Download 6.7/6.8 Analyzing Graphs of Polynomials

6.7/6.8 Analyzing Graphs of
Polynomials
• How do you find the local maximum and
minimum on a polynomial graph?
• What is the maximum number of turning points
based on the degree of polynomial?
• How do you find the equation of polynomial of
the least degree given the x-intercepts and
another point on the graph?
The Fundamental Theorem of
Algebra
If f(x) is a polynomial of degree n where n > 0,
then the equation has at least one root in the set
of complex numbers.
This means that the degree of polynomial will tell
you the number of solutions to look for. Some of
the solutions may be repeating solutions.
Zeros, Factors, Solutions, and
Intercepts
If f(x) is a polynomial function, then these
statements are equivalent.
• Zero: k is a zero of the polynomial.
• Factor: x – k is a factor of the polynomial.
• Solution: k is a solution of the polynomial
equation f(x).
• Intercept: If k is a real number then k is an
x-intercept of the graph of the polynomial.
Turning Points of Polynomial
Functions
The graph of every polynomial function of degree n
has at most n – 1 turning points. Moreover, if a
polynomial function has n distinct real zeros, then its
graph has exactly n -1 turning points.
f ( x)  x 3  3x 2  1x  3
f ( x)  x 3  x 2  x  1
Local Maximum and Minimum
• The y-coordinate of a turning point is a local maximum
of the function if the point is higher than all nearby
points.
• The y-coordinate of a turning point is a local minimum
of the funct if the point is lower than all nearby points.
Local maximum
Local minimum
Writing a Cubic Function
Remember:
• Two points determine a line.
• Three points determine a
parabola
• Four points determine a
cubic function
Given Three Intercepts and a
Fourth Point on the Graph
Example: Given x-intercepts (-2, 0), (-1, 0), (1, 0) and a fourth point on the
graph (0, 2), find the equation of a polynomial with the least degree.
1. Change the x-intercept to factor form. Write the polynomial
function in factor form with the leading coefficient as a.
f ( x )  a ( x  2)( x  1)( x  1)
2. Substitute the x and y coordinates of the fourth point into
the equation.
2  a (0  2)(0  1)(0  1)
3. Solve for a.
2  2a
2  a
4. Substitute a into the equation.
f ( x )  1( x  2)( x  1)( x  1)
f ( x)   x3  2 x 2  x  2