Writing Polynomial Function Formulas
Learning Outcomes
- Understanding how to write a polynomial function based on its graph and identifying its zeros.
- Recognizing that a polynomial in factored form has x-intercepts at the zeros of its factors.
- If a polynomial of the lowest degree p has zeros at x = x₁, x₂, …, xₙ, it can be expressed as:
- f(x) = a(x − x₁)²(x − x₂)²… (x − xₙ)² where pᵢ indicates the multiplicity of each factor.
- The stretch factor 'a' adjusts the polynomial vertically, determined by known points on the polynomial.
- Identify x-intercepts: Locate where the graph crosses the x-axis. These give initial factors for the polynomial.
- Examine graph behavior at intercepts: Determine if the polynomial bounces off (indicating even multiplicity) or passes through (indicating odd multiplicity).
- Construct polynomial of least degree: Combine identified factors based on their multiplicities.
- Locate stretch factor: Use another point from the graph, often the y-intercept, to solve for 'a'.
- Suppose a graph has x-intercepts at x = –3, x = 2, and x = 5.
- Y-intercept is at (0, -2).
- X-intercepts behavior:
- At x = -3 and x = 5, the graph crosses the axis linearly → linear factors.
- At x = 2, it bounces → quadratic factor.
- Polynomial resulting from these intercepts:
- f(x) = a(x + 3)(x - 2)²(x - 5)
Finding 'a'
- Plugging the y-intercept into the formula:
- f(0) = a(0 + 3)(0 - 2)²(0 - 5) = -2
- This simplifies to: -2 = a(3)(4)(-5) → -2 = -60a → a = 1/30
- Final Function:
- f(x) = (1/30)(x + 3)(x - 2)²(x - 5)
Local and Global Extrema
- Extrema Definition:
- Local Maximum: Highest point within a neighborhood.
- Local Minimum: Lowest point within a neighborhood.
- Global Maximum/Minimum: Highest/lowest point overall.
Conditions for Extrema
- A polynomial function of even degree has a global maximum/minimum.
- Example function f(x) = x (odd degree) has neither.
Applications of Local Extrema
- Example Problem: Determine box volume from cutting squares from a plastic sheet.
- Steps to Solve:
- Visualize the cut-outs with variable w.
- Calculate the volume function V(w) = (20 - 2w)(14 - 2w)(w)
- Identify realistic domain for w (should be between 0 and 7).
- Max volume approximately occurs when squares are cut 2.75 cm, refining down to about 2.7 cm for more accurate volume being 339 cubic cm.
Exploring Leading Coefficients
- Change the leading coefficient 'a' to see how it affects polynomial's shape (end behavior and intercepts).
- Observe effects of both positive and negative values of 'a'.
- Example outcomes: maximum occurs around (5.98, -398.8) and minimum around (0.02, 3.24).