Polynomial functions and their graphs

Lines and parabolas are graphs of polynomial functions of degree ${1}$ and ${2}$. This page shows what happens for higher degrees. You will learn to read the shape of the graph straight from the formula.

Polynomial functions

A polynomial function of degree ${n}$ has the form

$P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$,

where ${n}$ is a whole number and ${a_n\ne 0}$. The numbers ${a_n,\ldots,a_0}$ are the coefficients. The term ${a_nx^n}$ is the leading term, and ${a_n}$ is the leading coefficient.

For example, ${P(x)=-2x^3+5x^2-1}$ has degree ${3}$. Its leading term is ${-2x^3}$, and its leading coefficient is ${-2}$.

The graph of a polynomial is always smooth (no sharp corners) and continuous (no breaks or holes). You can draw it without lifting your pencil.

Power functions

The simplest polynomials are the power functions ${y=x^n}$. Their shape depends on whether ${n}$ is even or odd:

xy
Even powers: ${y=x^2}$ (blue) and ${y=x^4}$ (teal). Both ends go up.
xy
Odd powers: ${y=x^3}$ (blue) and ${y=x^5}$ (teal). The left end goes down, the right end up.

For even ${n}$, the graph looks like a U, because ${(-x)^n=x^n}$. For odd ${n}$, it looks like an S, because ${(-x)^n=-x^n}$. Higher powers are flatter near ${0}$ and steeper far from ${0}$.

End behavior

The end behavior of a function tells what its graph does at the far left and the far right. We write ${x\to\infty}$ for "${x}$ increases without bound", and ${x\to-\infty}$ for "${x}$ decreases without bound".

Far from ${0}$, the leading term is much larger than all the other terms. For example, take ${P(x)=x^3-20x}$ at ${x=100}$:

${x^3=1{,}000{,}000}$, but ${20x}$ is only ${2{,}000}$.

So far out, a polynomial behaves like its leading term:

The end behavior of ${P}$ is the same as the end behavior of its leading term ${a_nx^n}$.

So there are only four cases. They depend on the degree and the sign of the leading coefficient:

DegreeSign of ${a_n}$Left endRight end
evenpositiveupup
evennegativedowndown
oddpositivedownup
oddnegativeupdown

Example 1: Describe the end behavior of each polynomial.

(a) ${P(x)=-2x^3+5x^2-1}$

(b) ${Q(x)=3x^4-x^3+10x}$

Solution:

(a) The leading term is ${-2x^3}$. The degree ${3}$ is odd, and the coefficient is negative. So the graph goes up on the left and down on the right:

${P(x)\to\infty}$ as ${x\to-\infty}$, and ${P(x)\to-\infty}$ as ${x\to\infty}$.

(b) The leading term is ${3x^4}$. The degree is even, and the coefficient is positive. So both ends go up:

${Q(x)\to\infty}$ as ${x\to\pm\infty}$.

Zeros of a polynomial

A zero of ${P}$ is a number ${c}$ with ${P(c)=0}$. For a real number ${c}$, these four statements say the same thing:

${c}$ is a zero of ${P}$.

${x=c}$ is a solution of the equation ${P(x)=0}$.

${x-c}$ is a factor of ${P(x)}$.

The graph of ${P}$ crosses or touches the ${x}$-axis at ${x=c}$.

So if ${P}$ is factored, you can read off its zeros. Between two neighboring zeros, the graph stays on one side of the ${x}$-axis. It cannot change sides without passing through a zero. So one test value in each interval tells you whether the graph is above or below the axis there.

To sketch the graph of a polynomial:

  1. Factor ${P}$, and find its zeros. These are the ${x}$-intercepts.
  2. Find the ${y}$-intercept, ${P(0)}$.
  3. Use test values between the zeros to see where the graph is above or below the axis.
  4. Use the end behavior for the two ends.
  5. Draw a smooth curve through the points.

Example 2: Sketch the graph of ${P(x)=(x+2)(x-1)(x-3)}$.

Solution:

The zeros are ${-2}$, ${1}$, and ${3}$.

The ${y}$-intercept is

${P(0)=(2)(-1)(-3)=6}$.

The zeros split the ${x}$-axis into four intervals. Pick a test value in each one. The graph is above or below the ${x}$-axis:

${x}$${P(x)}$Graph
${-3}$${(-1)(-4)(-6)=-24}$below
${0}$${(2)(-1)(-3)=6}$above
${2}$${(4)(1)(-1)=-4}$below
${4}$${(6)(3)(1)=18}$above

Multiplying out, the leading term is ${x\cdot x\cdot x=x^3}$. So the left end goes down, and the right end goes up. This matches the test values at ${-3}$ and ${4}$.

−213(0, 6)xy
The graph of ${P(x)=(x+2)(x-1)(x-3)}$. It crosses the ${x}$-axis at each zero.

Multiplicity of a zero

A factor can appear more than once. If ${(x-c)^m}$ is a factor of ${P}$, and no higher power of ${x-c}$ is, then ${c}$ is a zero of multiplicity ${m}$. For example, in ${P(x)=(x-1)^2(x+2)}$, the zero ${1}$ has multiplicity ${2}$, and ${-2}$ has multiplicity ${1}$.

The multiplicity tells how the graph meets the ${x}$-axis:

Odd multiplicity: the graph crosses the ${x}$-axis at ${c}$.

Even multiplicity: the graph touches the ${x}$-axis at ${c}$ and turns back.

Why? Near ${x=1}$, the factor ${(x-1)^2}$ is never negative. So it does not change sign there, and neither does ${P}$. A factor with an odd power, like ${(x+2)}$, does change sign.

Example 3: Sketch the graph of ${P(x)=(x-1)^2(x+2)}$.

Solution:

The zero ${-2}$ has multiplicity ${1}$, so the graph crosses there. The zero ${1}$ has multiplicity ${2}$, so the graph only touches the axis there.

The ${y}$-intercept is

${P(0)=(-1)^2(2)=2}$.

The leading term is ${x^2\cdot x=x^3}$. So the left end goes down, and the right end goes up.

−21(0, 2)xy
The graph of ${P(x)=(x-1)^2(x+2)}$. It crosses at ${x=-2}$, but only touches the axis at ${x=1}$.

Turning points

A turning point is a point where the graph changes from rising to falling, or from falling to rising. A turning point that is higher than the nearby points is a local maximum. One that is lower than the nearby points is a local minimum.

A polynomial of degree ${n}$ has at most ${n-1}$ turning points.

For example, the cubic in Example 2 has two turning points. A cubic can have at most ${3-1=2}$. Finding the exact turning points needs calculus. For now, we only sketch them.

The intermediate value theorem

A polynomial graph has no breaks. So if it is below the axis at one point and above the axis at another, it must cross the axis in between.

If ${P(a)}$ and ${P(b)}$ have opposite signs, then ${P}$ has at least one zero between ${a}$ and ${b}$.

This is called the intermediate value theorem. "Intermediate" means "in between". We can use it to locate a zero that we cannot find by factoring.

Example 4: Show that ${P(x)=x^3-x-1}$ has a zero between ${1}$ and ${2}$. Then find it to one decimal place.

Solution:

Evaluate ${P}$ at both ends:

${P(1)=1-1-1=-1}$

${P(2)=8-2-1=5}$

The signs are opposite, so there is a zero between ${1}$ and ${2}$.

To get closer, try values in between:

${P(1.3)=2.197-1.3-1=-0.103}$

${P(1.4)=2.744-1.4-1=0.344}$

The sign changes between ${1.3}$ and ${1.4}$. So the zero is ${1.3\ldots}$, about ${1.3}$ to one decimal place.

Putting it together

Example 5: Sketch the graph of ${P(x)=x^4-4x^2}$.

Solution:

First factor. Take out ${x^2}$, then use the difference of squares:

$\begin{align*}&P(x)\\&=x^2(x^2-4)\\&=x^2(x-2)(x+2)\end{align*}$

  • Zeros: ${0}$ has multiplicity ${2}$, so the graph touches the axis there. The zeros ${2}$ and ${-2}$ have multiplicity ${1}$, so the graph crosses there.
  • ${y}$-intercept: ${P(0)=0}$.
  • End behavior: the leading term ${x^4}$ has even degree and a positive coefficient. Both ends go up.
  • Test values: ${P(1)=1-4=-3}$ and ${P(-1)=1-4=-3}$. So the graph is below the axis between ${-2}$ and ${2}$, except at ${0}$.

Also, ${P}$ is even, since only even powers appear. So the graph is symmetric about the ${y}$-axis.

−22xy
The graph of ${P(x)=x^4-4x^2}$. It crosses at ${\pm 2}$, and touches the axis at ${0}$.

The graph has ${3}$ turning points. That is the most a degree ${4}$ polynomial can have.

An application

Example 6: An open box is made from a square sheet of cardboard, ${12}$ inches on each side. A square of side ${x}$ inches is cut from each corner. Then the sides are folded up.

xx12 in12 − 2x
Cut an ${x}$ by ${x}$ square from each corner. Fold up the sides along the dashed blue lines.

(a) Write the volume ${V}$ of the box as a function of ${x}$.

(b) What values of ${x}$ make sense?

(c) Use the graph to estimate the size of the cut that gives the largest volume.

Solution:

(a) After folding, the height of the box is ${x}$. The bottom is a square. Each side of the bottom loses ${x}$ at both ends, so it is ${12-2x}$ long. The volume is length times width times height:

${V(x)=x(12-2x)^2}$

(b) The cut must have positive size, so ${x>0}$. The bottom must also have positive size, so ${12-2x>0}$, which gives ${x<6}$. So ${0<x<6}$.

(c) Here is the graph on ${[0,6]}$:

1234564080120(2, 128)xV
The volume of the box. The highest point of the graph is near ${x=2}$.

The highest point is near ${x=2}$, where

${V(2)=2(8)^2=128}$.

So a cut of about ${2}$ inches gives the largest volume, about ${128}$ cubic inches. Calculus can show that ${x=2}$ is exactly right.

Summary