NOTES Section 2.3 Day 1 Polynomial Functions and Their Graphs

Polynomial Functions –graphs are continuous without breaks, gaps, sharp corners, or overlap

Smooth and Continuous… Polynomial

Not continuous; has a gap… NOT Polynomial

Continuous but has sharp corners… NOT Polynomial

Zeros and Multiplicity: Let f be a polynomial function, and let a be a real number… If 0)( =af , then 1. a is a solution of the polynomial equation 0)( =xf 2. a is a root or zero of the function f 3. )0,(a is an x-intercept of the graph of f 4. )( ax - is a factor of the polynomial If )( ax - is a factor that occurs k times, then a is called a zero with multiplicity k. If k is odd, then the graph crosses the x- axis at a. If k is even, then the graph touches the x- axis at a. Example 1: Check out 34 )2()3)(1( +-- xxx End behavior & Leading Coefficient Test: The end behavior of the graph depends on the sign of the leading coefficient and the overall degree of the polynomial.

Odd Degree Even Degree Sign of Leading Coefficient Positive Negative Positive Negative End Behavior

Example 2: Describe (or sketch) right and left end behavior of the graphs of each of the following equations. A. 62 431)( xxxf --= B. xxxf 9)( 3 -= Intercepts: The graph of a polynomial function of degree n has one y-intercept and at most n x-intercepts. Turning Points/Local Extrema/peaks & valleys: A polynomial function of degree n has at most n – 1 turning points (one less than its degree).

o ft t

yexd 6 8a Ewe o 4I

a 3factorsX I 11 3 X 2 Zeros

y x2V2

y x

A3

off 93 J f

y X

quad V cubic µ quartic WN

Intermediate Value Theorem says if you can find a value x = a at which f is positive and another value x = b at which f is negative, you can conclude that f has at least one real zero between a and b.

In the case at the left, I went from a negative f(x) to a positive f (x) between –2 and –1 The only way I can go from a negative y to a positive y value is to cross the x-axis. Because of this, I know that there is an x value in the interval (–2, –1) that gives me a zero.

Example 6: Use the Intermediate Value Theorem to determine if f(x) = 4x3 – 7x2 – 21x + 17 has a real zero

a) between x = 0 and x = 1

b) between x = 1 and x = 2

Example 3: What is the maximum number of x–intercepts (zeros) and turning points for the functions below?

a) 549)( 23 ---= xxxxf

b) 19)( 2 += xxf

c) 59)( += xxf d) 5341756)( 24 -+-= xxxxf

Example 4: State the zeros in the graph below. Then give a possible factorization for the graph below. Use a leading coefficient of 1 or –1. Use the smallest degree possible. Leave in factored form.

Example 5: Write a possible equation in polynomial form for the situations below. A. degree of 2, zeros are 0 & 5 B. degree of 3, zeros of 4 (multiplicity 1) & -3 (multiplicity 2)

degree 3 2 I 4xint 3 2 I 4TP 2 I 0 3

www

zeros X 2 X I 11 3factors xt29 X 1 x 3 y

mum

Try4 0 4 5 21 4 X 3

2x o x s g x 4 its yx x 5 y x 4 Lx2t6xt9 y

6yX3t2n215x

3

y X2 y x3 62 9 4 2 244 3

f 1,1K int

4 3

flat Fan 21 yes since yvalues

go from pos tonegthere must be an X int

f II LIE Hmmmm mnbetween x otx I