
Session 04-01 - Polynomial Functions
Section 04: Advanced Functions
Entry Quiz - 10 Minutes
Review from Section 03
Work individually for 5 minutes, then discuss with the class (5 minutes)
Find the vertex of \(f(x) = 2x^2 - 8x + 3\)
If \(f(x) = 2x + 1\) and \(g(x) = x^2\), find \((f \circ g)(3)\)
Given the transformation \(h(x) = -2f(x - 3) + 4\), describe all transformations applied to \(f(x)\)
Find the inverse of \(f(x) = 3x - 5\)
Homework Discussion - 15 Minutes
Your questions from the Mock Exams
Focus on mock exam preparation and key concepts
- Which problems from the mock exam were most challenging?
- Common mistakes with composition and inverse functions
- Graph interpretation challenges
- Business application questions
. . .
Polynomials will extend these concepts to more complex business scenarios!
Learning Objectives
Today
By the end of this session, you will be able to:
- Identify polynomial functions and their key characteristics
- Analyze end behavior using degree and leading coefficient
- Find zeros and determine their multiplicities
- Sketch polynomial graphs from factored form
- Model business scenarios with polynomial functions
- Apply the Intermediate Value Theorem to locate zeros
Polynomial Basics
What is a Polynomial?
Building on our function knowledge
A polynomial function has the form:
\[P(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0\]
- \(a_n \neq 0\) (leading coefficient)
- \(n\) is a non-negative integer (degree)
- All exponents are whole numbers
. . .
- Linear: polynomials of degree 1, Quadratic: polynomials of degree 2
- Now we explore degree 3 and higher!
Polynomial Vocabulary
Key terminology you need to know
Structural Terms:
- Degree: highest power of \(x\)
- Leading: highest power coefficient
- Constant term: \(a_0\)
- Standard: descending powers
Examples:
- \(P(x) = 3x^4 - 2x^2 + x - 7\)
- Degree: 4
- Leading coefficient: 3
- Constant term: -7
. . .
Is \(f(x) = \frac{1}{x} + x^2\) a polynomial? No! The term \(\frac{1}{x} = x^{-1}\) has a negative exponent.
Degree and Leading Coefficient
These two values determine the big picture
The degree tells us:
- Maximum number of zeros
- Maximum number of turning points (degree - 1)
- Overall shape complexity
The leading coefficient determines:
- End behavior direction
- Vertical stretch/compression
Degree and Leading Coefficient II
End Behavior Analysis
Understanding End Behavior
What happens as \(x \to \pm\infty\)?
. . .
End behavior depends on:
- Degree (even or odd)
- Sign of leading coefficient (positive or negative)
. . .
Even Degree:
- Both ends go in same direction
- Positive \(a_n\): both up ↗↗
- Negative \(a_n\): both down ↘↘
Odd Degree:
- Ends go in opposite directions
- Positive \(a_n\): down-up ↘↗
- Negative \(a_n\): up-down ↗↘
End Behavior Patterns
The four fundamental patterns

Break - 10 Minutes
Zeros and Their Multiplicities
Finding Zeros
Where the polynomial crosses or touches the x-axis
A zero of \(P(x)\) is a value \(c\) where \(P(c) = 0\). To find zeros:
- Factoring (when possible)
- Quadratic formula (for degree 2 factors)
- Rational Root Theorem (for rational zeros)
- Graphing (approximate locations)
- Calculator (for precise values)
. . .
A polynomial of degree \(n\) has exactly \(n\) zeros (counting multiplicities and complex zeros).
Intermediate Value Theorem
The Theorem
A tool for finding zeros: Intermediate Value Theorem (IVT)
If \(P(x)\) is continuous on \([a, b]\) and \(P(a) \cdot P(b) < 0\), then there exists at least one \(c\) in \((a, b)\) where \(P(c) = 0\)
. . .
In simple terms:
- If the function is negative at one point
- And positive at another point
- It must cross zero somewhere in between!
. . .
IVT guarantees at least one zero exists but doesn’t tell us exactly where or how many!
Applying IVT
Locating zeros systematically
Example: Show that \(P(x) = x^3 - 2x - 5\) has a zero in \([2, 3]\)
. . .
Solution:
- \(P(2) = 8 - 4 - 5 = -1\) (negative)
- \(P(3) = 27 - 6 - 5 = 16\) (positive)
- Since \(P(2) < 0\) and \(P(3) > 0\), IVT guarantees a zero exists!
. . .
Business Application: Use IVT to prove break-even points exist when you know profit is negative at low production and positive at high production.
More complicated: Rational Zeros
Which “easy” numbers make our polynomial equal zero?
Consider any polynomial with integer coefficients, like:
\[P(x) = 2x^3 - 5x^2 + x + 2\]
. . .
Question: What values of \(x\) make \(P(x) = 0\)?
. . .
Some zeros might be “messy” (like \(\sqrt{2}\) or complex numbers), but some might be “easy” rational numbers (fractions).
. . .
- Rational zeros are exact and easy to work with
- They help us factor polynomials completely
The Rational Root Theorem
Instead of guessing randomly, use a systematic approach
The Rule: If our polynomial has integer coefficients, then any rational zero \(\frac{p}{q}\) (in lowest terms) must follow a pattern.
- \(p\) (numerator) must divide the constant term
- \(q\) (denominator) must divide the leading coefficient
. . .
- \(p\) (numerator) comes from the constant term - the “ending” of the polynomial
- \(q\) (denominator) comes from the leading coefficient - the “beginning” of the polynomial
Step 1: Find the p and q Options
Let’s use our example: \(P(x) = 6x^3 - 11x^2 + 6x - 1\)
Constant term (the number without \(x\)): -1
- Factors of -1: \(\pm 1\)
- Possible values for \(p\): \(\pm 1\)
Leading coefficient (number in front of highest power): 6
- Factors of 6: \(\pm 1, \pm 2, \pm 3, \pm 6\)
- Possible values for \(q\): \(\pm 1, \pm 2, \pm 3, \pm 6\)
. . .
See the Difference?
Step 2: Create All Combinations
Mix and match the p and q values
All possible \(\frac{p}{q}\) combinations:
From \(p \in \{\pm 1\}\) and \(q \in \{\pm 1, \pm 2, \pm 3, \pm 6\}\):
\[\frac{\pm 1}{\pm 1}, \frac{\pm 1}{\pm 2}, \frac{\pm 1}{\pm 3}, \frac{\pm 1}{\pm 6}\]
. . .
Start with simple fractions: \(\pm 1\), then try: \(\pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}\)
Step 3: Test the Candidates
Substitute each \(\frac{p}{q}\) candidate into the polynomial
\[P(1) = 6(1)^3 - 11(1)^2 + 6(1) - 1 = 0 \checkmark\]
\[P(\frac{1}{2}) = 6(\frac{1}{2})^3 - 11(\frac{1}{2})^2 + 6(\frac{1}{2}) - 1 = 0 \checkmark\]
\[P(\frac{1}{3}) = 6(\frac{1}{3})^3 - 11(\frac{1}{3})^2 + 6(\frac{1}{3}) - 1 = 0 \checkmark\]
. . .
Result: The rational zeros are \(x = 1\), \(x = \frac{1}{2}\), and \(x = \frac{1}{3}\)
. . .
The theorem doesn’t guarantee these will be zeros - it just tells us which ones are worth testing!
Visualizing Our Example

Understanding Multiplicity
How many times a zero appears
If \((x - c)^m\) is a factor of \(P(x)\), then \(c\) has multiplicity \(m\)
Behavior at zeros:
- Odd multiplicity: Graph crosses x-axis
- Even multiplicity: Graph touches x-axis (bounces off)
- Higher multiplicity: Flatter near the zero
. . .
- Linear \(y = x - 1\): The 0 at \(x = 1\) has multiplicity 1 (odd) → crosses the x-axis
- Quadratic \(y = (x - 1)^2\): 0 at \(x = 1\) has multiplicity 2 (even) → touches x-axis and bounces off
- This is the same concept! Polynomials just extend this pattern to higher multiplicities.
Understanding Multiplicity II

Factored Form Insights
Reading the story from the factors
Given: \(P(x) = -2(x + 3)(x - 1)^2(x - 4)\)
What can we determine?
- Zeros: \(x = -3\) (mult. 1), \(x = 1\) (mult. 2), \(x = 4\) (mult. 1)
- Degree: \(1 + 2 + 1 = 4\) (even)
- Leading coefficient: \(-2\) (negative)
- End behavior: Both down ↘↘
- y-intercept: \(P(0) = -2(3)(-1)^2(-4) = 24\)
Factored Form Insights II

Sketching from Factored Form
A systematic approach
Step-by-step process:
- Identify zeros and multiplicities
- Determine end behavior
- Find y-intercept
- Plot key points
- Connect smoothly, respecting multiplicities
. . .
Between zeros, the polynomial doesn’t cross the x-axis. Use test points to determine if the graph is above or below the x-axis in each interval.
Guided Practice - 25 Minutes
Individual Exercise Block I
Work alone for 5 minutes, then discuss for 5 minutes
- Analyze the polynomial \(P(x) = -x^4 + 5x^2 - 4\):
- Identify the degree and leading coefficient
- Describe the end behavior
- Factor completely and find all zeros with their multiplicities
- Sketch the graph showing all key features
Individual Exercise Block II
Work alone for 5 minutes, then discuss for 5 minutes
- Given \(Q(x) = (x + 2)^2(x - 1)(x - 3)^3\):
- List all zeros and their multiplicities
- Determine the degree
- If the leading coefficient becomes negative, how does the graph change?
Coffee Break - 15 Minutes
Business Applications
Real-World Polynomial Applications
Where do we see polynomials in business?
- Revenue functions: Price depends on quantity (non-linear demand)
- Cost functions: With economies and diseconomies of scale
- Market share models: Competition dynamics over time
- Multi-product interactions: How products affect each other
- Break-even analysis: Multiple equilibrium points
. . .
Let’s do this with an example!
TechCo Case Study - Part I
When products interact
TechCo produces three related products with profit function:
\[P(x) = -x^3 + 12x^2 - 35x + 24\]
where \(x\) is production level (thousands of units).
. . .
Business Question: At what levels does the company break even?
. . .
Mathematical Task: Solve \(P(x) = 0\) to find where profit equals zero!
Finding Break-Even Points I
Solving \(P(x) = 0\) to find where profit equals zero
Step 1: Factor out any common factors
\[-(x^3 - 12x^2 + 35x - 24) = 0\]
. . .
Step 2: Use the Rational Root Theorem to find possible rational roots
. . .
- Constant term: \(24\) → factors are \(\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24\)
- Leading coefficient: \(1\) → factors are \(\pm 1\)
- Possible rational roots: \(\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24\)
Finding Break-Even Points II
Solving \(P(x) = 0\) to find where profit equals zero
Step 3: Test \(x = 1\):
\[P(1) = -(1)^3 + 12(1)^2 - 35(1) + 24 = -1 + 12 - 35 + 24 = 0\]
. . .
Step 4: Factor out \((x - 1)\): \(P(x) = -(x - 1)(x^2 - 11x + 24)\)
. . .
Step 5: Factor the quadratic: \(x^2 - 11x + 24 = (x - 3)(x - 8)\)
. . .
Final form: \(P(x) = -(x - 1)(x - 3)(x - 8)\)
. . .
Break-even points: \(x = 1, 3, 8\) thousand units
Visualized Profit Function

Analysis: Break-even at 1,000, 3,000, and 8,000 units
Cost Function with Scale Effects
Complex cost structures
A manufacturing plant has cost function:
\[C(x) = 0.1x^4 - 2x^3 + 12x^2 + 50\]
- Fixed costs of €50,000 and variable costs
- Economies of scale (negative cubic)
- Diseconomies at high volume (positive quartic)
. . .
Polynomial cost functions capture the reality that unit costs often decrease initially (economies of scale) but may increase at very high production levels (capacity constraints).
Visualizing the Cost Function

Problem-Solving - 30 Minutes
TechCo Case Study - Part II
TechCo needs your help with additional questions:
End Behavior Analysis: Their competitor has profit function \(C(x) = -3x^5 + 2x^4 - 7x^2 + x - 9\). Describe the long-term behavior as production increases. What does this tell management?
Product Line Analysis: A subsidiary’s profit is modeled by \(S(x) = 2(x + 2)^2(x - 3)(x - 5)^3\). Find all break-even points and describe how the company enters/exits profitability at each point.
New Product Launch: TechCo’s profit (in thousands €) for a new product after \(x\) months is: \(P(x) = -x^3 + 9x^2 - 15x - 25\). What is the initial financial position at launch and how is the profit at months 5 and 7?
Spot the Error
Can you find the errors? Work with your neighbor
Time allocation: 5 minutes to find errors, 5 minutes to discuss
Student work:
“The polynomial \(P(x) = 3x^4 - 2x^2 + 1\) has degree 2 because there are two terms with \(x\)”
“If \((x - 2)^4\) is a factor, the graph crosses the x-axis at \(x = 2\)”
“A degree 5 polynomial always has 5 real zeros”
Wrap-Up
Key Takeaways
Today’s essential concepts
- Polynomials extend our function toolkit to more complex scenarios
- Degree and leading coefficient tell the big picture story
- Zeros and multiplicities reveal detailed behavior
- Business applications involve multiple equilibrium points
- Mathematical tools prove what’s possible in business
Final Assessment
5 minutes - Individual work
Given the polynomial \(P(x) = -2x^3 + 6x^2 + 8x\):
Factor completely and find all zeros with their multiplicities
Determine the end behavior
Describe the graph’s behavior at each zero
If this represents a company’s profit (in thousands €) where \(x\) is production in thousands of units, at what production levels does the company break even?
Next Session Preview
Session 04-02: Power Functions & Roots
Building on polynomial foundations
- Power functions with rational exponents
- Root functions and their domains
- Transformations of power functions
- Economic models with diminishing returns
- Production functions in economics
. . .
Complete Tasks 04-01!
