Math Core

Polynomials are the functions you can build from $x$ using only addition, subtraction and multiplication, and that makes them the workhorses of modeling. In this lesson you'll learn to read a polynomial's graph straight from its formula: where it heads at the far left and right, where it meets the $x$ -axis, and how it behaves there.

The parts of a polynomial function

Definition

Polynomial function

A polynomial function of degree $n$ has the form

f(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0,

where $n$ is a whole number and $a_n \ne 0$ . The term $a_n x^n$ is the leading term, $a_n$ is the leading coefficient, and $n$ is the degree.

Exponents must be whole numbers, so $f(x) = x^3 - \sqrt{x}$ and $g(x) = x^2 + \dfrac{1}{x}$ are not polynomials. The domain of every polynomial is all real numbers, and its graph is a single smooth, unbroken curve with no corners, gaps or jumps.

A polynomial is often given in factored form, like $f(x) = -2(x - 1)^2(x + 3)$ . You don't need to expand it to find the leading term: multiply the leading terms of the factors. Here that is $-2 \cdot x^2 \cdot x = -2x^3$ , so the degree is $3$ and the leading coefficient is $-2$ .

End behavior

End behavior describes what $f(x)$ does as $x \to \infty$ (far right) and $x \to -\infty$ (far left). For large $|x|$ , the leading term is so much bigger than all the other terms combined that it alone decides the direction. For example, at $x = 100$ , the polynomial $x^3 - 50x^2$ equals $1{,}000{,}000 - 500{,}000$ : the $x^3$ term still wins, and it wins by more and more as $x$ grows.

Leading term test

The end behavior of a polynomial matches the end behavior of its leading term $a_n x^n$ .

	$a_n > 0$	$a_n < 0$
$n$ even	up on both ends	down on both ends
$n$ odd	down on the left, up on the right	up on the left, down on the right

In symbols, "up on the right" is written $f(x) \to \infty$ as $x \to \infty$ . Even degree means the ends point the same way, like $y = x^2$ . Odd degree means they point in opposite directions, like $y = x^3$ .

Worked example: Reading end behavior

Describe the end behavior of each function.

$f(x) = 7x^3 - 2x^5 + x - 4$
$g(x) = (1 - x)(x + 2)^2(x - 3)$

Solutions.

The leading term is $-2x^5$ , not $7x^3$ : always look for the highest power, wherever it sits. The degree is odd and the leading coefficient is negative, so $f(x) \to \infty$ as $x \to -\infty$ and $f(x) \to -\infty$ as $x \to \infty$ .
Multiply the leading terms of the factors: $(-x)(x^2)(x) = -x^4$ . The degree is even and the coefficient is negative, so the graph falls on both ends: $g(x) \to -\infty$ as $x \to \pm\infty$ .

Zeros and multiplicity

The real zeros of $f$ are the $x$ -intercepts of its graph. In factored form you can read them off directly, and the exponent on each factor tells you how the graph meets the axis.

Multiplicity and the graph

If $(x - c)^k$ is a factor of $f(x)$ , and no higher power of $x - c$ is, then $c$ is a zero of multiplicity $k$ .

Odd $k$ : the graph crosses the $x$ -axis at $c$ . If $k \ge 3$ , it flattens out as it crosses, like $y = x^3$ at the origin.
Even $k$ : the graph touches the $x$ -axis at $c$ and turns back, like $y = x^2$ at the origin.

Why? Near $x = c$ , the other factors are close to a fixed nonzero number, so $f(x)$ behaves like a constant times $(x - c)^k$ . An odd power changes sign as $x$ passes $c$ , and an even power doesn't.

Look at $f(x) = (x + 2)(x - 1)^2$ . The zero $-2$ has multiplicity $1$ , so the graph cuts straight through the axis there. The zero $1$ has multiplicity $2$ , so the graph just touches the axis and bounces back up.

f(x) = (x + 2)(x − 1)². It crosses at x = −2 (multiplicity 1) and touches at x = 1 (multiplicity 2).Open in grapher →

Turning points

A turning point is a point where the graph changes from rising to falling or from falling to rising (a local maximum or minimum). In the graph above there are two: a peak at $(-1, 4)$ and a valley at $(1, 0)$ .

Counting zeros and turns

A polynomial of degree $n$ has at most $n$ real zeros and at most $n - 1$ turning points.

These are upper limits, not exact counts. The cubic $y = x^3$ has degree $3$ but only one real zero and no turning points at all. The bound is still useful as a check: if your sketch of a cubic has three turning points, something is wrong.

Sketching a polynomial

Put the pieces together:

Find the end behavior from the leading term.
Find the real zeros and their multiplicities; decide cross or touch at each.
Find the $y$ -intercept, $f(0)$ .
Connect the pieces with a smooth curve, keeping within $n - 1$ turning points. Plot an extra point between zeros if you need to know how high or low the curve goes.

Worked example: A complete sketch

Sketch $f(x) = -(x + 1)^2(x - 3)$ .

End behavior: the leading term is $-(x^2)(x) = -x^3$ . Odd degree, negative coefficient: up on the left, down on the right.
Zeros: $-1$ with multiplicity $2$ (touch) and $3$ with multiplicity $1$ (cross).
$y$ -intercept: $f(0) = -(1)^2(0 - 3) = 3$ .

Start high on the left, come down and touch the axis at $x = -1$ , then turn back up. The curve must pass through $(0, 3)$ , rise to a peak, and come back down to cross at $x = 3$ before falling forever. A test point shows the height: $f(1) = -(2)^2(-2) = 8$ . That's two turning points, the most a cubic can have.

f(x) = −(x + 1)²(x − 3): touches at x = −1, crosses at x = 3.Open in grapher →

Common mistake

Don't let a negative sign or a reversed factor trick you when you find the leading term. In $(1 - x)(x + 2)^2$ , the first factor's leading term is $-x$ , not $x$ . The sign of the leading coefficient flips the whole end behavior.

Working backward from a graph

Because the zeros and multiplicities determine the factors, you can often write the formula for a graph you're given. One extra point then fixes the constant in front.

Worked example: Writing a formula

A polynomial of degree $4$ touches the $x$ -axis at $x = -2$ , crosses it at $x = 1$ and $x = 3$ , and has $y$ -intercept $-12$ . Find it.

A touch means an even multiplicity, and the degree $4$ leaves room for exactly $2 + 1 + 1$ . So

f(x) = a(x + 2)^2(x - 1)(x - 3).

Use the $y$ -intercept: $f(0) = a(2)^2(-1)(-3) = 12a$ . Setting $12a = -12$ gives $a = -1$ , so $f(x) = -(x + 2)^2(x - 1)(x - 3)$ . As a check, the leading term is $-x^4$ , so this graph falls on both ends.

The intermediate value theorem

Since a polynomial's graph has no breaks, it can't get from below the axis to above it without crossing.

Intermediate value theorem for polynomials

If $f$ is a polynomial and $f(a)$ and $f(b)$ have opposite signs, then $f$ has at least one real zero between $a$ and $b$ .

Worked example: Locating a zero

Show that $f(x) = x^3 - 4x + 1$ has a zero between $0$ and $1$ .

$f(0) = 1$ and $f(1) = 1 - 4 + 1 = -2$ . One value is positive and the other negative, so by the intermediate value theorem there is a zero between $0$ and $1$ . You could narrow it further: $f(0.5) = 0.125 - 2 + 1 = -0.875$ , so the zero is actually between $0$ and $0.5$ .

Tip

The theorem only works one way. If $f(a)$ and $f(b)$ have the same sign, there might still be zeros in between (an even number of crossings, or a touch). No sign change doesn't mean no zero.

Practice

Practice 1

Which describes the end behavior of $f(x) = -3x^4 + x^3 + 5$ ?

Practice 2

What is the leading coefficient of $f(x) = (2x - 1)^3(x + 4)^2$ when it is expanded?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

What is the greatest number of turning points the graph of a degree- $6$ polynomial can have?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

Find all real zeros of $f(x) = 2x^3 + x^2 - 18x - 9$ .

Separate answers with commas, e.g. 2, -5

Practice 5

Which function has a graph that falls on the left, rises on the right, touches the $x$ -axis at $x = 3$ , and crosses it at $x = -1$ ?

Practice 6

The polynomial $f(x) = a(x + 1)(x - 2)^2$ has $y$ -intercept $8$ . Find $a$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

For $f(x) = x^3 - 3x - 1$ , which interval is guaranteed by the intermediate value theorem to contain a zero?

Practice 8

A cubic polynomial has zeros $-3$ , $1$ and $2$ (each with multiplicity $1$ ) and passes through $(0, -6)$ . When it is written as $f(x) = ax^3 + bx^2 + cx + d$ , what is $a$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.