Tasks 04-01 - Polynomial Functions

Section 04: Advanced Functions

Problem 1: Polynomial Identification (x)

Determine which of the following are polynomial functions. For those that are, state the degree and leading coefficient:

  1. \(f(x) = 3x^4 - 2x^2 + 7x - 1\)

  2. \(g(x) = \frac{2}{x} + x^3 - 5\)

  3. \(h(x) = \sqrt{2}x^5 - \pi x^2 + e\)

  4. \(p(x) = (x - 3)(x + 2)(x - 1)\)

  5. \(q(x) = x^{2.5} + 3x - 1\)

  6. \(r(x) = \frac{x^3 - 2x + 1}{2}\)

  1. Yes, polynomial
    • Degree: 4 (highest power of x)
    • Leading coefficient: 3
  2. No, not a polynomial
    • The term \(\frac{2}{x} = 2x^{-1}\) has a negative exponent
    • Polynomials only have non-negative integer exponents
  3. Yes, polynomial
    • Degree: 5
    • Leading coefficient: \(\sqrt{2}\) (coefficients can be any real number)
  4. Yes, polynomial
    • When expanded: \((x - 3)(x + 2)(x - 1)\)
    • First: \((x - 3)(x + 2) = x^2 + 2x - 3x - 6 = x^2 - x - 6\)
    • Then: \((x^2 - x - 6)(x - 1) = x^3 - x^2 - x^2 + x - 6x + 6 = x^3 - 2x^2 - 5x + 6\)
    • Degree: 3, Leading coefficient: 1
  5. No, not a polynomial
    • The exponent 2.5 is not a whole number
    • Polynomials require whole number exponents
  6. Yes, polynomial
    • Can be written as: \(r(x) = \frac{1}{2}x^3 - x + \frac{1}{2}\)
    • Degree: 3, Leading coefficient: \(\frac{1}{2}\)

Problem 2: End Behavior Analysis (x)

Without graphing, describe the end behavior of each polynomial:

  1. \(P(x) = 4x^6 - 3x^4 + 2x - 7\)

  2. \(Q(x) = -2x^5 + 8x^3 - x + 10\)

  3. \(R(x) = -\frac{1}{3}x^8 + 5x^5 - 2x^2\)

  4. \(S(x) = 7x^3 - 4x^2 + x - 9\)

For end behavior, focus on the leading term (highest degree term):

  1. \(P(x) = 4x^6 - ...\)
    • Degree: 6 (even)
    • Leading coefficient: 4 (positive)
    • End behavior: Both ends up (↗↗)
    • As \(x \to -\infty\): \(P(x) \to +\infty\)
    • As \(x \to +\infty\): \(P(x) \to +\infty\)
  2. \(Q(x) = -2x^5 + ...\)
    • Degree: 5 (odd)
    • Leading coefficient: -2 (negative)
    • End behavior: Up-Down (↗↘)
    • As \(x \to -\infty\): \(Q(x) \to +\infty\)
    • As \(x \to +\infty\): \(Q(x) \to -\infty\)
  3. \(R(x) = -\frac{1}{3}x^8 + ...\)
    • Degree: 8 (even)
    • Leading coefficient: \(-\frac{1}{3}\) (negative)
    • End behavior: Both ends down (↘↘)
    • As \(x \to -\infty\): \(R(x) \to -\infty\)
    • As \(x \to +\infty\): \(R(x) \to -\infty\)
  4. \(S(x) = 7x^3 - ...\)
    • Degree: 3 (odd)
    • Leading coefficient: 7 (positive)
    • End behavior: Down-Up (↘↗)
    • As \(x \to -\infty\): \(S(x) \to -\infty\)
    • As \(x \to +\infty\): \(S(x) \to +\infty\)

Problem 3: Zeros and Multiplicities (xx)

For each factored polynomial:

  • List all zeros and their multiplicities
  • Describe whether the graph crosses or touches at each zero
  • Find the y-intercept
  • State the degree and end behavior

  1. \(P(x) = 3(x - 2)^2(x + 1)(x - 4)\)

  2. \(Q(x) = -2x^3(x + 3)^2\)

  3. \(R(x) = \frac{1}{2}(x + 2)^3(x - 1)^2(x - 3)\)

  1. \(P(x) = 3(x - 2)^2(x + 1)(x - 4)\)

    Zeros and behavior:

    • \(x = 2\) (multiplicity 2, even → touches x-axis)
    • \(x = -1\) (multiplicity 1, odd → crosses x-axis)
    • \(x = 4\) (multiplicity 1, odd → crosses x-axis)

    y-intercept: Set \(x = 0\) \(P(0) = 3(0-2)^2(0+1)(0-4) = 3 \cdot 4 \cdot 1 \cdot (-4) = -48\)

    Degree and end behavior:

    • Degree: \(2 + 1 + 1 = 4\) (even)
    • Leading coefficient: \(3 \cdot 1 \cdot 1 \cdot 1 = 3\) (positive)
    • End behavior: Both ends up (↗↗)

  2. \(Q(x) = -2x^3(x + 3)^2\)

    Zeros and behavior:

    • \(x = 0\) (multiplicity 3, odd → crosses x-axis, flattened)
    • \(x = -3\) (multiplicity 2, even → touches x-axis)

    y-intercept: \(Q(0) = 0\)

    Degree and end behavior:

    • Degree: \(3 + 2 = 5\) (odd)
    • Leading coefficient: \(-2 \cdot 1 \cdot 1 = -2\) (negative)
    • End behavior: Up-Down (↗↘)

  3. \(R(x) = \frac{1}{2}(x + 2)^3(x - 1)^2(x - 3)\)

    Zeros and behavior:

    • \(x = -2\) (multiplicity 3, odd → crosses x-axis, flattened)
    • \(x = 1\) (multiplicity 2, even → touches x-axis)
    • \(x = 3\) (multiplicity 1, odd → crosses x-axis)

    y-intercept: Set \(x = 0\) \(R(0) = \frac{1}{2}(2)^3(-1)^2(-3) = \frac{1}{2} \cdot 8 \cdot 1 \cdot (-3) = -12\)

    Degree and end behavior:

    • Degree: \(3 + 2 + 1 = 6\) (even)
    • Leading coefficient: \(\frac{1}{2} \cdot 1 \cdot 1 \cdot 1 = \frac{1}{2}\) (positive)
    • End behavior: Both ends up (↗↗)

Problem 4: Manufacturing Cost Analysis (xx)

A factory producing electronic components has a monthly cost function:

\[C(x) = 0.5x^3 - 12x^2 + 90x + 200\]

where \(x\) is production in thousands of units and \(C(x)\) is cost in thousands of euros.

  1. What are the fixed costs (costs when production is zero)?

  2. Calculate the cost of producing 5,000 units.

  3. Find \(C(10) - C(8)\) and interpret this difference.

  4. The company breaks even when revenue equals cost. If revenue is \(R(x) = 100x\), show that there’s a break-even point at exactly 4,000 units.

  1. Fixed costs: \(C(0) = 0.5(0)^3 - 12(0)^2 + 90(0) + 200 = 200\)

    Fixed costs are €200,000

  2. Cost for 5,000 units (x = 5): \(C(5) = 0.5(5)^3 - 12(5)^2 + 90(5) + 200\) \(C(5) = 0.5(125) - 12(25) + 450 + 200\) \(C(5) = 62.5 - 300 + 450 + 200 = 412.5\)

    Cost is €412,500

  3. Cost difference: \(C(10) = 0.5(1000) - 12(100) + 900 + 200\) \(C(10) = 500 - 1200 + 900 + 200 = 400\)

    \(C(8) = 0.5(512) - 12(64) + 720 + 200\) \(C(8) = 256 - 768 + 720 + 200 = 408\)

    \(C(10) - C(8) = 400 - 408 = -8\)

    Interpretation: Increasing production from 8,000 to 10,000 units decreases total cost by €8,000, showing economies of scale in this production range.

  4. Break-even analysis: Profit function: \(P(x) = R(x) - C(x) = 100x - (0.5x^3 - 12x^2 + 90x + 200)\) \(P(x) = -0.5x^3 + 12x^2 + 10x - 200\)

    Check \(x = 4\): \(P(4) = -0.5(64) + 12(16) + 10(4) - 200\) \(P(4) = -32 + 192 + 40 - 200 = 0\) ✓

    The company breaks even at exactly 4,000 units of production.

Problem 5: Market Share Analysis (xxx)

Three companies compete in a market. Company A’s market share over time is modeled by:

\[S(t) = -t^3 + 6t^2\]

where \(t\) is years since product launch \((0 \leq t \leq 4)\) and \(S(t)\) is market share in percentage.

  1. Find \(S(0)\), \(S(2)\), and \(S(4)\). Interpret each value.

  2. Factor the market share function completely.

  3. Find all times when Company A has zero market share.

  4. Based on the factored form and a sign analysis, determine when Company A has positive market share.

  1. Market share values:

    • \(S(0) = -(0)^3 + 6(0)^2 = 0\) Interpretation: No market share at launch

    • \(S(2) = -(2)^3 + 6(2)^2 = -8 + 24 = 16\) Interpretation: 16% market share after 2 years

    • \(S(4) = -(4)^3 + 6(4)^2 = -64 + 96 = 32\) Interpretation: 32% market share after 4 years

  2. Factoring: \(S(t) = -t^3 + 6t^2\) \(S(t) = t^2(-t + 6)\) \(S(t) = -t^2(t - 6)\)

  3. Zeros: From the factored form \(S(t) = -t^2(t - 6)\):

    • \(t = 0\) (multiplicity 2)
    • \(t = 6\)

    Within the domain \([0, 4]\), Company A has zero market share only at \(t = 0\) (launch).

  4. Sign analysis: From \(S(t) = -t^2(t - 6)\):

    • For \(0 < t < 6\): \(t^2 > 0\) and \((t - 6) < 0\)
    • So \(S(t) = -(positive)(negative) = positive\)

    Company A has positive market share for \(0 < t \leq 4\) (entire period after launch).

Problem 6: Polynomial Sketching (xx)

Sketch the graph of \(P(x) = -\frac{1}{4}(x + 3)(x - 1)^2(x - 4)\) by following these steps:

  1. Identify all zeros and their multiplicities
  2. Determine the end behavior
  3. Find the y-intercept
  4. Determine the sign of \(P(x)\) in each interval between zeros
  5. Sketch the complete graph

  1. Zeros and multiplicities:

    • \(x = -3\) (multiplicity 1 → crosses x-axis)
    • \(x = 1\) (multiplicity 2 → touches x-axis)
    • \(x = 4\) (multiplicity 1 → crosses x-axis)

  2. End behavior:

    • Degree: \(1 + 2 + 1 = 4\) (even)
    • Leading coefficient: \(-\frac{1}{4}\) (negative)
    • End behavior: Both ends down (↘↘)

  3. y-intercept: Set \(x = 0\) \(P(0) = -\frac{1}{4}(3)(-1)^2(-4) = -\frac{1}{4} \cdot 3 \cdot 1 \cdot (-4) = 3\)

  4. Sign analysis: Test a point in each interval:

    • Interval \((-\infty, -3)\): Test \(x = -4\) Signs: \((-)(-)(-)(-)\) = \((+)\) → Negative (due to \(-\frac{1}{4}\))

    • Interval \((-3, 1)\): Test \(x = 0\) \(P(0) = 3\) → Positive

    • Interval \((1, 4)\): Test \(x = 2\) \(P(2) = -\frac{1}{4}(5)(1)(-2) = \frac{10}{4} = 2.5\) → Positive

    • Interval \((4, \infty)\): Test \(x = 5\) Signs: \((+)(+)(+)(+)\) = \((+)\) → Negative (due to \(-\frac{1}{4}\))

  5. Graph characteristics:

    • Starts from \(-\infty\) (left side, going down)
    • Crosses up through \(x = -3\)
    • Positive between \(-3\) and \(4\)
    • Touches (bounces off) x-axis at \(x = 1\)
    • Crosses down through \(x = 4\)
    • Ends at \(-\infty\) (right side, going down)
    • y-intercept at \((0, 3)\)

Problem 7: Business Application - Profit Analysis (xxx)

A company’s quarterly profit (in thousands of euros) is modeled by:

\[P(x) = (x - 1)(x - 3)(x - 5)\]

where \(x\) represents the quarter number \((1 \leq x \leq 8)\).

  1. Expand the polynomial into standard form.

  2. Find all break-even quarters (when profit = 0).

  3. Calculate the profit for quarters 2, 4, and 6.

  4. Determine during which quarters the company experiences losses.

  5. Without using calculus, estimate which quarter likely has the highest profit.

  1. Expansion to standard form: Step 1: \((x - 1)(x - 3) = x^2 - 3x - x + 3 = x^2 - 4x + 3\)

    Step 2: \((x^2 - 4x + 3)(x - 5)\) \(= x^3 - 5x^2 - 4x^2 + 20x + 3x - 15\) \(= x^3 - 9x^2 + 23x - 15\)

  2. Break-even quarters: From the factored form, \(P(x) = 0\) when:

    • \(x = 1\) (Quarter 1)
    • \(x = 3\) (Quarter 3)
    • \(x = 5\) (Quarter 5)

  3. Specific quarter profits:

    • Quarter 2: \(P(2) = (2-1)(2-3)(2-5) = (1)(-1)(-3) = 3\) thousand euros
    • Quarter 4: \(P(4) = (4-1)(4-3)(4-5) = (3)(1)(-1) = -3\) thousand euros
    • Quarter 6: \(P(6) = (6-1)(6-3)(6-5) = (5)(3)(1) = 15\) thousand euros

  4. Loss periods: Sign analysis of \((x-1)(x-3)(x-5)\):

    • \(x \in [1, 3)\): \((+)(-)(-) = (+)\) Profit
    • \(x \in (3, 5)\): \((+)(+)(-) = (-)\) Loss
    • \(x \in (5, 8]\): \((+)(+)(+) = (+)\) Profit

    The company experiences losses during quarters 4 (between Q3 and Q5).

  5. Maximum profit estimate:

    • The polynomial has degree 3 with positive leading coefficient
    • After the last zero (x = 5), the function increases rapidly
    • Checking boundary: \(P(8) = (7)(5)(3) = 105\) thousand euros

    The highest profit likely occurs in Quarter 8.

Problem 8: Intermediate Value Theorem (xxxx)

Consider the polynomial \(P(x) = x^4 - 5x^2 + 2x + 3\).

  1. Calculate \(P(-2)\), \(P(-1)\), \(P(0)\), \(P(1)\), and \(P(2)\).

  2. Use the Intermediate Value Theorem to identify at least two intervals that must contain zeros.

  3. Show that \(P(x)\) can be rewritten as \((x^2 - 3)(x^2 - 2) + 2x - 3\) by expanding this expression.

  4. Based on the degree and complexity of \(P(x)\), explain why finding its exact zeros requires advanced techniques beyond what we’ve studied.

  1. Function values:

    • \(P(-2) = 16 - 5(4) + 2(-2) + 3 = 16 - 20 - 4 + 3 = -5\)
    • \(P(-1) = 1 - 5(1) + 2(-1) + 3 = 1 - 5 - 2 + 3 = -3\)
    • \(P(0) = 0 - 0 + 0 + 3 = 3\)
    • \(P(1) = 1 - 5(1) + 2(1) + 3 = 1 - 5 + 2 + 3 = 1\)
    • \(P(2) = 16 - 5(4) + 2(2) + 3 = 16 - 20 + 4 + 3 = 3\)

  2. Intervals containing zeros (by IVT): Looking at our calculated values:

    • \(P(-1) = -3 < 0\) and \(P(0) = 3 > 0\) → There’s at least one zero in \((-1, 0)\)

    To find another interval, check \(P(-3)\): \(P(-3) = 81 - 5(9) + 2(-3) + 3 = 81 - 45 - 6 + 3 = 33 > 0\)

    • \(P(-3) = 33 > 0\) and \(P(-2) = -5 < 0\) → There’s at least one zero in \((-3, -2)\)

  3. Verification of the alternate form: Expand \((x^2 - 3)(x^2 - 2) + 2x - 3\):

    Step 1: \((x^2 - 3)(x^2 - 2) = x^4 - 2x^2 - 3x^2 + 6 = x^4 - 5x^2 + 6\)

    Step 2: \(x^4 - 5x^2 + 6 + 2x - 3 = x^4 - 5x^2 + 2x + 3\) ✓

    This matches our original polynomial \(P(x)\).

  4. Why exact zeros are challenging:

    • \(P(x)\) is a degree 4 polynomial (quartic)
    • While quadratic formulas exist for degree 2, and some methods work for special cubics, general quartics require complex formulas
    • The presence of the \(2x\) term means this isn’t a biquadratic (which would have only even powers)
    • From part (c), we see \(P(x)\) mixes quadratic terms with a linear adjustment
    • Without the quartic formula or numerical methods, we can only approximate the zeros using IVT

    This demonstrates that polynomials of degree 4 and higher often have zeros that cannot be expressed using simple radicals.