Math Problem Statement

Given the function P(x)=x(x-8)(x+7)

Solution

The given function is:

P(x)=x(x8)(x+7)P(x) = x(x - 8)(x + 7)

Key features of this polynomial:

  1. Degree of the polynomial: The degree is 3, as it is the product of three linear factors xx, (x8)(x-8), and (x+7)(x+7).
  2. Roots of the polynomial:
    • x=0x = 0 (from the factor xx)
    • x=8x = 8 (from the factor x8x - 8)
    • x=7x = -7 (from the factor x+7x + 7)
  3. End behavior:
    • As xx \to \infty, P(x)P(x) \to \infty, since the leading term is x3x^3.
    • As xx \to -\infty, P(x)P(x) \to -\infty.
  4. Y-intercept:
    • The y-intercept is the value of P(0)P(0): P(0)=0(08)(0+7)=0P(0) = 0(0 - 8)(0 + 7) = 0 Hence, the y-intercept is at (0,0)(0, 0).
  5. Intervals of increase and decrease:
    • To determine where the function is increasing or decreasing, we calculate the first derivative P(x)P'(x).

First Derivative of P(x)P(x):

Using the product rule:

P(x)=x(x8)(x+7)P(x) = x(x-8)(x+7)

Let u=xu = x, v=(x8)(x+7)v = (x-8)(x+7). Then,

P(x)=uv+uvP'(x) = u'v + uv'

Step 1: Differentiate uu:

u=x    u=1u = x \implies u' = 1

Step 2: Differentiate vv:

v=(x8)(x+7)v = (x - 8)(x + 7) Using the product rule: v=(x8)(x+7)+(x8)(x+7)v' = (x-8)'(x+7) + (x-8)(x+7)' v=1(x+7)+(x8)(1)=(x+7)+(x8)=2x1v' = 1(x+7) + (x-8)(1) = (x+7) + (x-8) = 2x - 1

Step 3: Combine results:

P(x)=(1)((x8)(x+7))+x(2x1)P'(x) = (1)((x-8)(x+7)) + x(2x-1) P(x)=(x8)(x+7)+x(2x1)P'(x) = (x-8)(x+7) + x(2x-1)

Expand and simplify P(x)P'(x):

(x8)(x+7)=x2x56(x-8)(x+7) = x^2 - x - 56 P(x)=(x2x56)+(2x2x)P'(x) = (x^2 - x - 56) + (2x^2 - x) P(x)=3x22x56P'(x) = 3x^2 - 2x - 56

Critical Points:

To find the critical points, solve P(x)=0P'(x) = 0: 3x22x56=03x^2 - 2x - 56 = 0

This is a quadratic equation. Solve using the quadratic formula: x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} Here, a=3a = 3, b=2b = -2, c=56c = -56: x=(2)±(2)24(3)(56)2(3)x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(3)(-56)}}{2(3)} x=2±4+6726x = \frac{2 \pm \sqrt{4 + 672}}{6} x=2±6766x = \frac{2 \pm \sqrt{676}}{6} x=2±266x = \frac{2 \pm 26}{6}

Solutions: x=2+266=4.67,x=2266=4x = \frac{2 + 26}{6} = 4.67, \quad x = \frac{2 - 26}{6} = -4

Summary of critical points:

  • x=4x = -4
  • x4.67x \approx 4.67

These divide the x-axis into intervals for testing P(x)P'(x).

Would you like further analysis, such as the intervals of increase/decrease or a sketch of the graph?

Questions:

  1. How would the behavior of P(x)P(x) change if the degree were even instead of odd?
  2. What are the implications of a triple root in a polynomial function?
  3. How does the y-intercept change if a constant is added to P(x)P(x)?
  4. Can you determine the maximum/minimum values using the second derivative test?
  5. What is the relationship between the signs of the roots and the graph's orientation?

Tip:

Always check the multiplicity of roots—odd multiplicities indicate the graph crosses the x-axis, while even multiplicities mean it just touches the x-axis.

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Math Problem Analysis

Mathematical Concepts

Polynomial functions
Roots of polynomials
End behavior of functions
Critical points and derivatives

Formulas

Product rule for derivatives: (uv)' = u'v + uv'
Quadratic formula: x = (-b ± √(b² - 4ac)) / 2a

Theorems

Fundamental Theorem of Algebra
Behavior of polynomials based on degree and leading coefficient

Suitable Grade Level

Grades 10-12