339 — End Behavior of Polynomial Functions
Picture it
Odd and even degree, positive and negative leading coefficient.
Figure (svg): Four small graphs showing the end behaviour of polynomial functions by degree parity and sign of the leading coefficient
The middles differ from one polynomial to the next; the ends do not. Any odd-degree polynomial with a positive leading coefficient has the first picture's ends, whatever happens in between.
Worked example
Example 4, run in reverse.
\[ \text{A graph falls to the left and rises to the right. What is true of its degree and leading coefficient?} \]
Describe the left end
Why: As x approaches negative infinity, the function approaches negative infinity.
Describe the right end
Why: As x approaches positive infinity, the function approaches positive infinity.
Decide the parity
Why: The two ends disagree, so the degree is odd.
Decide the sign
Why: For an odd degree, rising on the right means a positive leading coefficient.
\[ a > 0 \]
Figure (svg): The solution to Worked example read the degree from a graph shown as a ladder of expressions, one row per algebraic move
\[ \text{degree odd}, \quad a_n > 0 \]
Verify: test it against a function you know
Why: The simplest odd-degree function with a positive leading coefficient is y equals x, which falls to the left and rises to the right. So does y equals x cubed, and so does any cubic with a positive leading coefficient however much it wiggles in the middle. The rule does not care about the middle at all.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.2 Evaluate and Graph Polynomial Functions §5.2, pp. 339-339
Sorting
Read the degree's parity and the leading coefficient's sign.
Sort into buckets
Sort each function by its end behaviour.
Parity settles agreement; the sign settles direction. Two independent readings, exactly as with a and h in Lesson 4.2.
Worked example
Example 5's two functions, before graphing either.
\[ \text{State the end behaviour of } -x^3 + x^2 + 3x - 3 \text{ and } x^4 - x^3 - 4x^2 + 4. \]
First: read the leading term
Why: The degree is 3, which is odd, and the leading coefficient is negative 1.
\[ o d d, a < 0 \]
First: state both ends
Why: Odd degree means the ends disagree, and a negative coefficient means it rises on the left and falls on the right.
Second: read the leading term
Why: The degree is 4, which is even, and the leading coefficient is 1.
Second: state both ends
Why: Even degree means the ends agree, and a positive coefficient sends both upward.
Figure (svg): The solution to Worked example state the end behaviour of two functions shown as a ladder of expressions, one row per algebraic move
\[ f(x)\to+\infty, -\infty; \qquad f(x)\to+\infty, +\infty \]
Verify: check with a large value
Why: For the cubic at x equal to 10, the leading term is negative 1000 while the other terms total 127, so the value is deeply negative — falling on the right, as predicted. The lower terms are not negligible near the origin, but by x equal to 10 the leading term is already eight times everything else combined.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.2 Evaluate and Graph Polynomial Functions §5.2, pp. 340-340
Error analysis
A student describes the end behaviour of a cubic.
Annotate
On: \( f(x) = -x^3 + x^2 + 3x - 3 \text{ has four terms, an even number, so its ends agree} \)
Only the leading term matters. The number of terms, and everything about the lower ones, affects the middle of the graph and nothing else.
Matching
Four cases, four shapes.
Match the pairs
Why: The two odd cases are reflections of each other, as are the two even cases. A useful anchor: degree 1 with positive slope is the first case, and an upward parabola is the third — both already familiar from Chapters 2 and 4.
Fill the middle
Example 5a.
Fill in the blanks
f(x) = -x^3 + x^2 + 3x - 3: \; f(x) \to +infinity \text___ x \to -\infty
Why: For large negative x, x cubed is a large negative number and the leading coefficient of negative 1 flips it positive, so the function rises without bound on the left. The lower terms cannot compete: at x equal to negative 10 the leading term alone is 1000 while everything else totals 67.
Prediction
Commit before reasoning.
Predict first
Why do the lower-degree terms stop mattering far from the origin?
Correct: Their ratio to the leading term shrinks toward zero as x grows.
\[ \frac{3x}{x^3} = \frac{3}{x^2} \to 0 \text{ as } x \to \infty \]
Why: Compare 3x with x cubed: the ratio is 3 over x squared, which shrinks as x grows. The lower terms do not become zero — at x equal to 100 the term 3x is still 300 — but the leading term is a million, so the graph's shape is set by the leading term alone. Near the origin the reverse can hold, which is exactly why the middle of the graph still has to be plotted.
Section
Section 5
Concept
To graph a polynomial function, make a table of values around the origin to fix the middle of the curve, plot the points, join them smoothly, and then use the end-behaviour rule to draw the two ends.
\[ f(x) = -x^3 + x^2 + 3x - 3 \]
Five to seven integer inputs is usually enough. A polynomial of degree n turns at most n minus 1 times, so a cubic has at most two turning points and a quartic at most three.
Figure (svg): A cubic and a quartic plotted from tables of values, with their end behaviour marked
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.2 Evaluate and Graph Polynomial Functions §5.2, pp. 340-340 — Graph polynomial functions
Picture it
Example 5, both parts, with their tabulated points marked.
Figure (svg): A cubic and a quartic plotted from tables of values, with their end behaviour marked
The cubic turns twice and the quartic turns three times, which is the most each degree allows. The plotted points caught every turn, which is why the tables were long enough.
Worked example
Example 5. Table, plot, join, then the ends.
\[ \text{Graph } f(x) = -x^3 + x^2 + 3x - 3 \text{ and } f(x) = x^4 - x^3 - 4x^2 + 4. \]
Tabulate the first from negative 3 to 3
Why: The values are negative 24, 3, negative 4, negative 3, 0, negative 1, negative 12.
Plot and join smoothly
Why: The curve rises to a peak near negative 1.5, dips, and rises again to a smaller peak near 1.
Add the ends
Why: Odd degree with a negative leading coefficient: up on the left, down on the right.
Repeat for the quartic
Why: The values from negative 3 to 3 are 76, 12, 2, 4, 0, negative 4, negative 2, and the ends both go up.
Figure (svg): The solution to Worked example graph two polynomial functions shown as a ladder of expressions, one row per algebraic move
\[ \text{a cubic and a quartic} \]
Verify: count the turning points
Why: The cubic turns twice and the quartic three times, which is the maximum each degree permits. If a sketch showed a cubic turning four times, the plotted points would have to be wrong — the degree caps the number of turns, and that cap is a free check on every graph you draw.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.2 Evaluate and Graph Polynomial Functions §5.2, pp. 340-340
Fill the middle
Example 6, at 20 knots.
Fill in the blanks
E = 0.0029(20)^4 = 0.0029 \times 160464000 = ___
Why: Twenty to the fourth is 160,000, and 0.0029 times that is 464 foot-pounds. The tiny coefficient and the huge power are typical of physical models: the coefficient sets the units and the exponent sets how fast the quantity grows.
Worked example
Example 6. Only positive inputs make sense.
\[ \text{Graph } E = 0.0029s^4 \text{ and estimate the wind speed giving } E = 1000. \]
Tabulate over meaningful inputs
Why: Wind speed is non-negative, so tabulate at 0, 10, 20, 30 and 40 knots, giving 0, 29, 464, 2349 and 7424.
Plot and join
Why: The leading coefficient is positive and the degree even, so the curve rises steeply to the right.
Find where the curve reaches 1000
Why: Draw the horizontal line at 1000 and read where it meets the curve.
\[ \text{between } 20\text{ and } 30 \]
Estimate
Why: The crossing is at about 24 knots.
\[ \text{about } 24\text{ knots} \]
Figure (svg): The solution to Worked example the wave energy model shown as a ladder of expressions, one row per algebraic move
\[ s \approx 24 \text{ knots} \]
Verify: check algebraically
Why: Setting 0.0029 s to the fourth equal to 1000 gives s to the fourth about 344,828, whose fourth root is about 24.2 — matching the graphical estimate. Note how steep the growth is: doubling the wind speed from 20 to 40 knots multiplies the energy by 16, because energy depends on the fourth power.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.2 Evaluate and Graph Polynomial Functions §5.2, pp. 340-340
Error analysis
A student plots seven points of a cubic and connects them.
Annotate
On: \( \text{plot } (-3,-24), (-2,3), (-1,-4), (0,-3), (1,0), (2,-1), (3,-12) \text{ and join with segments} \)
The plotted points are a scaffold, not the graph. Draw through them smoothly, and expect the turns to fall between them rather than at them.
Ranking
Sketching a polynomial function by hand.
Put in order
Why: Reading the leading term first tells you what the finished sketch should look like, so a table that disagrees with it signals an arithmetic error before any drawing happens. The ends go last because they are the part the table cannot show.
Two truths and a lie
All three are about graphs of polynomial functions.
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. A quartic has AT MOST four x-intercepts and may have fewer — Example 5b crosses the axis only twice, and a quartic such as x to the fourth plus 1 never crosses at all. The degree caps the number of intercepts and of turning points without guaranteeing either, which is the same distinction as the discriminant in Lesson 4.8.
Prediction
Commit before reasoning.
Predict first
The wave model was tabulated up to 40 knots. What does it give at 100, and should it be trusted?
Correct: About 290,000, but the model was never tested that far out.
\[ 0.0029(100)^4 = 290{,}000 \text{ ft-lb, far outside the tabulated range} \]
Why: Substituting gives 0.0029 times 100 million, or 290,000 foot-pounds — an enormous figure produced by extrapolating a fourth-power model well beyond its data. It may be right, but nothing in the lesson supports it, and a fourth power amplifies any inaccuracy in the coefficient dramatically. The same caution applied to the regression model of Lesson 4.10.
Comparison
Fill the blanks. Everything is capped by the degree.
Comparison matrix
| Degree | Type | At most this many x-intercepts / turns |
|---|---|---|
| 1 | linear | 1 intercept, 0 turns |
| 2 | quadratic | 2 intercepts, 1 turn |
| 3 | cubic | 3 intercepts, 2 turns |
| 4 | quartic | 4 intercepts, 3 turns |
At most, never exactly. A quartic may cross the axis four times, twice or not at all, and the caps are what make a sketch checkable.
Pattern
One routine for meeting a new polynomial function.
For a model, tabulate only over inputs the situation allows, and do not extend the curve beyond the range the data support.
OpenStax Algebra and Trigonometry 2e, §5.2 Power Functions and Polynomial Functions §5.2
Check
Classification. Reorder first.
Check your understanding
Describe f(x) = 6 + 2x^2 - 5x^4.
Answer: A
Why: Standard form is -5x^4 + 2x^2 + 6, so the degree is 4 and the leading coefficient is -5.
Check
Synthetic substitution. Mind the missing term.
Check your understanding
Use synthetic substitution to find f(3) for f(x) = 2x^4 - 5x^3 - 4x + 8.
Answer: A
Why: With coefficients 2, -5, 0, -4, 8 the bottom row is 2, 1, 3, 5, 23.
Check
End behaviour. Parity, then sign.
Check your understanding
What is the end behaviour of f(x) = -x^3 + x^2 + 3x - 3?
Answer: A
Why: The degree is odd so the ends disagree, and the negative leading coefficient makes it rise on the left.
Real world
An open box is made from a 20 by 30 centimetre sheet of card by cutting a square of side x from each corner and folding up the sides.
Discussion prompt
Write the volume as a polynomial function of x, state its degree and end behaviour, say which values of x make sense, and estimate the x giving the largest volume.
Hint: The base measures 20 minus 2x by 30 minus 2x, and the height is x.
Answer:
\[ V(x) = x(20-2x)(30-2x) = 4x^3 - 100x^2 + 600x \]
It is a cubic with a positive leading coefficient, so it falls to the left and rises to the right. But only 0 less than x less than 10 makes sense, because a cut larger than 10 would remove more than the shorter side.
\[ V(3) = 1008, \; V(4) = 1056, \; V(5) = 1000 \;\Longrightarrow\; \text{maximum near } x \approx 3.9 \]
Two things are worth noticing. The end behaviour is real but physically meaningless: the model rising to infinity on the right describes cuts that cannot be made. And the maximum falls between the tabulated integers, exactly as the turning points did in Example 5, which is why a table locates a turn without pinning it down.
Commit first
Answer, then rate your confidence honestly.
Predict first
Can a polynomial function of degree 5 have no x-intercepts at all?
Correct: No — an odd degree forces at least one crossing.
\[ \text{odd degree} \;\Longrightarrow\; \text{at least one real zero} \]
Why: An odd-degree polynomial has ends that disagree, so it takes both very large positive and very large negative values, and being an unbroken curve it must pass through zero somewhere in between. Every odd-degree polynomial therefore has at least one real zero. Even degrees carry no such guarantee: x to the fourth plus 1 never reaches zero. This is why Lesson 5.7's fundamental theorem counts complex roots rather than real ones.
Explain it
They can graph lines and parabolas and have just been shown a cubic.
Discussion prompt
In four sentences or fewer, explain how to know what the two ends of a polynomial graph do without plotting a single point.
Hint: Only one term matters out there.
Answer:
Far from the origin the term with the highest power is so much larger than all the others that they stop mattering, so the ends of the graph behave like that one term. If the highest power is odd the two ends go opposite ways, like a line; if it is even they go the same way, like a parabola.
Then the sign of that term's coefficient says which way: positive means the right end goes up, negative means it goes down. The middle still has to be plotted, because near the origin the smaller terms are the ones doing the work.
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 standard form, reorder before reading anything at all. For the missing zero, count that a degree-n polynomial has n plus 1 coefficients before starting the tableau. For end behaviour, anchor on two graphs you already know — a line for odd and a parabola for even — and flip for a negative coefficient. For smooth curves, remember that turning points fall between plotted values, not at them. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Take the single function f of x equals 2x to the fourth minus 5x cubed minus 4x plus 8 and work it four ways. Top left: write it in standard form, mark the degree, type, leading coefficient and constant term, and write out its full list of coefficients including the zero. Top right: evaluate it at x equal to 3 twice, once by direct substitution and once with a synthetic tableau, and count the operations each took. Bottom left: state its end behaviour and sketch just the two ends, leaving the middle blank. Bottom right: tabulate it from x equal to negative 2 to 3 and fill that middle in, joining smoothly. In a margin, write the maximum number of turning points and x-intercepts a quartic can have, and how many yours actually has.
If your sketch shows four turning points, recheck the table: a quartic can turn at most three times, and the cap is a hard limit rather than a rule of thumb.
Recap
Five things, and the third and fourth are what make a graph sketchable by hand.
| If you see | Then |
|---|---|
| A negative or fractional exponent | Not a polynomial function |
| A constant base with a variable exponent | Not a polynomial function |
| Terms out of order | Reorder before reading the degree |
| A missing power in a tableau | Write a zero for it |
| An odd degree | The two ends disagree |
| A negative leading coefficient | Both ends flip |
| Degree n | At most n intercepts and n - 1 turns |
Lesson 5.3 returns to arithmetic: adding, subtracting and multiplying polynomials, which is what the factoring of Lesson 5.4 has to undo.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.2 Evaluate and Graph Polynomial Functions §5.2, pp. 337-345 — everything on these slides traces back here
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