The degree of a polynomial is its highest exponent, and far from the origin the leading term decides everything. An even degree sends both ends the same way, an odd degree opposite ways, and a negative leading coefficient flips both. Degree n allows at most n − 1 turning points and exactly n roots, counting repeats and complex ones.
Step 1: Picture It
Start with something you already know.
A bike racing a car
- A bike and a car start a race side by side.
- The bike gets off the line faster and leads for a moment.
- The bike keeps a steady speed, but the car keeps speeding up.
- Soon the car is ahead, and the gap only grows.
- A mile down the road, nobody remembers the bike ever led.
Each term of a polynomial is a racer, and its exponent says how fast it grows.
Near zero a lower power can lead. Far from zero the highest power always wins, so it alone decides where the ends of the graph go.
Where the picture stops: a race says which term is biggest, not which way it points. The sign of the leading coefficient, and whether x is negative, still decide up or down.
The two terms of x³ − 4x, racing as x grows.
| x | x³ against 4x |
|---|---|
| 1 | 1 against 4 |
| 2 | 8 against 8 |
| 10 | 1,000 against 40 |
| 100 | 1,000,000 against 400 |
At 1 the smaller power leads. By 10 it is not close, and by 100 the 4x hardly matters at all.
Step 2: Let's Learn
Read it, or press Listen and follow the words.
Try it first
f(x) = −x³ + 20x. Work out f(10) and f(−10). Far out to the right and far out to the left, is the graph high or low?
Stuck is fine. Let’s Learn shows the way next. Got it first time? You can jump ahead to Watch an Example.
Why the leading term wins
At x = 100, x³ − 4x is 1,000,000 − 400. The lower terms are noise. So x³ − 4x behaves like x³ at the ends: down on the left, up on the right, because a negative cubed is negative.
Four cases
Even and positive: both up, like x². Even and negative: both down, like −x². Odd and positive: left down, right up, like x³. Odd and negative: left up, right down, like −x³. Two questions, four shapes.
Roots from factors
x³ − 4x = x(x − 2)(x + 2), so the real roots are 0, 2 and −2, three crossings for a degree of 3. A factored polynomial shows its roots directly; an unfactored one hides them.
Touch or cross
g(x) = (x − 1)²(x + 3): the root 1 has multiplicity 2, even, so the graph touches the axis at 1 and turns back. The root −3 has multiplicity 1, odd, so it crosses. Total 2 + 1 = 3, the degree.
Between the roots
For x³ − 4x, f(1) = 1 − 4 = −3 and f(−1) = 3. Between roots 0 and 2 the graph dips below the axis; between −2 and 0 it rises above. Two turning points, the most a cubic allows.
Sketching in order
Ends first, from the leading term. Then the roots, marked touch or cross. Then one test value between each pair of roots. That is enough to draw the shape without a calculator.
Step 3: Try It Yourself
Tap and try it out.
Sketch f(x) = x³ − 4x: its ends, its roots and how it meets the axis at each, and where it sits between them
- 1degree 3, odd, leading coefficient +1: left end down, right end upthe leading term x³
Step 0 of 4
Step 4: In Real Life
A roller coaster’s profile
A coaster track is designed as a polynomial. Its degree fixes how many humps it can have, and the leading coefficient whether the ends rise or fall. The designer knows the limits before drawing.
Step 5: Watch an Example
One step at a time.
Watch Rosa Sketch a Cubic
Rosa sketches f(x) = x³ − x² − 6x in the lesson’s order: ends, roots, test values, with a check at the end.
- Step 1
The ends come first, from the leading term alone. The highest exponent is 3, so the degree is odd and the ends go opposite ways. The coefficient of x³ is +1, positive, so the left end goes down and the right end goes up, like x³.
Watch Leo Read a Factored Cubic
Leo describes g(x) = −(x − 2)²(x + 1) without multiplying it out. He finds the ends and the roots, and you finish the reading.
- Step 1
The leading term comes from multiplying the x in every factor, along with the minus in front: −1 · x · x · x = −x³. The degree is 3, odd, and the sign is negative, so the left end goes up and the right end goes down, like −x³.
Step 6: Your Turn
Practice makes it stick.
The Tunnel
Problem 1 of 15
A coaster track’s height above the station, in meters, is h(x) = x(x − 3)(x − 6), where x is the distance along the ground in tens of meters. Between x = 3 and x = 6 the track runs through a tunnel below the station. How far below the station is the track at x = 4, in meters?
meters
The Profit Model
Problem 2 of 15
A small company’s yearly profit, in thousands of dollars, is modeled by P(t) = −t³ + 12t², where t is the number of years since it opened. Factored, P(t) = −t²(t − 12). After how many years does the model say the profit falls back to $0?
years
Read the Polynomial
Ends from the leading term, roots from the factors, and a test value where you need one.
Problem 3 of 15
f(x) = 3x⁴ − 5x + 2. Which way do the ends of the graph go?
Problem 4 of 15
f(x) = 7 + x³ − 2x⁵. Which way do the ends of the graph go?
Problem 5 of 15
A polynomial’s graph has 4 turning points. What is the smallest degree the polynomial can have?
Problem 6 of 15
g(x) = (x − 4)²(x + 1)³. At which root does the graph touch the axis and turn back?
Problem 7 of 15
h(x) = x(x − 5)². What is the test value h(2)?
Whose Step?
One line in each reading is a slip. Find it, then fix it.
Problem 8 of 15
Priya described the ends of f(x) = 6x³ + 1 − x⁴. One line is a slip. Which?
Priya’s reading of f(x) = 6x³ + 1 − x⁴
- 1f(x) = 6x³ + 1 − x⁴
- 2leading term 6x³
- 3degree 3, odd: the ends go opposite ways
- Answercoefficient 6, positive: left end down, right end up
Problem 9 of 15
Which way do the ends of f(x) = 6x³ + 1 − x⁴ really go?
Problem 10 of 15
Dev read the roots of g(x) = (x + 2)²(x − 1). One line is a slip. Which?
Dev’s reading of g(x) = (x + 2)²(x − 1)
- 1g(x) = (x + 2)²(x − 1)
- 2roots: x = −2 and x = 1
- 3x = −2 has multiplicity 2; x = 1 has multiplicity 1
- 4at x = −2 the graph crosses the axis
- Answerso the graph crosses the axis twice
Problem 11 of 15
How many times does the graph of g(x) = (x + 2)²(x − 1) really cross the x-axis?
Pick the Move
Decide which move answers the question. There is no need to sketch the whole graph.
Problem 12 of 15
Ana wants to know whether the graph of f(x) = (x − 1)(x − 4)(x + 2) is above or below the axis between the roots 1 and 4. Which x should she test?
Problem 13 of 15
For f(x) = 2x³ − 100x², which term decides what the graph does far out to the right?
Same Behavior?
Judge from the equations, without sketching.
Problem 14 of 15
The graph of y = −x² has both ends going down. Select every polynomial whose graph does the same.
Problem 15 of 15
Select every polynomial whose graph touches the x-axis at x = 3 and turns back.
Step 7: Quick Check
Show what you know.
Question 1 of 2
A degree-4 polynomial. At most how many turning points?
Question 2 of 2
What decides end behavior?
Ready for more?Optional. Past the lesson, for anyone who wants it.
Stretch 1 of 2
Write the polynomial of smallest degree, with leading coefficient 1, whose graph touches the x-axis at x = 2, crosses it at x = −1, and meets it nowhere else. Type it in factored form.
Stretch 2 of 2
A polynomial f has degree 4, a negative leading coefficient, and only real roots: x = −3 and x = 2 with multiplicity 1 each, and x = 0 for the rest. Without a formula for f, is its graph above or below the axis at x = 1?
What You Learned
- Far from the origin the leading term decides the ends: even degree the same way, odd degree opposite ways.
- A negative leading coefficient flips both ends.
- Degree n gives at most n − 1 turning points and exactly n roots, counting repeats and complex ones.
- Odd multiplicity crosses the axis; even multiplicity touches and turns back.
For the grown-up
The misconception this lesson targets is that the term that looks biggest decides the ends: the one written first, or the one with the largest coefficient. Far from the origin only the highest exponent matters, so 6x³ + 1 − x⁴ falls at both ends, and 2x³ − 100x² rises on the right once x passes 50. Its cousin is treating every root as a crossing; a root whose factor is squared (or raised to any even power) is a touch, because that factor never changes sign. A test value between two roots, worked out in one line, settles either doubt.