1. More Nonlinear Functions and Equations
Sec 5.2 & 5.3
More Nonlinear Functions and Equations

2. Polynomial Functions
A polynomial function f of degree n in the variable x can be represented by
f(x) = anxn + an–1xn–1 +…+ a2x2 + a1x + a0,
where each coefficient ak is a real number, an ≠ 0, and n is a nonnegative integer. The leading coefficient is an and the degree is n.

3. Polynomial Functions
The domain of a polynomial function is all real numbers, and its graph is continuous and smooth without breaks or sharp edges.

4. Finding Extrema
Graphs of polynomial functions often have “hills” or “valleys”.
The “highest hill” on the graph is located at (–2, 12.7). This is the absolute maximum of g.
There is a smaller peak located at the point (3, 2.25). This is called the local maximum.

5. Finding Extrema
Maximum and minimum values that are either absolute or local are called extrema.
A function may have several local extrema, but at most one absolute maximum and one absolute minimum.

6. Finding Extrema
It is possible for a function to assume an absolute extremum at two values of x.
The absolute maximum is 11. It is a local maximum as well, because near x = –2 and x = 2 it is the largest y-value.

7. Absolute and Local Extrema
Let c be in the domain of f.
f(c) is an absolute (global) maximum if f(c) ≥ f(x) for all x in the domain of f.
f(c) is an absolute (global) minimum if f(c) ≤ f(x) for all x in the domain of f.
f(c) is an local (relative) maximum if f(c) ≥ f(x) when x is near c.
f(c) is an local (relative) minimum if f(c) ≤ f(x) when x is near c.

8. Even Symmetry
If a graph was folded along the y-axis, and the right and left sides match, then the graph would be
symmetric with respect to the y-axis. A function whose graph satisfies this characteristic is called an even function.

9. Even Function
A function f is an even function if f(–x) = f(x) for every x in its domain. The graph of an even function is symmetric with respect to the y-axis.

10. Odd Symmetry
Another type of of symmetry occurs in respect to the origin. If the graph could rotate, the original
graph would reappear after half a turn. This represents an odd function.

11. Odd Function
A function f is an odd function if f(–x) = –f(x) for every x in its domain. The graph of an odd function is symmetric with respect to the origin.

12. Identifying Odd and Even Functions
The terms odd and even have special meaning when they are applied to a polynomial function f. If f(x) contains terms that have only odd powers of x, then f is an odd function. Similarly, if f(x) contains terms that have only even powers of x (and possibly a constant term), then f is an even function.

13. Example: Identifying odd and even functions
Identify whether the function is odd, even, or neither.
Solution
The function defined by the table has domain D = {–3, –2, –1, 0, 1, 2, 3}. Notice that
f(–3) = 10.5 = f(3), f(–2) = 2 = f(2), and
f(–1) = –0.5 = f(1). The function f satisfies the statement f(–x) = f(x) for every x in D. Thus f is an even function. x –3 –2 –1 1 2 3
f(x)
10.5
–0.5

14. Example: Identifying odd and even functions
Identify whether the function is odd, even, or neither.
Solution
Since f is a polynomial containing only
odd powers of x, it is an odd function. This also can be shown symbolically as follows.

15. Graphs of Polynomial Functions
A polynomial function f of degree n can be expressed as
f(x) = anxn + … + a2x2 + a1x + a0,
where each coefficient ak is a real number, an  0, and n is a nonnegative integer.

16. Graphs of Polynomial Functions
A turning point occurs whenever the graph of a polynomial function changes from increasing to decreasing or from decreasing to increasing. Turning points are associated with “hills” or “valleys” on a graph:
(–2, 8) and (2, –8)

17. Constant Polynomial Function
f(x) = a
Has no x-intercepts or turning points

18. Linear Polynomial Function
f(x) = ax + b
Degree 1 and one x-intercept and no turning points.

19. Quadratic Polynomial Functions
f(x) = ax2 + bx + c
Degree 2, parabola that opens up or down. Can have zero, one or two x-intercepts. Has exactly one turning point, which is also the vertex.

20. Cubic Polynomial Functions
f(x) = ax3 + bx2 + cx + d
Degree 3, can have up to 3 x-intercepts and zero or two turning points.

21. Quartic Polynomial Functions
f(x) = ax4 + bx3 + cx2 + dx + e
Degree 4, can have up to four x-intercepts and three turning points.

22. Quintic Polynomial Functions
f(x) = ax5 + bx4 + cx3 + dx2 + ex + k
Degree 5, may have up to five x-intercepts and four turning points.

23. Degree, x-intercepts, and Turning Points
The graph of a polynomial function of degree n, with n ≥ 1, has at most n
x-intercepts and at most n – 1 turning points.

24. Example: Analyzing the graph of a polynomial function
Use the graph of the polynomial function f
shown.
a) How many turning points and x-intercepts are there?
b) Is the leading coefficient
a positive or negative? Is the degree odd or even?
c) Determine the minimum degree of f.

25. Example: Analyzing the graph of a polynomial function
Solution
a) There are three turning points corresponding to the one “hill” and two “valleys”.
There appear to be 4 x-intercepts.

26. Example: Analyzing the graph of a polynomial function
b) The left side rises and the right side falls. Therefore, a < 0 and the polynomial function has odd degree.
c) The graph has four turning points. A polynomial of degree n can have at most n  1 turning points. Therefore, f must be at least degree 5.

27. Example: Analyzing the graph of a polynomial function
Graph f(x) = x3  2x2  5x + 6, and then
complete the following.
a) Identify the x-intercepts.
b) Approximate the coordinates of any turning points to the nearest hundredth.
c) Use the turning points to approximate any local extrema.

28. Example: Analyzing the graph of a polynomial function
Solution
a) The graph appears to intersect the x-axis at the points (–2,0), (1, 0), and (3, 0). The x-intercepts are –2, 1, and 3.

29. Example: Analyzing the graph of a polynomial function
b) There are two turning points. One has coordinates approximately (–0.79, 8.21).

30. Example: Analyzing the graph of a polynomial function
b) The other has coordinates approximately (2.12, –4.06).

31. Example: Analyzing the graph of a polynomial function
c) There is a local maximum of about 8.21 and a local minimum of about –4.06.

32. Example: Analyzing the end behavior of a graph
Let f(x) = 2 + 3x – 3x2  2x3.
a) Give the degree and leading coefficient.
b) State the end behavior of the graph of f.
Solution
a) The term with the highest degree is –2x3 so the degree is 3 and the leading coefficient is –2.

33. Example: Analyzing the end behavior of a graph
b) The degree is odd and the leading
coefficient is negative. The graph of f rises to the left and falls to the right. More formally,

34. Example: Evaluating a piecewise-defined polynomial function
Evaluate f(x) at 3, –2, 1, and 2.
Solution
To evaluate f(3) we use the formula x2 – x because 3 is the interval 5 ≤ x < –2.
f(3) = (3)2 – (3) = 12

35. Example: Evaluating a piecewise-defined polynomial function
To evaluate f(2) we use the formula x3 because 2 is in the interval –2 ≤ x < 2.
f(2) = (2)3 = –(–8) = 8
Similarly, f(1) = –13 = –1
To evaluate f(2) we use the formula 4  4x because 2 is in the interval 2 ≤ x ≤ 5.
f(2) = 4  4(2) = 4  8 = 4

36. Example: Graphing a piecewise-defined polynomial function
Complete the following.
a) Sketch the graph of f.
b) Determine if f is continuous on its domain.
c) Evaluate f(1).

37. Example: Graphing a piecewise-defined polynomial function
Solution
a) Sketch the graph of f.
Graph
on the interval –4 ≤ x ≤ 0
Graph y = 2x – 2 between
the points (0, –2) and (2, 2).
Graph y = 2 from the points (2, 2) to (4, 2).

38. Example: Graphing a piecewise-defined polynomial function
b) The domain is –4 ≤ x ≤ 4
c) The x-coordinates for the points of
intersection can be found by solving 2x – 2 = 1 and
Solutions are:

39. Factor Theorem
A polynomial f(x) has a factor x – k if and only if f(k) = 0.

40. Example: Applying the factor theorem
Use the graph and the factor theorem to list the factors of f(x).
Solution
The zeros or x-intercepts of f are 2, 1 and 3. Since f(2) = 0, the factor theorem states that (x + 2) is a factor, and f(1) = 0 implies that
(x  1) is a factor and f(3) = 0 implies (x  3) is a factor. Thus the factors are (x + 2), (x  1), and (x  3).

41. Zeros with Multiplicity
The polynomial f(x) = x2 + 4x + 4 can be written as f(x) = (x + 2)2. Since the factor (x + 2) occurs twice in f(x), the zero –2 is called a zero of multiplicity 2.
The polynomial g(x) = (x + 1)3(x – 2) has zeros –1 and 2 with multiplicities 3 and 1, respectively: x-intercepts coincide with zeros of g.

42. Complete Factored Form
Suppose a polynomial
f(x) = anxn + … + a2x2 + a1x + a0
has n real zeros c1, c2, c3, …, cn, where a distinct zero is listed as many times as its multiplicity. Then f(x) can be written in complete factored form as
f(x) = an(x – c1)(x – c2)(x – c3) …(x – cn)

43. Example: Finding a complete factorization
Write the complete factorization for the polynomial 7x3 – 21x2 – 7x + 21 with given zeros –1, 1 and 3.
Solution
Leading coefficient is 7. Zeros are –1, 1 and 3
The complete factorization:
f(x) = 7(x + 1)(x – 1)(x – 3).

44. Example: Factoring a polynomial graphically
Use the graph of f to factor f(x) = 2x3 – 4x2 – 10x + 12.
Solution
Leading coefficient is 2
Zeros are –2, 1 and 3
The complete factorization:
f(x) = 2(x + 2)(x – 1)(x – 3).

45. Graphs and Multiple Zeros
The polynomial f(x) = 0.02(x + 3)3(x – 3)2 has zeros –3 and 3 with multiplicities 3 and 2, respectively. At the zero of even
multiplicity the graph does not cross the x-axis (3, 0), whereas the graph does cross the x-axis at the zero of odd multiplicity (–3, 0).

46. Polynomial Equations
Factoring can be used to solve polynomial equations with degree greater than 2.
We can also solve the graphically.

47. Example: Solving a polynomial equation
Find all real solutions to each equation symbolically.
4x4 – 5x2 – 9 = 0
Solution

48. Example: Solving a polynomial equation
The only real solutions are

49. Example: Solving a polynomial equationb.
The solutions are –2, , and 2.

50. Example: Finding the solution graphically
Solve the equation x3 – 2x – 4 = 0 graphically. Round any solutions to the nearest hundredth.
Solution
Since there is only one x-intercept the equation has one real solution: x  2.65