Degree, end behavior and the shape of a polynomial graph
Without plotting a single point, what can the formula of a polynomial tell you about the shape of its graph?
Where you are 6 in Polynomial functions
Why this exists
A table of values shows only the window you chose. Something has to say what happens outside it: whether the graph comes back down, how often it can cross the axis, how many hills it can have.
For a polynomial, the degree answers most of this at once. The term with the highest power decides where both ends go, the degree caps the number of zeros and of turning points, and an odd or even degree decides whether the outputs can be every real number.
Definition
A polynomial function has the form
where n is a whole number, the coefficients are real numbers, and a_n \neq 0. As in Degree, terms and standard form, n is the degree, a_n x^n the leading term and a_n the leading coefficient. The leading term is the one with the highest power wherever it is written: in 8x - x^2 + 2x^3 it is 2x^3.
- A real zero of p is a real number r with p(r) = 0, an x-intercept of the graph.
- A turning point is a peak or a valley, where the graph switches between rising and falling, as in Turning points, maximums and minimums.
- The end behavior is what the graph does far to the left and far to the right, in the arrows of End behavior: “as x \to \infty, p(x) \to -\infty” says that far to the right the graph falls without bound.
Intuition
Compare p(x) = 3x^5 - x^3 with its leading term alone.
| x | p(x) | \displaystyle \frac{p(x)}{3x^5} |
|---|---|---|
| 2 | 88 | 0.917 |
| 10 | 299{,}000 | 0.997 |
| 100 | 29{,}999{,}000{,}000 | 0.99997 |
At x = 2 the term -x^3 still matters; at x = 100 it changes the total by one part in 30{,}000. The ratio is \displaystyle 1 - \frac{1}{3x^2}, which closes in on 1 as x grows, and the left side mirrors the right: p(-100) = -29{,}999{,}000{,}000. Far out, p goes where 3x^5 goes: down on the left, up on the right.
A power x^n does one of two things. If n is even, x^n is positive for every x \neq 0, so both ends of y = x^n rise, like y = x^2. If n is odd, x^n has the sign of x, so the left end falls and the right end rises, like y = x^3 (both in The squaring and cubing functions). A negative leading coefficient turns the picture upside down. That gives four pictures and no others.
Results
End behavior. The ends of a polynomial of degree n \geq 1 match the ends of its leading term a_n x^n.
| a_n > 0 | a_n < 0 | |
|---|---|---|
| n even | both ends rise | both ends fall |
| n odd | left falls, right rises | left rises, right falls |
Zeros and turning points. A polynomial of degree n \geq 1 has at most n real zeros and at most n - 1 turning points.
Domain and range. Every polynomial has domain all real numbers. If the degree is odd, the range is all real numbers too. If the degree is even and at least 2, the range is bounded on one side: y \geq m for a least value m when a_n > 0, and y \leq M for a greatest value M when a_n < 0.
Symmetry. A polynomial with only even powers (a constant counts, as a_0 x^0) is an even function, one with only odd powers is an odd function, and one with both is neither, by the test in Even and odd functions. The degree alone does not decide it: x^3 + 1 has odd degree and is neither.
Why it is true
The leading term wins. For x \neq 0, factor it out: p(x) = a_n x^n \cdot B(x), where
Each fraction has a fixed top and a bottom that grows without bound, so it shrinks toward 0 and B(x) closes in on 1. Once B(x) is between \displaystyle \frac{1}{2} and \displaystyle \frac{3}{2}, p(x) has the sign of a_n x^n and at least half its size, so it goes wherever a_n x^n goes.
At most n real zeros. By the The remainder and factor theorems, later in this unit, each zero r gives a factor x - r of p. Degrees add when polynomials are multiplied, so a degree n polynomial has room for at most n such factors.
At most n - 1 turning points. The proof needs calculus: the steepness of a polynomial graph is given by a polynomial of degree n - 1 (The power rule and basic differentiation rules), the steepness is 0 at every turn, and that polynomial has at most n - 1 zeros.
Odd degree, every output. A polynomial graph has no breaks (Continuous graphs, jumps and holes). With odd degree one end falls without bound and the other rises without bound, so for any height c the graph is below c somewhere and above it somewhere, and passes through c in between.
Even degree, one side bounded. Take a_n > 0. Both ends rise, so outside some interval -R \leq x \leq R every output exceeds p(0). On that closed interval the graph has a lowest point (see Turning points, maximums and minimums), whose output m is at most p(0) and so is the least output anywhere. Every height above m is passed on the way out to a rising end. For a_n < 0, turn everything upside down.
Even and odd. (-x)^k is x^k for even k and -x^k for odd k, so replacing x by -x leaves an all-even polynomial unchanged and flips the sign of an all-odd one.
Worked examples
Ends from the leading term
Describe the end behavior of (a) f(x) = -2x^4 + 5x^2 + 1, (b) g(x) = -x^3 + 2x^2 + 5x - 6, (c) r(x) = 8x - x^2 + 2x^3.
Solution(a) The leading term -2x^4 has even degree, so x^4 is positive and huge at both ends, and the factor -2 makes it negative. As x \to \pm\infty, f(x) \to -\infty.
(b) The leading term -x^3 has odd degree: far right x^3 is huge and positive, so -x^3 is huge and negative, and far left the reverse. As x \to -\infty, g(x) \to \infty, and as x \to \infty, g(x) \to -\infty. Check: g(100) = -1{,}000{,}000 + 20{,}000 + 500 - 6, which is -979{,}506.
(c) The highest power is written last; the leading term is 2x^3, odd and positive. As x \to -\infty, r(x) \to -\infty, and as x \to \infty, r(x) \to \infty.
One valley and two zeros
Find the real zeros, the turning points, the domain and range, and the symmetry of q(x) = x^4 - 1.
SolutionZeros. q(x) = 0 means (x^2)^2 = 1, so x^2 = 1 or x^2 = -1. A square is never negative, so x = -1 or x = 1: two real zeros, where degree 4 allows four.
Turning points. For x < 0, moving right shrinks |x| and so x^4: the graph falls. For x > 0 it rises. One turning point, a valley at (0, -1), within the limit of 3.
Domain and range. The domain is all real numbers. Since x^4 \geq 0, q(x) \geq -1, and each height y \geq -1 is reached at \displaystyle x = \sqrt[4]{y + 1}. The range is y \geq -1, bounded below.
Symmetry. Both powers are even (the constant is -1 \cdot x^0), so q(-x) = (-x)^4 - 1 = x^4 - 1: an even function, symmetric about the y-axis.
Every zero and every turn degree 3 allows
Show that c(x) = x^3 - 4x has three real zeros and exactly two turning points.
SolutionTake out the common factor, then use the difference of squares: x^3 - 4x = x(x^2 - 4) = x(x - 2)(x + 2). A product is 0 only when a factor is, so the zeros are -2, 0 and 2, the most degree 3 allows.
Test one input in each gap: c(-1) = -1 + 4 = 3 and c(1) = 1 - 4 = -3. Between -2 and 0 the graph leaves the axis, reaches 3 and comes back, so it switches from rising to falling at least once. Between 0 and 2 it goes down to -3 and back: another turn. At least two turns, and degree 3 allows at most two, so there are exactly two.
The range is all real numbers, as for any odd degree. Both powers are odd, so c is an odd function with half-turn symmetry about the origin, which is why the peak and the valley mirror each other.
A quartic bounded above
Find the range of f(x) = -2x^4 + 5x^2 + 1, its number of real zeros, and its turning points.
SolutionEven degree, negative leading coefficient: both ends fall and the range has a greatest value. Treat x^2 as the variable and complete the square:
The square is never negative, so \displaystyle f(x) \leq \frac{33}{8}, with equality at \displaystyle x = \pm\frac{\sqrt{5}}{2}. The ends fall without bound and there are no breaks, so every lower height is reached too. The range is \displaystyle y \leq \frac{33}{8}.
For zeros, \displaystyle \left(x^2 - \frac{5}{4}\right)^2 = \frac{33}{16} gives \displaystyle x^2 = \frac{5 \pm \sqrt{33}}{4}. Since \displaystyle \sqrt{33} > 5, the minus sign makes x^2 negative, which no real x allows. The plus sign gives \displaystyle x = \pm\frac{\sqrt{5 + \sqrt{33}}}{2} \approx \pm 1.64: two real zeros out of four allowed.
The two peaks are turning points, and between them the graph must fall and rise again: a valley. Three turns is the limit for degree 4, so there is exactly one valley. Since f has only even powers its graph is symmetric about the y-axis, so the lone valley lies on the axis (its mirror image would be a second one): it is (0, f(0)) = (0, 1).
Reading the degree off a graph
The graph below is of a polynomial. What can you say about its degree and its leading coefficient?
The left end falls and the right end rises: odd degree, positive leading coefficient. The graph crosses the x-axis 5 times, at -2, -1, 0, 1 and 2, so the degree is at least 5; it turns 4 times, so n - 1 \geq 4, again n \geq 5. The least possible degree is 5. Degree 7 or 9 is not ruled out.
Check. The graph was drawn from y = x^5 - 5x^3 + 4x = x(x^2 - 1)(x^2 - 4), of degree 5 with zeros 0, \pm 1 and \pm 2.
A hang glider’s flight
Yuval launches his hang glider from a ridge. His altitude in meters above the landing field, t minutes after launch, is modeled by
until he lands at t = 12. (a) What is the end behavior of h, and how much of it describes the flight? (b) Show that h(t) = 432 - t(t - 6)^2 and h(t) = (12 - t)(t^2 + 36), and use them to find the range of h during the flight. (c) Ulani says the model could show Yuval climbing in three separate thermals (columns of rising air). Can it?
Solution(a) The leading term is -t^3: as t \to -\infty, h(t) \to \infty, and as t \to \infty, h(t) \to -\infty. Neither end is the flight: a negative t is before launch, and after t = 12 the formula goes below the field. In context the domain is 0 \leq t \leq 12.
(b) Since t(t - 6)^2 = t^3 - 12t^2 + 36t, the first form expands to h(t); so does (12 - t)(t^2 + 36), which is 12t^2 + 432 - t^3 - 36t. In the first form t(t - 6)^2 \geq 0 for t \geq 0, so h(t) \leq 432, equal at t = 0 and t = 6. In the second, 12 - t \geq 0 for t \leq 12 and t^2 + 36 > 0, so h(t) \geq 0, equal at t = 12. Coming down from t = 6 to t = 12 he passes every height between. The range during the flight is 0 \leq h \leq 432 meters.
(c) No. Three climbs with sinking between them need a turn at the top of each climb and at the bottom of each sink, at least 5 turns, and degree 3 allows 2. The model gives h(0) = 432, h(2) = 400, h(6) = 432 and h(12) = 0: one dip and one climb, using both turns, then the glide to land.
Boundaries
“At most” is not “exactly”. x^4 - 1 has two real zeros, x^2 + 1 has none, and x^3 has no turning points. Counting complex zeros, and repeated zeros as often as they repeat, a degree n polynomial has exactly n zeros: the The fundamental theorem of algebra.
Degree 0 and the zero polynomial. A nonzero constant has degree 0 and no zeros or turns. The zero polynomial has no degree, and every number is a zero of it.
Combine terms first. Top powers can cancel in a sum, so read the leading term only after collecting like terms. In a product, the leading term is the product of the factors’ leading terms.
Restricted domains. When a model applies only between two inputs (Example 6), its ends describe the formula, not the situation, and its range is found on that interval.
The ends say nothing about the middle. They fix neither where the zeros are nor whether the graph crosses the axis at a zero or only touches it.
Where this goes next
Cubic, quartic and higher degree functions studies degree 3 and higher graphs closely, Zeros and multiplicity explains what the graph does at each zero, and Graphing polynomial functions combines zeros, ends and the turning-point limit into a full sketch. In Modeling with polynomial functions the degree must allow as many turns as the data shows. The The remainder and factor theorems and Polynomial long division turn known zeros into factors. And far out, every polynomial is overtaken by exponential growth (Linear, polynomial and exponential growth compared).
Exercises
End behavior
Describe the end behavior in arrow notation.
- f(x) = 4x^3 - 9x^2 + 2
- g(x) = -5x^6 + x^5 + 100
- h(x) = 3 - 7x + x^2 - 2x^5
- k(x) = (2 - x)(x + 3)^2
- m(x) = (x^2 - x)^2 - x^4
Zeros and turning points
- What is the greatest number of real zeros, and of turning points, that q(x) = x^2(5 - x^3) can have?
- Show that x^4 - 10x^2 + 9 has four real zeros and exactly three turning points.
- Show that x^3 + x has exactly one real zero and no turning points.
Domain and range
Give the domain and range.
- f(x) = 2x^5 - x + 4
- g(x) = 3 - x^4
- p(x) = x^4 - 8x^2 + 3, by completing the square in x^2
Even and odd
- Which is an even function, which odd, and which neither? (a) 5x^4 - x^2 + 7 (b) -x^7 + 3x^3 (c) x^5 + x^2
Reading graphs
For each graph, say whether the degree is even or odd, give the sign of the leading coefficient, and give the least possible degree.
- Graph A.
- Graph B.
- Graph C.
- Graph D.
In context
- Sefa’s flight log shows her hang glider climbing straight after launch, then sinking, climbing again, sinking, climbing a third time, and gliding down to land. What is the least degree of a polynomial that could model the whole flight?
- A reservoir holds V(t) = -t^3 + 18t^2 + 800 megaliters of water t weeks after the thaw, until it runs dry at t = 20. (a) Show that V(t) = (20 - t)(t^2 + 2t + 40) and 1664 - V(t) = (t - 12)^2(t + 6). (b) Find the range of V for 0 \leq t \leq 20. (c) What does the end behavior of V say after t = 20?
Explaining
- Explain why every polynomial of odd degree has at least one real zero.
- Explain why the graph of a polynomial of degree 4 cannot have exactly two turning points.
- Yuval says that since x^6 + 2 has degree 6, its graph must cross the x-axis somewhere. Is he right? What does the result about at most n zeros say here?
Answers
- Leading term 4x^3, odd and positive: as x \to -\infty, f(x) \to -\infty; as x \to \infty, f(x) \to \infty.
- Leading term -5x^6, even and negative: as x \to \pm\infty, g(x) \to -\infty.
- Leading term -2x^5, odd and negative: as x \to -\infty, h(x) \to \infty; as x \to \infty, h(x) \to -\infty.
- Leading term (-x)(x^2) = -x^3: as x \to -\infty, k(x) \to \infty; as x \to \infty, k(x) \to -\infty.
- (x^2 - x)^2 = x^4 - 2x^3 + x^2, so the x^4 terms cancel and m(x) = -2x^3 + x^2. As x \to -\infty, m(x) \to \infty; as x \to \infty, m(x) \to -\infty.
- q(x) = 5x^2 - x^5 has degree 5: at most 5 real zeros and at most 4 turning points.
- x^4 - 10x^2 + 9 = (x^2 - 1)(x^2 - 9), with zeros \pm 1 and \pm 3. The values at -2, 0 and 2 are -15, 9 and -15, so in each of the three gaps between zeros the graph leaves the axis and returns, turning at least once. Degree 4 allows at most 3 turns, so there are exactly three.
- x^3 + x = x(x^2 + 1) and x^2 + 1 \geq 1, so the only zero is x = 0. Both x^3 and x grow as x grows, so their sum always rises and never turns.
- Domain all real numbers; odd degree, so the range is all real numbers.
- Domain all real numbers. x^4 \geq 0, so g(x) \leq 3, and each y \leq 3 is reached at \displaystyle x = \sqrt[4]{3 - y}. Range y \leq 3.
- Domain all real numbers. p(x) = (x^2 - 4)^2 - 13; the square is never negative and is 0 at x = \pm 2, and both ends rise. Range y \geq -13.
- (a) Even: only even powers, the constant included. (b) Odd: only odd powers. (c) Neither: p(-x) = -x^5 + x^2 matches neither p(x) nor -p(x).
- A rises on the left and falls on the right: odd degree, negative leading coefficient. Three zeros and two turns: least degree 3.
- B falls at both ends: even degree, negative leading coefficient. Four zeros and three turns: least degree 4.
- C rises at both ends: even degree, positive leading coefficient. No zeros and one turn: least degree 2.
- D falls on the left and rises on the right: odd degree, positive leading coefficient. One zero but two turns, so n - 1 \geq 2: least degree 3.
- The turns are peak, valley, peak, valley, peak: 5 turning points, so n - 1 \geq 5 and the degree is at least 6.
- (a) (20 - t)(t^2 + 2t + 40) expands to 20t^2 + 40t + 800 - t^3 - 2t^2 - 40t, which is -t^3 + 18t^2 + 800. And (t - 12)^2(t + 6) is (t^2 - 24t + 144)(t + 6), which expands to t^3 - 18t^2 + 864, that is 1664 - V(t). (b) For t \geq 0 the second form gives V(t) \leq 1664, equal at t = 12. For t \leq 20, 20 - t \geq 0 and t^2 + 2t + 40 = (t + 1)^2 + 39 > 0, so V(t) \geq 0, equal at t = 20. Range 0 \leq V \leq 1664 megaliters. (c) As t \to \infty, V(t) \to -\infty: past t = 20 the formula goes negative, so the model no longer applies.
- One end falls without bound and the other rises without bound, so the graph is below the x-axis somewhere and above it somewhere. It has no breaks, so it crosses the axis in between, at a real zero.
- With a_n > 0 both ends rise, so reading left to right the graph starts out falling and ends rising. Each turning point switches direction, and getting from falling to rising takes an odd number of switches: 1 or 3, never 2. With a_n < 0, the same with directions swapped.
- No. x^6 \geq 0, so x^6 + 2 \geq 2 for every x, and the graph never reaches the axis. “At most 6” is an upper limit and promises no zeros; only an odd degree forces one (exercise 19).