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B. Derivatives and Their Applications · Curve Sketching: Connecting f, f′ and f″

Curve Sketching II: Quartics, Flat Points and Unfriendly Zeros

B1.5 B1.4About 75 minutes
Learning goals
  • I can sketch quartic and quintic polynomials with repeated factors, horizontal inflection points and several inflection points.
  • I can use the heights of turning points to count the zeros of a polynomial and locate them between integers.
  • I can describe the family of polynomials that fits partial derivative information.

Words to know

horizontal (flat) inflection point
A point where f′ = 0 and the concavity changes; the graph flattens without turning.
rational root theorem
Any rational zero p/q of a polynomial with integer coefficients has p dividing the constant term and q dividing the leading coefficient.

Harder polynomials

In the last lesson every feature fell out neatly. Real problems are messier: zeros that repeat, two or three inflection points, turning points that hide in a tiny region, and zeros that will not factor. This lesson collects the strategies for those cases.

Three tips before we start:

  • Zoom where it matters. Choose your scale so the interesting points are spread out (Example 1 uses steps of 0.5).
  • Use exact values, then decimals. Keep fractions and radicals while calculating; convert to decimals only to plot.
  • Let f′ count the zeros. The heights of the turning points tell you how many times the graph can cross the x-axis.
Worked example

Example 1: A quartic with a flat inflection point

Sketch f(x) = 3x⁴ − 4x³.

  1. Intercepts: f = x³(3x − 4): zeros 0 (order 3, crosses while flattening) and 4/3. y-int 0.
  2. Symmetry: neither. Ends: up on both sides (3x⁴).
  3. f′ = 12x³ − 12x² = 12x²(x − 1). Critical numbers 0 and 1. 12x² ≥ 0, so f′ has the sign of x − 1: negative for x < 1 (x ≠ 0), positive for x > 1. Min (1, −1); no extremum at 0.
  4. f″ = 36x² − 24x = 12x(3x − 2). Zeros 0 and 2/3; signs +, −, +. Both are inflection points.
  5. f(2/3) = 3(16/81) − 4(8/27) = 48/81 − 96/81 = −48/81 = −16/27 ≈ −0.59.
  6. Chart: x < 0 falling ∪; 0 < x < 2/3 falling ∩; 2/3 < x < 1 falling ∪; x > 1 rising ∪.

Min (1, −1); IPs (0, 0) (horizontal) and (2/3, −16/27); zeros 0 and 4/3.

y = 3x⁴ − 4x³xy−0.50.511.5−1.5−0.512flat IP (0, 0)IP (2/3, −16/27)min (1, −1)(4/3, 0)
Zoomed in to show the details: a horizontal inflection point at the origin, a second inflection point at (2/3, −16/27), and the only turning point, a minimum at (1, −1).
Worked example

Example 2: An even quartic

Sketch f(x) = x⁴ − 8x² + 7.

  1. Intercepts: treat as a quadratic in x²: (x² − 1)(x² − 7) = 0, so x = ±1, ±√7. y-int 7.
  2. Symmetry: only even powers: even. Work out the right half and mirror it.
  3. f′ = 4x³ − 16x = 4x(x − 2)(x + 2). Signs −, +, −, +: mins at x = ±2, max at 0. f(±2) = 16 − 32 + 7 = −9; f(0) = 7.
  4. f″ = 12x² − 16 = 0 when x² = 4/3, x = ±2/√3 ≈ ±1.15, with sign changes. f = 16/9 − 32/3 + 7 = (16 − 96 + 63)/9 = −17/9.
  5. Concave up for |x| > 2/√3, down for |x| < 2/√3.

Max (0, 7); mins (±2, −9); IPs (±2/√3, −17/9); zeros ±1, ±√7. A "W" shape.

y = x⁴ − 8x² + 7xy−3−2−1123−10−6−248max (0, 7)(−2, −9)(2, −9)IPIP
An even quartic: the left half is the mirror image of the right half. The zeros are ±1 and ±√7 ≈ ±2.65, the minimums are (±2, −9), and the inflection points are (±2/√3, −17/9) ≈ (±1.15, −1.89).
Worked example

Example 3: Zeros that do not factor

Sketch f(x) = x³ − 6x² + 9x + 1 and determine how many zeros it has.

  1. The rational root theorem allows only ±1 as rational zeros, and f(1) = 5, f(−1) = −15, so there are no rational zeros. Use calculus instead.
  2. f′ = 3x² − 12x + 9 = 3(x − 1)(x − 3). Max (1, 5); min (3, 1).
  3. f″ = 6x − 12: IP (2, 3). Concave down for x < 2, up for x > 2.
  4. Counting zeros: to the right of x = 1 the lowest the graph goes is 1 > 0, so it never reaches the axis there. To the left, f falls toward −∞ as x → −∞ and f(0) = 1 > 0, so there is exactly one zero, and it is negative.
  5. Estimating it: f(−0.1) = −0.001 − 0.06 − 0.9 + 1 = 0.039 > 0 and f(−0.2) = −0.008 − 0.24 − 1.8 + 1 = −1.048 < 0. The zero is between −0.2 and −0.1, very close to −0.1 (technology gives about −0.104).

Exactly one zero, at about x ≈ −0.10. Max (1, 5), min (3, 1), IP (2, 3).

y = x³ − 6x² + 9x + 1xy1234−2−112345max (1, 5)IP (2, 3)min (3, 1)zero ≈ −0.10
The minimum value is 1, which is above the x-axis, so the graph crosses the x-axis only once, just left of the origin. Calculus told us how many zeros there are before we estimated where.

Common pitfalls

  • Even-power factors in f′ or f″ (like 12x²) do not change sign: no turn, or no change of concavity, there.
  • Inflection points are not halfway between turning points except for cubics; always solve f″ = 0.
  • Check the count: a polynomial of degree n has at most n − 1 turning points and at most n − 2 inflection points.
  • Every key point needs a y-value from the original f.

Possible graphs from partial information

Sometimes you are given only some facts, for example "a quartic with a horizontal inflection point at the origin and one local minimum at x = 3". Use what you know about degrees:

  • f′ is a cubic with a double zero at 0 and a simple zero at 3: f′ = ax²(x − 3).
  • If a > 0 the minimum is real (f′ goes − to + at 3) and the graph opens up at both ends.
  • f = a(x⁴/4 − x³) + C for any a > 0 and any C: infinitely many graphs, all with the same shape as Example 1, scaled and shifted.
Just for fun

Riddle: Which polynomial is the best at keeping secrets?

A quartic with a flat inflection point. Even its own derivative says "0" when asked if anything is happening there.

Hands-on activity

Paper-strip curve builder

You need: 6 strips of paper (about 2 cm by 10 cm), tape, a pencil and a large sheet of paper with axes drawn on it.

  1. On the strips, draw the four basic shapes from the second-derivative lesson: rising ∪, rising ∩, falling ∪, falling ∩ (make extra copies of the ones you need).
  2. Choose f(x) = x⁴ − 8x² + 7. Use its combined chart: x < −2 falling ∪; −2 to −1.15 rising ∪; −1.15 to 0 rising ∩; 0 to 1.15 falling ∩; 1.15 to 2 falling ∪; x > 2 rising ∪.
  3. Plot the key points on the big sheet, then tape one strip per interval between them, in order.
  4. Trace a smooth curve along your strips.

What to look for: the strips should join smoothly at every key point, with no corners. If two neighbouring strips meet at an angle, one of them has the wrong shape; recheck the chart for that interval.

Your turn

1

Sketch f(x) = x⁴ − 2x² (all key points).

2

Sketch f(x) = x⁴ + 4x³ (all key points).

3

How many zeros does f(x) = x³ − 3x + 3 have?

4

How many zeros does f(x) = x³ − 12x + 1 have?

5

Find the inflection points of f(x) = −x⁴ + 6x² − 5.

6

What is the greatest number of turning points and inflection points a degree-5 polynomial can have?

More practice: warm-up

1

Factor x⁴ − 5x² + 4.

2

f′ = 8x²(x − 2). Does f turn at x = 0?

3

Max (−1, 4) and min (2, −3) for a cubic with positive leading coefficient. How many zeros?

4

Is x⁴ − 8x² + 7 even?

5

Most inflection points for a quartic?

More practice: on your own

1

Sketch f(x) = (x − 1)²(x + 2) (all key points).

2

Sketch f(x) = x⁴ − 4x³ + 4x² = x²(x − 2)².

3

Show that f(x) = x⁴ + x² + 1 has no zeros and no inflection points.

4

Locate the zeros of f(x) = x³ − 12x + 1 between consecutive integers.

5

A quartic has f′(x) = 4(x + 1)(x − 1)(x − 3) and f(1) = 0. Classify its critical points.

Challenge

1

Sketch f(x) = x⁵ − 5x⁴ (all key points).

2

For which k does f(x) = x⁴ − 6x² + k have exactly two zeros? Exactly four?

3

Prove that a polynomial of even degree with positive leading coefficient has an absolute minimum.

Check your understanding

Quiz

1.

For f(x) = 3x⁴ − 4x³, what is the local minimum?

2.

A cubic has a local max at (−2, 3) and a local min at (1, 1). How many real zeros does it have?

3.

For f(x) = x⁴ − 8x² + 7, where are the minimums?

4.

The greatest number of inflection points a quartic can have is:

5.

f′(x) = 5x³(x − 4). At x = 0, f has:

6.

Which statement is true for every quartic with positive leading coefficient?

For parents and teachers

Big idea (B1.5, B1.4): the same sketching routine works for harder polynomials, but your teen must watch for repeated factors (no sign change), more than one inflection point, and zeros that only calculus can count and locate.

How to help: the paper-strip activity makes the idea "each interval has one of four shapes" physical. Ask your teen to predict how many zeros a function has from its turning points before checking on a graphing tool.

Common difficulties:

  • Assuming every critical number is a turning point.
  • Arithmetic with fractions at points like x = 2/3.
  • Choosing a scale that squashes the interesting part of the graph.
  • Stopping when zeros will not factor, instead of using turning-point heights and sign changes.