lecture 5: Polynomial functions, their graphs, their domain and range, and their infinity behavior

What Are Polynomial Functions?

1. Real-World Motivation

Polynomials help model real-life situations with more complexity than just straight lines or parabolas.

Examples include:

A polynomial is a function made by adding powers of xx with constant coefficients:

p(x)=anxn+an1xn1++a1x+a0p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0

Where:


Why Study Polynomials?


Domain and Range

1. Domain

All polynomials are defined for all real numbers.

Domain of any polynomial: R\text{Domain of any polynomial: } \mathbb{R}

2. Range

The range depends on:

We usually estimate range using the graph or calculus (later in the course).


Roots of Polynomials

1. What are Roots?

A root of a polynomial is a value of xx that makes the polynomial equal to zero.

2. How to find roots?

3. Example

Let:

f(x)=x36x2+11x6f(x) = x^3 - 6x^2 + 11x - 6


Graphing Polynomial Functions

1. Shape and Degree

The degree of a polynomial controls how many “turns” the graph can have.

Degree Common Shape Max Turning Points
1 Line 0
2 Parabola 1
3 S-shaped 2
4 W- or M-shaped 3

Example of polynomials of different degrees:

Polynomial degree

A degree-nn polynomial has at most n1n-1 turning points.


2. Infinity Behavior (a.k.a. End Behavior)

As x±x \to \pm \infty, the graph of the polynomial behaves like its leading term.

Degree Leading Coefficient End Behavior
Even Positive f(x)+f(x) \to +\infty as x±x \to \pm\infty
Even Negative f(x)f(x) \to -\infty as x±x \to \pm\infty
Odd Positive f(x)f(x) \to -\infty as xx \to -\infty,
f(x)+f(x) \to +\infty as x+x \to +\infty
Odd Negative f(x)+f(x) \to +\infty as xx \to -\infty,
f(x)f(x) \to -\infty as x+x \to +\infty

An infographic picture of polynomials behaviour at infinities:

General Infinity behavior

Example with functions:

  1. f(x)=x2+2x3f(x) = x^2 + 2x - 3
  2. f(x)=x2+2x+3f(x) = -x^2 + 2x + 3
  3. f(x)=x32xf(x) = x^3 - 2x
  4. f(x)=x3+2xf(x) = -x^3 + 2x
Infinity behavior

3. Example

Let:

f(x)=x36x2+11x6f(x) = x^3 - 6x^2 + 11x - 6

Turning Points: Up to 2
End Behavior:
f(x)f(x) \to -\infty as xx \to -\infty
f(x)+f(x) \to +\infty as x+x \to +\infty


Regular Exercises

1. Classify Polynomials

For each function, write the degree, leading coefficient, and whether it has an absolute maximum or minimum:

2. Match the Graph to the Function

Each of the following is a sketch of a polynomial. Without using a calculator:

Polynomial graphs

3. Real-World Modeling

A company's profit over time is modeled by:

P(t)=2t3+9t211t+5P(t) = -2t^3 + 9t^2 - 11t + 5

Solution
a) The function is:

P(t)=2t3+9t211t+5P(t) = -2t^3 + 9t^2 - 11t + 5

The degree of a polynomial is the highest power of the variable. Here, the highest power is t3t^3, so the degree is:
Degree = 3

b) The leading coefficient is the coefficient of the highest degree term, which is −2.
Since it is negative and the degree is odd, the graph will rise to the left and fall to the right.
In other words, as tt \to -\infty, P(t)P(t) \to \infty
and as tt \to \infty, P(t)P(t) \to -\infty

This means that in the long run, the company's profits will decrease (possibly losses grow).

c) The company is losing money when P(t)<0P(t) < 0.
Without graphing exactly, we can estimate by plugging in a few values:

P(0)=5(positive)P(1)=2(1)3+9(1)211(1)+5=2+911+5=1(positive)P(2)=16+3622+5=3(positive)P(3)=54+8133+5=1(negative)P(4)=128+14444+5=23(negative)\begin{aligned} P(0) &= 5 \quad (\text{positive}) \\ P(1) &= -2(1)^3 + 9(1)^2 - 11(1) + 5 = -2 + 9 - 11 + 5 = 1 \quad (\text{positive}) \\ P(2) &= -16 + 36 - 22 + 5 = 3 \quad (\text{positive}) \\ P(3) &= -54 + 81 - 33 + 5 = -1 \quad (\text{negative}) \\ P(4) &= -128 + 144 - 44 + 5 = -23 \quad (\text{negative}) \end{aligned}

So the company begins losing money around t=3t = 3 and continues downward afterward.

4. Design a Polynomial

Create a polynomial function that meets all of the following conditions:

Solution
a) We want a degree 3 polynomial, with positive leading coefficient, that passes through the origin (f(0)=0f(0) = 0), and has a turning point between x=1x = 1 and x=3x = 3.

One possible function is:

f(x)=x(x1)(x3)f(x) = x(x - 1)(x - 3)

c)

5. Polynomial Passing Through Points

You are told a cubic polynomial passes through the following points:

Solution
a) We're told the function is cubic and passes through three x-intercepts (three roots):
f(1)=0f(1) = 0, f(2)=0f(2) = 0, f(3)=0f(3) = 0, and f(0)=12f(0) = 12

This suggests we can write the function in factored form:

f(x)=a(x1)(x2)(x3)f(x) = a(x - 1)(x - 2)(x - 3)

Use f(0)=12f(0) = 12 to solve for aa:

f(0)=a(1)(2)(3)=6a=12a=2f(0) = a(-1)(-2)(-3) = -6a = 12 \Rightarrow a = -2

So, the function is:

f(x)=2(x1)(x2)(x3)f(x) = -2(x - 1)(x - 2)(x - 3)

b) Expand the expression:
First expand two of the factors:

(x1)(x2)=x23x+2(x - 1)(x - 2) = x^2 - 3x + 2

Multiply with the third:

(x23x+2)(x3)=x33x2+2x3x2+9x6=x36x2+11x6(x^2 - 3x + 2)(x - 3) = x^3 - 3x^2 + 2x - 3x^2 + 9x - 6 = x^3 - 6x^2 + 11x - 6

Multiply by 2-2:

f(x)=2x3+12x222x+12f(x) = -2x^3 + 12x^2 - 22x + 12

c) The leading coefficient is −2

This tells us the function falls to the right and rises to the left.

d)


Extra Advanced Exercises

1. Multiplicity Investigation

Consider the polynomial f(x)=(x+2)2(x1)(x4)3f(x) = (x+2)^2(x-1)(x-4)^3.

2. Turning Points Estimation

Suppose h(x)=x4+4x2h(x) = -x^4 + 4x^2

3. Difference of polynomials

Let f(x)=a2x2+a1x+a0f(x) = a_2x^2 + a_1x + a_0.

Consider the polynomial g(x)=bnxn+bn1xn1++b1x+b0g(x) = b_nx^n + b_{n-1}x^{n-1} + \cdots + b_1x + b_0.