A polynomial of degree n has at most n − 1 turning points and at most n real zeros. Far from the origin every other term is negligible, so the leading term alone decides the ends. Even degree: both ends the same way. Odd: opposite ways.
Step 1: Picture It
Start with something you already know.
A long race
- Two runners start a race together.
- One jogs, and one sprints, faster and faster.
- In the first few seconds, the jogger can be in front.
- A few minutes later, the sprinter is far ahead.
- From then on the gap only grows, and nobody catches up.
A polynomial is a race between its terms.
Near 0 any term can lead, but far out the highest power is so far ahead that the rest hardly count.
So the ends of the graph copy the leading term alone. Its sign says which way it runs, up or down.
Where the picture falls short: the graph has two far ends, left and right. Far to the left an even power is still positive, but an odd power has turned negative.
The same idea with numbers. Here 2x⁴ races 50x as x grows.
| x | 2x⁴ against 50x |
|---|---|
| 1 | 2 against 50 |
| 2 | 32 against 100 |
| 3 | 162 against 150 |
| 10 | 20,000 against 500 |
Near 0 the 50x term leads. By x = 3 the x⁴ term has passed it, and from there the gap only grows.
Step 2: Let's Learn
Read it, or press Listen and follow the words.
Try it first
q(x) = x³ − 25x. Near the origin the −25x term is in charge: q(1) = 1 − 25 = −24. Work out q(−10) and q(10). Which pair is right?
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, −2x⁴ is −200,000,000 and 7x is 700. The ratio is nearly three hundred thousand to one. Whatever the middle terms do near the origin, the ends copy the highest power.
Even degree
x², x⁴, x⁶: both ends up. Negate the coefficient and both ends go down. The graph starts and finishes on the same side.
Odd degree
x³ falls on the left and rises on the right. −x³ does the reverse. Because the ends are on opposite sides, an odd-degree polynomial always crosses the axis: at least one real zero.
Turning points
A cubic bends at most twice; a quartic at most three times. It may bend fewer: x³ has no turning point at all. The degree gives a ceiling, not a count.
Zeros
At most n real zeros. x⁴ − 5x² + 4 = (x² − 1)(x² − 4) has four: ±1, ±2. x⁴ + 1 has none. An even-degree polynomial can miss the axis entirely; an odd one cannot.
A rough sketch
From the ends and the zeros, a shape follows. −2x⁴ + 7x − 1: both ends down, so it rises somewhere in the middle. f(0) = −1 and f(1) = 4: a zero between 0 and 1, and another beyond 1 where it comes back down.
Step 3: Try It Yourself
Tap and try it out.
For g(x) = 3x⁵ − x³ + 2x² − 8, state the end behavior, the maximum turning points, and whether a real zero is guaranteed
- 1leading term 3x⁵: degree 5, coefficient 3read the leading term
Step 0 of 4
Step 4: In Real Life
A roller coaster’s profile
A designer wants three humps in the track. That needs at least a degree-4 polynomial, and the leading term decides whether the ends rise or dive. The degree sets the limits before any drawing.
Step 5: Watch an Example
One step at a time.
Watch Maya Read a Cubic
A fresh polynomial: p(x) = −x³ + 6x² − 4. Maya reads its end behavior, its two ceilings, and whether a real zero is certain, then checks with numbers.
- Step 1
Maya looks for the highest power first. It is x³, carried by −x³, so the leading term is −x³: degree 3, coefficient −1. The 6x² and the − 4 matter near the origin and nowhere else.
Watch Leo Start, Then Finish It Yourself
f(x) = 2x⁴ − 9x² + 3. Leo finds the leading term; the ends, the ceiling on zeros and one zero are yours.
- Step 1
Leo finds the highest power: x⁴, in the term 2x⁴. So the degree is 4 and the leading coefficient is 2. The − 9x² and the + 3 shape the middle of the graph but not its ends.
Step 6: Your Turn
Practice makes it stick.
Hills on the Track
Problem 1 of 16
A roller coaster designer models the height of a track, in meters, with a polynomial. The ride needs 4 hilltops with 3 dips between them. That is 7 turning points. What is the lowest degree the polynomial can have?
A Model That Runs Out
Problem 2 of 16
A new food truck models its weekly profit, in dollars, by P(t) = −t³ + 12t² + 30t, where t is the number of weeks since opening. The model fits the first eight weeks well. What does its leading term say about weeks far in the future?
Read the Leading Term
Find the highest power first, wherever it is written. Then read the degree and the sign.
Problem 3 of 16
g(x) = 6x² − x⁵ + 2x. At most how many turning points can its graph have?
Problem 4 of 16
h(x) = −x⁶ + 5x² + 1. Which way do the two ends of its graph go?
Problem 5 of 16
p(x) = 7 − 2x + 3x⁵. Which way do the two ends of its graph go?
Problem 6 of 16
k(x) = x⁴ − 10x² + 9, which factors as (x² − 1)(x² − 9). How many real zeros does k have?
Problem 7 of 16
q(x) = x⁴ + 2. How many real zeros does q have?
Whose Step?
One line on each page is a slip. Find it, then put it right.
Problem 8 of 16
Ana read the end behavior of f(x) = 3x² − 2x⁵ + 4. One line is a slip. Which?
Ana’s working for the ends of f(x) = 3x² − 2x⁵ + 4
- 1f(x) = 3x² − 2x⁵ + 4
- 2leading term 3x²
- 3degree 2, even: both ends the same way
- Answercoefficient 3, positive: both ends rise
Problem 9 of 16
Which way do the two ends of f(x) = 3x² − 2x⁵ + 4 really go?
Problem 10 of 16
Dev read h(x) = x⁴ − 4x² from its degree. One line is a slip. Which?
Dev’s working for h(x) = x⁴ − 4x²
- 1h(x) = x⁴ − 4x²
- 2leading term x⁴: degree 4
- 3even degree, positive: both ends rise
- Answerat most 4 turning points
Problem 11 of 16
At most how many turning points can the graph of h(x) = x⁴ − 4x² have?
Problem 12 of 16
Sam counted the real zeros of p(x) = x² − 9. One line is a slip. Which?
Sam’s working for the zeros of p(x) = x² − 9
- 1p(x) = x² − 9
- 2degree 2, even: both ends rise
- 3even degree, so the graph never crosses the axis
- Answerno real zeros
Problem 13 of 16
How many real zeros does p(x) = x² − 9 have?
Pick the Move
Choose the move that settles the question. Neither one needs a full graph.
Problem 14 of 16
f(x) = 0.01x⁵ − 300x³ + 2. Which is the sound way to decide its end behavior?
Problem 15 of 16
g is a polynomial of degree 4. You want to show that g has a real zero between 1 and 2. Which move does it?
Same Ends?
Judge each one from its leading term. No values needed.
Problem 16 of 16
Select every polynomial whose two ends go the same ways as the ends of −x³.
Step 7: Quick Check
Show what you know.
Question 1 of 1
Degree 8. Maximum turning points?
Ready for more?Optional. Past the lesson, for anyone who wants it.
Stretch 1 of 2
Build a polynomial of degree 3 with leading coefficient −1 whose real zeros are exactly −2, 0 and 2. Type it with no brackets.
Stretch 2 of 2
Ravi wants a polynomial whose graph is high on the far left, low on the far right, and turns exactly 3 times. What is true?
What You Learned
- A degree n polynomial has at most n zeros and at most n − 1 turning points.
- End behavior is decided entirely by the leading term, because it swamps the rest far out.
- Even degree sends both ends the same way; odd degree sends them opposite ways.
- Odd degree guarantees at least one real zero.
For the grown-up
The misconception this lesson targets is taking the leading term to be the term written first, or the one with the biggest coefficient. It is the term with the highest power, wherever it sits, and far out it beats every other term however small its coefficient is. Two quieter slips ride along: reading the degree as a count rather than a ceiling (x⁴ + 2 has degree 4 and no real zeros; x³ has degree 3 and no turns), and mixing the two ceilings (n for real zeros, n − 1 for turning points). Asking "what happens at x = 10?" settles the ends every time, and a parabola, which turns once and crosses at most twice, settles which ceiling is which.