Writing a cubic from its graph and a fourth point, computing finite differences, the two properties linking constant nth-order differences to degree n, building a function from a system of four equations, and polynomial regression on measured data.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 5 — Polynomials and Polynomial Functions
Write Polynomial Functions and Models
Objectives
Five outcomes. The chapter ends where Lesson 4.10 did, one degree higher.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.9 Write Polynomial Functions and Models §5.9, pp. 393-399 — the lesson these objectives are drawn from
Warm-up
Lesson 4.10 built a quadratic from three points. The pattern continues.
Discussion prompt
Two points determine a line and three determine a parabola. How many points should determine a cubic, and why?
Hint: Count the unknown coefficients.
Answer:
\[ f(x) = ax^3 + bx^2 + cx + d: \; \text{four unknowns} \]
Four. Each point gives one equation, and four unknowns need four equations. The pattern is exact: a polynomial of degree n has n plus 1 coefficients, so n plus 1 points determine it — which is why a line needs two and a parabola three.
Concept
If equally spaced data come from a polynomial of degree n, then the nth-order differences of the values are constant and non-zero. That runs both ways, so computing differences until they go constant reveals the degree, and the coefficients then follow from a system of equations.
finite differences — The differences of consecutive values of a function evaluated at equally spaced inputs. First-order differences are differences of the values; second-order differences are differences of those, and so on.
\[ \deg f = n \;\Longleftrightarrow\; n\text{th-order differences constant and non-zero} \]
The equally spaced condition is essential. Differences of values at unevenly spaced inputs say nothing at all about degree.
Figure (svg): Three difference triangles showing constant differences at the first, second and third order
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.9 Write Polynomial Functions and Models §5.9, pp. 393-394
Section
Section 1
Concept
If a cubic's three x-intercepts are visible, write it in factored form with an unknown leading coefficient. Substituting any fourth point on the curve gives one equation in that coefficient.
\[ f(x) = a(x-p)(x-q)(x-r) \]
This is Lesson 4.10's intercept-form method one degree higher. The only change is that three factors are needed instead of two, so three intercepts must be given rather than two.
Figure (svg): A cubic graph with three x-intercepts and a fourth point used to find the leading coefficient
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.9 Write Polynomial Functions and Models §5.9, pp. 393-393 — Write a cubic function
Picture it
Example 1: intercepts at negative 4, 1 and 3, through (0, negative 6).
Figure (svg): A cubic graph with three x-intercepts and a fourth point used to find the leading coefficient
The fourth point gave a equal to negative one half, and the negative sign matches what the graph shows: odd degree falling to the right.
Worked example
Example 1. Factors first, then the coefficient.
\[ \text{Write the cubic with x-intercepts } -4, 1, 3 \text{ passing through } (0, -6). \]
Write the factored form
Why: Each intercept k gives a factor x minus k, by the factor theorem.
\[ f(x) = a(x + 4) (x - 1) (x - 3) \]
Substitute the fourth point
Why: At x equal to 0 the three brackets are 4, negative 1 and negative 3.
\[ -6 = a(4) (-1) (-3) \]
Solve for a
Why: The product of the brackets is 12, so negative 6 equals 12a.
\[ a = -\frac{1}{2} \]
Write the function
Why: Substituting a completes it.
\[ f(x) = -(\frac{1}{2}) (x + 4) (x - 1) (x - 3) \]
Figure (svg): The solution to Worked example write the cubic shown as a ladder of expressions, one row per algebraic move
\[ f(x) = -\tfrac{1}{2}(x+4)(x-1)(x-3) \]
Verify: check the end behaviour against the picture
Why: Three linear factors give degree 3, which is odd, and a is negative, so the curve rises on the left and falls on the right. The graph does exactly that. Checking the end behaviour catches a sign error in a instantly, and it costs nothing.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.9 Write Polynomial Functions and Models §5.9, pp. 393-393
Fill the middle
Example 1, at the substitution.
Fill in the blanks
-6 = a(0+4)(0-1)(0-3) = 12a \;\Longrightarrow\; a = -\tfrac______
Why: Four times negative 1 is negative 4, times negative 3 is positive 12. Two negatives cancelled, which is why a came out negative rather than positive — losing one of them would have given a equal to positive one half and a graph pointing the wrong way.
Worked example
Guided Practice 1 and 2.
\[ \text{Write cubics through } (-4,0),(0,10),(2,0),(5,0) \text{ and through } (-1,0),(0,-12),(2,0),(3,0). \]
First: identify the three intercepts
Why: The points with y equal to zero are at negative 4, 2 and 5.
\[ f(x) = a(x + 4) (x - 2) (x - 5) \]
First: use the fourth point
Why: At x equal to 0 the brackets are 4, negative 2 and negative 5, whose product is 40.
\[ 10 = 40 a, a = \frac{1}{4} \]
Second: identify the intercepts
Why: The zeros are negative 1, 2 and 3.
\[ f(x) = a(x + 1) (x - 2) (x - 3) \]
Second: use the fourth point
Why: At x equal to 0 the brackets are 1, negative 2 and negative 3, whose product is 6.
\[ -12 = 6 a, a = -2 \]
Figure (svg): The solution to Worked example two more cubics shown as a ladder of expressions, one row per algebraic move
\[ \tfrac{1}{4}(x+4)(x-2)(x-5); \quad -2(x+1)(x-2)(x-3) \]
Verify: check the y-intercept of each
Why: Substituting zero into the first gives a quarter of 40, which is 10 — matching the given point. For the second, negative 2 times 6 is negative 12. The y-intercept is always the easiest point to check because all three brackets reduce to constants.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.9 Write Polynomial Functions and Models §5.9, pp. 394-394
Trap
\[ \text{intercepts } -4, 1, 3 \]
Write a factor for each, copying the number
Why: The factors are taken as x minus 4, x minus 1 and x minus 3.
\[ a(x-4)(x-1)(x-3) \quad \text{(wrong)} \]
This function has intercepts at 4, 1 and 3, not at negative 4. Substituting negative 4 gives a times negative 8 times negative 5 times negative 7, which is not zero.
\[ a(x+4)(x-1)(x-3) \]
Write the factor that vanishes at each intercept
Why: An intercept at negative 4 needs a factor that is zero there, which is x plus 4.
\[ \text{at } x=-4: \; (-4+4) = 0 \;\Longrightarrow\; f(-4) = 0 \]
Same sign trap as intercept form in Lesson 4.2 and the divider value in Lesson 5.5. Asking what makes the bracket zero avoids it every time.
Matching
The zero-valued points give the factors.
Match the pairs
Why: The last has an intercept at the origin, so one factor is simply x and the fourth point cannot be the y-intercept — it has to be somewhere else, here at x equal to 1. Choosing the fourth point away from the intercepts is essential, since an intercept would give 0 equals 0.
Prediction
Commit before reasoning.
Predict first
Two points determine a line and four a cubic. How many determine a quartic?
Correct: Five, one for each of its five coefficients.
\[ \deg n \;\Longrightarrow\; n+1 \text{ coefficients} \;\Longrightarrow\; n+1 \text{ points} \]
Why: A quartic is a x to the fourth plus bx cubed plus cx squared plus dx plus e, with five unknowns, so five points give five equations. The rule is n plus 1 points for degree n, and it is the reason a line needs two: a linear function has two coefficients. Counting coefficients rather than memorising cases makes every case obvious.
Comparison
Fill the blanks. Lesson 4.10 against this lesson.
Comparison matrix
| Question | Quadratic, Lesson 4.10 | Cubic, here |
|---|---|---|
| Intercept form | a(x - p)(x - q) | a(x - p)(x - q)(x - r) |
| Intercepts needed | two | three |
| Extra points needed | one | one |
| Total points | three | four |
Only the number of factors changes. The method — write the factors, substitute one more point, solve for a — is identical at every degree.
Section
Section 2
Concept
For equally spaced inputs, the first-order differences are the differences of consecutive function values. The second-order differences are the differences of those, and so on. Higher-order differences are computed by repeating one subtraction step.
\[ f(1), f(2), f(3), \dots \;\to\; \text{first differences} \;\to\; \text{second differences} \]
The inputs must be equally spaced. Differences taken at 1, 2, 5 and 9 mean nothing, because a larger gap naturally produces a larger difference.
Figure (svg): A difference triangle for the triangular numbers, with constant second-order differences
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.9 Write Polynomial Functions and Models §5.9, pp. 394-394 — Find finite differences
Picture it
Example 2: the triangular numbers 1, 3, 6, 10, 15, 21, 28.
Figure (svg): A difference triangle for the triangular numbers, with constant second-order differences
The first differences are the counting numbers and the second differences are all 1. Since the formula is quadratic, its second-order differences going constant is exactly what the property predicts.
Worked example
Example 2. Two rows of subtraction.
\[ \text{Show that } f(n) = \tfrac{1}{2}(n^2 + n) \text{ has constant second-order differences.} \]
Write out several values
Why: For n from 1 to 7 the values are 1, 3, 6, 10, 15, 21 and 28.
Subtract consecutive values
Why: The first-order differences are 2, 3, 4, 5, 6 and 7.
Subtract again
Why: The second-order differences are 1, 1, 1, 1 and 1.
State the conclusion
Why: They are constant and non-zero, which matches the degree of 2.
\[ \text{constant at order } 2 \]
Figure (svg): The solution to Worked example build a difference triangle shown as a ladder of expressions, one row per algebraic move
\[ \Delta^2 f = 1 \text{ throughout} \]
Verify: check the constant against the formula
Why: For a quadratic a n squared plus bn plus c, the second differences are always 2a. Here a is one half, so 2a is 1 — exactly what the triangle shows. That relationship gives the leading coefficient straight from the difference table, without solving anything.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.9 Write Polynomial Functions and Models §5.9, pp. 394-394
Fill the middle
Example 2, at the first row.
Fill in the blanks
1, \; 3, \; 6, \; 10, \; 15 \;\Longrightarrow\; 2, \; 3, \; 4, \; 5
Why: Fifteen minus 10 is 5. Five values give four first-order differences, and in general n values give n minus 1 differences — which is why a long enough table is needed before the constant row appears.
Worked example
Guided Practice 3.
\[ \text{Show that } f(n) = \tfrac{1}{2}n(3n-1) \text{ has constant second-order differences.} \]
Compute several values
Why: For n from 1 to 6 the values are 1, 5, 12, 22, 35 and 51.
First differences
Why: Subtracting consecutive values gives 4, 7, 10, 13 and 16.
Second differences
Why: Subtracting again gives 3, 3, 3 and 3.
Confirm with the formula
Why: Expanding gives three halves n squared minus one half n, so a is three halves and 2a is 3.
Figure (svg): The solution to Worked example the pentagonal numbers shown as a ladder of expressions, one row per algebraic move
\[ \Delta^2 f = 3 \text{ throughout} \]
Verify: compare with the triangular numbers
Why: Both are quadratic, so both have constant second differences, but the constants differ — 1 against 3 — because their leading coefficients differ. The order at which the differences go constant gives the degree, and the constant itself gives the leading coefficient.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.9 Write Polynomial Functions and Models §5.9, pp. 394-394
Error analysis
A student computes differences from a table with uneven inputs.
Annotate
On: \( x: 1, 2, 5, 9; \; f(x): 1, 4, 25, 81 \;\Longrightarrow\; \text{differences } 3, 21, 56 \)
Check the spacing of the inputs before subtracting anything. On unevenly spaced data the whole method says nothing at all.
Ranking
Building a difference triangle.
Put in order
Why: Step one is the hypothesis check and the only one that can invalidate everything after it. Step five's condition has two parts: constant and non-zero — a row of zeros means the previous row was already constant and the degree is one lower.
Sorting
Each row is shorter than the one above it.
Sort into buckets
To see a constant row, sort each situation by whether the table is long enough.
The cubic with four values is the borderline case: it produces exactly one third-order difference, which is a number but not yet evidence of a pattern.
Prediction
Commit before reasoning.
Predict first
For a quadratic a n squared plus bn plus c with inputs one apart, what is the constant second difference?
Correct: 2a.
\[ \Delta^n f = n! \cdot a_n \quad \text{for inputs one apart} \]
Why: Working it out for a general quadratic gives second differences of 2a every time — for the triangular numbers a is one half and the constant is 1, and for the pentagonal numbers a is three halves and the constant is 3. Both match. In general the nth-order difference of a degree-n polynomial is n factorial times its leading coefficient, so a cubic's third differences are 6a.
Section
Section 3
Concept
If a polynomial function has degree n, then its nth-order differences at equally spaced inputs are non-zero and constant. Conversely, if the nth-order differences of equally spaced data are non-zero and constant, the data fit a polynomial of degree n.
\[ \deg f = n \;\Longleftrightarrow\; \Delta^n f \text{ constant, non-zero} \]
The converse is what makes the property useful: it turns a table of numbers into a statement about the model, before any coefficients are found.
Figure (svg): Three difference triangles showing constant differences at the first, second and third order
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.9 Write Polynomial Functions and Models §5.9, pp. 394-394 — Properties of Finite Differences
Picture it
A linear, a quadratic and a cubic function, differenced.
Figure (svg): Three difference triangles showing constant differences at the first, second and third order
The linear function's first differences go constant, the quadratic's second and the cubic's third. Counting down the triangle to the constant row reads the degree straight off the data.
Worked example
Guided Practice 4. Difference until it goes constant.
\[ \text{For } x = 1\dots 6 \text{ with } f(x) = 6, 15, 22, 21, 6, -29, \text{ find the degree.} \]
Check the spacing
Why: The inputs are 1 through 6, one apart, so the method applies.
First-order differences
Why: Subtracting consecutive values gives 9, 7, negative 1, negative 15 and negative 35.
Second-order differences
Why: Subtracting again gives negative 2, negative 8, negative 14 and negative 20.
Third-order differences
Why: Subtracting once more gives negative 6, negative 6 and negative 6.
Conclude
Why: Third-order differences are constant and non-zero, so the data fit a cubic.
\[ ^\circ 3 \]
Figure (svg): The solution to Worked example find the degree from data shown as a ladder of expressions, one row per algebraic move
\[ \deg = 3 \]
Verify: read the leading coefficient from the constant
Why: For a cubic the third differences equal 6a, and they are negative 6 here, so a is negative 1. That is one coefficient obtained for free, before any system is solved, and it is a useful check on whatever the system produces.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.9 Write Polynomial Functions and Models §5.9, pp. 395-395
Matching
Count how many rows down the constant appears.
Match the pairs
Why: Each differencing step lowers the degree by one, so a degree-n function reaches a non-zero constant after exactly n steps and reaches zero after n plus 1. That is why the constant row is at the nth order and not somewhere else.
Worked example
Example 3, first half.
\[ \text{Find the degree of } 1, 4, 10, 20, 35, 56, 84. \]
First-order differences
Why: Subtracting gives 3, 6, 10, 15, 21 and 28.
Second-order differences
Why: Subtracting again gives 3, 4, 5, 6 and 7.
Third-order differences
Why: Subtracting once more gives 1, 1, 1 and 1.
Conclude
Why: The data fit a cubic.
Figure (svg): The solution to Worked example the triangular pyramidal numbers shown as a ladder of expressions, one row per algebraic move
\[ \deg = 3 \]
Verify: notice the first differences
Why: The first-order differences are exactly the triangular numbers from Example 2, which are quadratic — and differencing a quadratic once gives something linear, then constant. So the pyramidal numbers being one degree above the triangular ones is not a coincidence: each differencing step lowers the degree by exactly one.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.9 Write Polynomial Functions and Models §5.9, pp. 395-395
Error analysis
A student differences a data set and concludes too early.
Annotate
On: \( \text{first differences } 9, 7, -1, -15, -35: \text{ no constant, so try again} \)
Only an exactly repeated value counts. On exact data the constant row is unmistakable; on measured data it never appears at all, which is the signal to use regression.
Fill the middle
Guided Practice 4.
Fill in the blanks
\text-1 -6 \;\Longrightarrow\; 6a = -6 \;\Longrightarrow\; a = ___
Why: For a cubic the third-order difference is 3 factorial, or 6, times the leading coefficient, so a is negative 1. Getting one coefficient free from the difference table is worth doing before the system, both as a shortcut and as a check on the answer.
Two truths and a lie
All three are about the properties.
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. The properties are stated for equally spaced inputs, and on unevenly spaced data the differences confuse the gap with the function. They also require exact data: measured values with error will never give a perfectly constant row, which is why the last idea uses regression instead.
Prediction
Commit before reasoning.
Predict first
The third-order differences of a data set are all zero. What does that say?
Correct: The data fit a polynomial of degree 2 or less.
\[ \Delta^3 f = 0 \;\Longrightarrow\; \Delta^2 f \text{ was already constant} \]
Why: The property requires the constant row to be non-zero. A row of zeros at order 3 means the row above it was already constant, so the degree is 2 — or lower if that row was zero too. Differencing one step too far always produces zeros, so the rule is to stop at the first non-zero constant row.
Section
Section 4
Concept
Once the differences give the degree, write the general polynomial of that degree and substitute enough data points to get one equation per unknown coefficient. Solving the system gives the function.
\[ f(n) = an^3 + bn^2 + cn + d \]
A cubic needs four equations, which four data points supply. The system can be solved by elimination as in Lesson 3.4, or as a matrix equation on a calculator.
Figure (svg): Four data points turned into four linear equations and solved for the coefficients of a cubic
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.9 Write Polynomial Functions and Models §5.9, pp. 395-395 — Model with finite differences
Picture it
Example 3: the triangular pyramidal numbers.
Figure (svg): Four data points turned into four linear equations and solved for the coefficients of a cubic
The solution is a equal to one sixth, b to one half, c to one third and d to zero. Fractional coefficients are normal here and are not a sign of an error.
Worked example
Example 3, second half.
\[ \text{Find a polynomial giving } 1, 4, 10, 20, 35, 56, 84. \]
Use the differences to fix the degree
Why: Third-order differences are constant, so the model is cubic.
Substitute the first four values
Why: Each substitution gives one linear equation in the four unknowns.
Solve the system
Why: By elimination or as a matrix equation, the solution is a one sixth, b one half, c one third and d zero.
\[ a = \frac{1}{6}, b = \frac{1}{2}, c = \frac{1}{3}, d = 0 \]
Write the function
Why: The constant term is zero, so no constant appears.
\[ f(n) = (\frac{1}{6}) n ^{3} + (\frac{1}{2}) n ^{2} + (\frac{1}{3}) n \]
Figure (svg): The solution to Worked example the pyramidal numbers shown as a ladder of expressions, one row per algebraic move
\[ f(n) = \tfrac{1}{6}n^3 + \tfrac{1}{2}n^2 + \tfrac{1}{3}n \]
Verify: test a value not used in the system
Why: At n equal to 5: a sixth of 125 is about 20.83, plus half of 25 is 12.5, plus a third of 5 is about 1.67 — totalling 35, which matches the fifth pyramidal number. Testing a point that was not used to build the model is the only check worth trusting, since the first four are guaranteed to fit.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.9 Write Polynomial Functions and Models §5.9, pp. 395-395
Fill the middle
Example 3, at the second substitution.
Fill in the blanks
f(2) = 4 \;\Longrightarrow\; 8a + 4b + 2c + d = 4
Why: Substituting n equal to 2 gives 2 cubed for a, 2 squared for b, 2 for c and 1 for d. Each row of the system is just the powers of that input, which is why the four equations look so systematic.
Worked example
Guided Practice 4, completed.
\[ \text{Fit a polynomial to } f(1)=6, f(2)=15, f(3)=22, f(4)=21, f(5)=6, f(6)=-29. \]
Use the differences
Why: Third-order differences are all negative 6, so the model is cubic with a equal to negative 1.
\[ a = -1 \]
Write three more equations
Why: Substituting n equal to 1, 2 and 3 gives three equations in b, c and d.
Eliminate
Why: Subtracting consecutive equations gives 3b plus c equal to 16 and 5b plus c equal to 26.
Solve
Why: Subtracting gives 2b equal to 10, so b is 5, then c is 1 and d is 1.
\[ b = 5, c = 1, d = 1 \]
Figure (svg): The solution to Worked example a cubic from a table shown as a ladder of expressions, one row per algebraic move
\[ f(x) = -x^3 + 5x^2 + x + 1 \]
Verify: test the last two values
Why: At x equal to 5: negative 125 plus 125 plus 5 plus 1 is 6, matching. At x equal to 6: negative 216 plus 180 plus 6 plus 1 is negative 29, also matching. Neither was used in the system, so the model really does describe the whole table rather than just the part it was built from.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.9 Write Polynomial Functions and Models §5.9, pp. 395-395
Trap
\[ f(n) = an^3 + bn^2 + cn + d \]
Substitute all seven data points
Why: Every value in the table is used, giving seven equations.
\[ \text{seven equations, four unknowns: much extra work} \]
Four of the seven determine the answer completely, and the other three are then automatically satisfied — so solving all seven is three times the labour for no gain.
\[ \text{substitute } f(1), f(2), f(3), f(4) \text{ only} \]
Use exactly as many points as there are unknowns
Why: Four unknowns need four equations; the remaining data become a check.
\[ f(5) = 35 \quad \checkmark \quad \text{(a genuine test)} \]
The unused values are worth more as a check than as equations. If one of them fails, the data do not fit a cubic after all and the difference table should be re-examined.
Ranking
Building a polynomial from exact data.
Put in order
Why: Step two must come before step three, since the degree decides how many unknowns there are and therefore how many points step four needs. The test at the end is the only real verification, because the points used are fitted exactly by construction.
Comparison
Fill the blanks. Coefficients and equations match.
Comparison matrix
| Degree | Coefficients | Data points needed |
|---|---|---|
| 1 | two | two |
| 2 | three | three |
| 3 | four | four |
| n | n + 1 | n + 1 |
The last row is the general statement, and every earlier lesson in this course is a case of it — including two points for a line back in Lesson 2.4.
Prediction
Commit before reasoning.
Predict first
The pyramidal numbers are all whole, yet the model has coefficients one sixth, one half and one third. Is that a problem?
Correct: No — fractional coefficients can still give whole values at whole inputs.
\[ \tfrac{1}{6}n^3 + \tfrac{1}{2}n^2 + \tfrac{1}{3}n = \tfrac{n(n+1)(n+2)}{6} \]
Why: At every whole n the three fractions combine to a whole number, because the expression equals n times the quantity n plus 1, times n plus 2, over 6 — a product of three consecutive integers, which is always divisible by 6. Whole outputs do not require whole coefficients, and insisting on them would make the correct model impossible to write.
Section
Section 5
Concept
Measured data almost never lie exactly on a polynomial, so their finite differences never go constant. Instead a scatter plot suggests a degree and a regression feature produces the best-fitting polynomial of that degree.
polynomial regression — A calculator procedure that finds the polynomial of a chosen degree coming as close as possible to a set of data points. It generally passes through none of them.
\[ y = 0.00650x^3 - 0.739x^2 + 49.0x - 236 \]
The choice of degree is a judgement about the shape of the scatter, not a deduction. A higher degree always fits the data more closely and usually describes them worse.
Figure (svg): A scatter plot of shuttle speed against time with a cubic model and the four thousand four hundred line
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.9 Write Polynomial Functions and Models §5.9, pp. 396-396 — Solve a multi-step problem
Picture it
Example 4: eight measurements and a cubic model.
Figure (svg): A scatter plot of shuttle speed against time with a cubic model and the four thousand four hundred line
The model reaches 4400 feet per second at about 106 seconds, which is beyond the last measurement at 80 seconds — an extrapolation, and therefore a prediction rather than a reading.
Worked example
Example 4, in the book's four steps.
\[ \text{Fit the shuttle data and find when the speed reaches } 4400 \text{ ft/s.} \]
Make a scatter plot
Why: The points curve upward with an increasing slope, suggesting a cubic rather than a line.
Run cubic regression
Why: The calculator returns coefficients of about 0.00650, negative 0.739, 49.0 and negative 236.
Check the fit
Why: Graphing the model with the data shows it passing close to every point.
Find where the model reaches 4400
Why: Graphing the horizontal line y equals 4400 and using the intersect feature gives about 106.
\[ \text{about } 106\text{ seconds} \]
Figure (svg): The solution to Worked example model the shuttle data shown as a ladder of expressions, one row per algebraic move
\[ x \approx 106 \text{ seconds} \]
Verify: evaluate the model at 106
Why: The cubic term contributes about 7740, the quadratic about negative 8300, the linear about 5190, and the constant negative 236 — totalling about 4400 feet per second. The model checks, though its answer lies well beyond the data, which is worth stating explicitly whenever it happens.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.9 Write Polynomial Functions and Models §5.9, pp. 396-396
Sorting
Ask whether the data are counted or measured.
Sort into buckets
Sort each data set by the right method.
Counted quantities are exact; measured ones are not. That distinction, rather than the size of the table, decides the method.
Worked example
Guided Practice 5 and 6. The degree is a judgement each time.
\[ \text{Fit polynomials to } (1,5),(2,13),(3,17),(4,11),(5,11),(6,56) \text{ and } (0,8),(2,0),(4,15),(6,69),(8,98),(10,87). \]
First: plot and inspect
Why: The values rise, dip, and then rise sharply, which needs at least two turning points.
First: fit and check
Why: A quartic follows the dip and the sharp rise far better than a cubic does.
Second: plot and inspect
Why: The values fall, rise steeply, and then level off and drop, again needing two turns.
Second: fit and check
Why: A cubic follows the shape closely across the whole range.
Figure (svg): The solution to Worked example two more regressions shown as a ladder of expressions, one row per algebraic move
\[ \text{degree } 4; \qquad \text{degree } 3 \]
Verify: count the turns in each scatter
Why: The first has two visible turns, which needs degree 3 at least, and its sharp final rise fits a quartic better. The second has two turns as well and a cubic handles both. Counting turning points in the scatter is the practical way to choose a degree, using Lesson 5.8's rule in reverse.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.9 Write Polynomial Functions and Models §5.9, pp. 396-396
Error analysis
A student tries finite differences on the shuttle table.
Annotate
On: \( 202.4, 463.3, 748.2, 979.3 \;\Longrightarrow\; \text{differences } 260.9, 284.9, 231.1: \text{ not constant, so no model exists} \)
Exact data and measured data need different tools. The failure of the difference table is a signal about the data, not about whether a model exists.
Fill the middle
Example 4, at the prediction.
Fill in the blanks
y = 4400 \text106 x \approx ___ \text___
Why: The intersect feature gives about 106.03 seconds, reported as 106 because the model's coefficients carry only three significant digits. Note that 106 is beyond the last data point at 80 seconds, so the answer is an extrapolation and should be presented as such.
Prediction
Commit before reasoning.
Predict first
A degree-7 polynomial would pass exactly through all eight shuttle data points. Is that a better model?
Correct: No — it would follow the measurement errors and predict badly.
\[ 8 \text{ points} \;\Longrightarrow\; \text{a degree-7 exact fit exists, and is useless} \]
Why: Eight points always determine a degree-7 polynomial exactly, but that curve would swing wildly between the points and beyond them, chasing noise rather than the underlying trend. The cubic ignores the small errors and captures the shape, which is what makes its prediction at 106 seconds worth anything. Fitting more closely and describing better are different things.
Comparison
Fill the blanks. Each suits a different kind of data.
Comparison matrix
| Question | Finite differences | Regression |
|---|---|---|
| Data must be | exact and equally spaced | anything, but usually measured |
| Degree is found by | the order of the constant row | judging the scatter plot |
| The curve passes | through every point | near the points |
| Extra points are | a check on the model | part of what the fit balances |
The second row is where the judgement lives. One method deduces the degree; the other asks you to choose it, and choosing badly is the commonest way a regression model goes wrong.
Comparison
Fill the blanks. The same idea at every degree.
Comparison matrix
| Degree | Points needed | Lesson |
|---|---|---|
| 1, a line | two | 2.4 |
| 2, a parabola | three | 4.10 |
| 3, a cubic | four | 5.9 |
| n | n + 1 | the general rule |
One point per coefficient, every time. The methods differ only in how the resulting system is solved.
Pattern
One routine for exact data, one for measured data.
Finite differences need exact and equally spaced data. If either condition fails, go straight to a scatter plot.
OpenStax Algebra and Trigonometry 2e, §5.3 Graphs of Polynomial Functions §5.3
Check
From a graph. Mind the sign of each intercept.
Check your understanding
Write the cubic with x-intercepts -4, 1 and 3 passing through (0, -6).
Answer: A
Why: The factors vanish at -4, 1 and 3, and substituting (0, -6) gives -6 = 12a, so a = -1/2.
Check
Finite differences. Difference until constant.
Check your understanding
For f(1) = 6, f(2) = 15, f(3) = 22, f(4) = 21, f(5) = 6, f(6) = -29, what degree fits?
Answer: A
Why: The third-order differences are -6, -6, -6, constant and non-zero, so the data fit a cubic.
Check
Choosing a method.
Check your understanding
Eight measured shuttle speeds at equally spaced times. Which method fits a model?
Answer: A
Why: Measured data carry error, so no order of differences is exactly constant and no polynomial passes through every point.
Real world
A stack of oranges is built as a square pyramid: one orange on top, then a 2 by 2 layer, then 3 by 3, and so on. The totals for 1 through 6 layers are 1, 5, 14, 30, 55 and 91.
Discussion prompt
Find a polynomial giving the total for n layers, and use it to predict the total for 10 layers.
Hint: Difference the data until a row goes constant.
Answer:
\[ \text{first differences } 4, 9, 16, 25, 36; \; \text{second } 5, 7, 9, 11; \; \text{third } 2, 2, 2 \]
Third-order differences are constant, so the model is a cubic, and the constant of 2 gives 6a equal to 2, so a is one third.
\[ f(n) = \tfrac{1}{3}n^3 + \tfrac{1}{2}n^2 + \tfrac{1}{6}n = \tfrac{n(n+1)(2n+1)}{6} \]
For 10 layers the total is 385 oranges.
Two things are worth noticing. The first differences are the perfect squares, which makes sense because each new layer adds n squared oranges — so differencing undid the summing. And the coefficients are again fractional while every output is a whole number, exactly as with the pyramidal numbers.
Commit first
Answer, then rate your confidence honestly.
Predict first
A data set's fourth-order differences are all zero. What degree fits it?
Correct: At most three.
\[ \Delta^4 f = 0 \;\Longrightarrow\; \Delta^3 f \text{ constant} \;\Longrightarrow\; \deg \le 3 \]
Why: The property requires the constant row to be non-zero, and a row of zeros at order 4 means the third-order row was already constant — so the degree is 3, or lower if that row was zero as well. Differencing one step past the degree always gives zeros, which is why the rule is to stop at the first non-zero constant row rather than the first constant one.
Explain it
They can fit a line to two points and have been handed a table of six numbers.
Discussion prompt
In four sentences or fewer, explain how a table of numbers can tell you what degree of polynomial to fit.
Hint: Subtract, and subtract again.
Answer:
If the inputs are equally spaced, subtract each value from the next to get a row of differences, then do the same to that row, and keep going. For a polynomial of degree n, the row you get after n steps is constant and every row after that is zeros.
So you count how many steps it took to reach a constant row, and that number is the degree. Then you write the general polynomial of that degree and substitute enough points to solve for its coefficients — one point per coefficient.
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 factors, ask what makes each bracket zero rather than copying the number. For difference tables, write each row directly beneath the gaps of the one above so nothing gets misaligned. For systems, look first for a substitution that eliminates a variable outright, and use the constant row to get the leading coefficient free. For choosing, ask whether the data were counted or measured. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Take the sequence 1, 4, 10, 20, 35, 56, 84 and build the whole model on one page. Top left: write the difference triangle in full, aligning each row beneath the gaps of the one above, and circle the constant row. Top right: state the degree it implies and use the constant to find the leading coefficient before doing anything else. Middle: write the four equations from the first four values and solve them, showing the elimination. Bottom left: write the finished function and test it at n equal to 5, 6 and 7 — values that were not used. Bottom right: write down what you would have done instead if the data had been eight measured speeds rather than seven counted totals. In a margin, note how many data points a degree-5 model would need.
If your leading coefficient does not match the constant row divided by 6, one of the two is wrong — and the difference table is usually the more reliable of the pair.
Recap
Five things, and together they close the chapter.
| If you see | Then |
|---|---|
| All the x-intercepts on a graph | Write factored form and find a |
| Exact, equally spaced data | Difference until a non-zero constant row |
| A constant row at order n | Fit a polynomial of degree n |
| A row of zeros | The degree is one lower |
| Degree n to determine | Use n + 1 data points |
| Measured data | Scatter plot and regression |
| An answer beyond the data | Say that it is an extrapolation |
Chapter 5 began by evaluating polynomials and ends by producing them from evidence. Chapter 6 changes direction entirely: it undoes powers rather than building them, starting with nth roots and rational exponents.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.9 Write Polynomial Functions and Models §5.9, pp. 393-399 — everything on these slides traces back here
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