Mathematics: Polynomial and Rational Functions

Polynomial Functions

  • Function notation:
    f(x)=a<em>nxn+a</em>n1xn1++a<em>1x+a</em>0f(x) = a<em>n x^n + a</em>{n-1} x^{n-1} + … + a<em>1 x + a</em>0
  • Degree (n):
    • If n is odd:
    • Leading coefficient a_n > 0 indicates the graph rises to the right
    • Leading coefficient a_n < 0 indicates the graph falls to the right
    • If n is even:
    • Leading coefficient a_n > 0 indicates the graph rises to the right
    • Leading coefficient a_n < 0 indicates the graph falls to the right

Intervals and Sign Analysis

  • Consider logarithmic aspects and sign intervals of the functions:

    • Example intervals:

    • (ext,3)(- ext{∞}, -3), (3,3)(-3, 3)

    • Analyze behavior as x approaches critical points:

    • When evaluating points such as x approaching -3 or 3, check the sign of the function or factors (e.g. (x3)(x+3)(x-3)(x+3)) to determine intervals where the function is positive or negative.

Analyzing Functions with Asymptotes

  • Vertical Asymptotes: Locations where the function becomes undefined, for example:
    • At x=3x = -3, x=3x = 3
  • Horizontal Asymptote: Determines the value the function approaches as xx approaches infinity, example:
    • At y=0y = 0

Function Evaluation

  • To find specific points of interest such as intercepts:
    • f(0)=9f(0) = -9 helps identify the value of the function at zero
    • Roots and intercepts are crucial; for example, setting f(x)=0f(x) = 0 gives roots

Rational Functions and Their Behavior

  • Analyzing rational functions of the form f(x)=N(x)D(x)f(x) = \frac{N(x)}{D(x)} where N and D are polynomials. Considerations:
    • Finding Roots: Set numerator to zero
    • Finding Vertical Asymptotes: Set denominator to zero
  • Example polynomial factoring:
    • x22x15=(x5)(x+3)x^2 - 2x - 15 = (x - 5)(x + 3)

Partial Fraction Decomposition

  • Breaking down complex rational functions to simpler components:

    • For example, find constants A and B if
      5x1(x5)(x+3)=Ax5+Bx+3\frac{5x - 1}{(x - 5)(x + 3)} = \frac{A}{x - 5} + \frac{B}{x + 3}
  • Solve for A and B by substituting and forming a system of equations resulting from matching coefficients:

    1. A+B=5A + B = 5
    2. 5A+3B=1-5A + 3B = -1
  • Example solution yielding:

    • A = 2, B = 3