Math Problem Statement

ate a polynomial p with the following attributes. • As x → −∞, p(x) → ∞.

• The point (−1 , 0) yields a local maximum. • The degree of p is 5. • The point (9, 0) is one of the x-intercepts of the graph of p.

Solution

To create a polynomial p(x)p(x) that satisfies the given conditions, we need to consider each requirement step by step:

  1. Degree of the polynomial is 5: This means the polynomial will have the general form p(x)=a(xr1)(xr2)(xr3)(xr4)(xr5)p(x) = a(x - r_1)(x - r_2)(x - r_3)(x - r_4)(x - r_5) where r1,r2,r3,r4,r5r_1, r_2, r_3, r_4, r_5 are the roots of the polynomial, and aa is the leading coefficient.

  2. As xx \to -\infty, p(x)p(x) \to \infty: For this behavior, the leading coefficient aa must be negative since a polynomial of odd degree (5 in this case) will go to opposite infinities on either end. Specifically, if a<0a < 0, then p(x)p(x) will tend to \infty as xx \to -\infty.

  3. The point (1,0)(-1, 0) yields a local maximum: This implies that x=1x = -1 is both a root and a critical point where the derivative changes sign from positive to negative. Thus, (x+1)2(x + 1)^2 should be a factor of the polynomial, ensuring that x=1x = -1 is a double root and hence a local maximum.

  4. The point (9,0)(9, 0) is one of the x-intercepts: This implies x=9x = 9 is a root of the polynomial.

Given these conditions, the polynomial can be expressed as: p(x)=a(x+1)2(x9)(xr1)(xr2)p(x) = -a(x + 1)^2(x - 9)(x - r_1)(x - r_2) where r1r_1 and r2r_2 are the remaining roots.

To fully specify p(x)p(x), we could assume additional roots or constraints, but based on the information given, a simple version with the remaining roots at x=0x = 0 and x=2x = 2 (just as an example) could be: p(x)=1(x+1)2(x9)x(x2)p(x) = -1(x + 1)^2(x - 9)x(x - 2)

This polynomial has the following properties:

  • Degree 5 (since it is the product of five factors).
  • As xx \to -\infty, p(x)p(x) \to \infty because of the negative leading coefficient.
  • x=1x = -1 is a double root, ensuring a local maximum at this point.
  • x=9x = 9 is an x-intercept.

Final Polynomial:

p(x)=(x+1)2(x9)x(x2)p(x) = -(x + 1)^2(x - 9)x(x - 2)

Do you want any more details or have any questions?

Related Questions:

  1. How does the sign of the leading coefficient affect the end behavior of a polynomial?
  2. What is the significance of a double root in a polynomial?
  3. How can you determine the number of local maxima and minima in a polynomial?
  4. What happens to the graph of a polynomial at a point where it has an odd multiplicity root?
  5. How do we derive the derivative of a polynomial and use it to find critical points?

Tip:

When constructing polynomials with specific behaviors, it's crucial to consider the multiplicity of roots and the sign of the leading coefficient, as these significantly influence the shape and behavior of the graph.

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Math Problem Analysis

Mathematical Concepts

Polynomials
End Behavior
Roots and Intercepts
Critical Points

Formulas

General form of a polynomial p(x) = a(x - r_1)(x - r_2)...(x - r_n)

Theorems

Fundamental Theorem of Algebra
Multiplicity of Roots

Suitable Grade Level

Grades 10-12