X squared asks x cubed if he is

• X squared asks x cubed if he is a religious variable • I do believe in higher powers, if that’s what you mean. • student notes MADE for 2 -2 and 2 -3 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 1

Section 2. 2 Polynomial Functions of Higher Degree

Warm up - Graph this business ! h c Psy Copyright © by Houghton

A polynomial function is a function of the form where n is a nonnegative integer and each ai (i = 0, , n) is a real number. The polynomial function has a leading coefficient an and degree n. Examples: Find the leading coefficient and degree of each polynomial function. Polynomial Function Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Leading Coefficient Degree -2 5 1 3 14 0 4

Graphs of polynomial functions are continuous. That is, they have no breaks, holes, or gaps. f (x) = x 3 – 5 x 2 + 4 x + 4 y y x continuous smooth polynomial y x not continuous not polynomial x continuous not smooth not polynomial Polynomial functions are also smooth with rounded turns. Graphs with points or cusps are not graphs of polynomial functions. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 5

Polynomial functions of the form f (x) = x n, n 1 are called power functions. 5 f (x) = x 4 y f (x) = x 2 f (x) = x 3 x x What do you think even functions look like? If n is even, their graphs resemble the graph of f (x) = x 2. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. What do you think odd functions look like? If n is odd, their graphs resemble the graph of f (x) = x 3. 6

Ex 1: Sketch the graph of f (x) = – (x + 2)4. Reflect the graph of y = x 4 in the x-axis. Then shift the graph two units to the left. y y = x 4 x f (x) = – (x + 2)4 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. y = – x 4 7

Leading Coefficient Test As x grows positively or negatively without bound, the value f (x) of the polynomial function f (x) = anxn + an – 1 xn – 1 + … + a 1 x + a 0 (an 0) grows positively or negatively without bound depending upon the sign of the leading coefficient an and whether the degree n is odd or even. y y an positive x x n odd an negative Copyright © by Houghton Mifflin Company, Inc. All rights reserved. n even 8

Ex 2: Describe the right-hand left-hand behaviour for the graph of f(x) = – 2 x 3 + 5 x 2 – x + 1. Degree Leading Coefficient As , 3 Odd -2 Negative and as , y x f (x) = – 2 x 3 + 5 x 2 – x + 1 Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 9

A real number a is a zero of a function f (x) if and only if f (a) = 0. Real Zeros of Polynomial Functions If a is a zero of a function f(x), then the following are true: 1. a is a solution of the polynomial equation f (x) = 0. 2. x – a is a factor of the polynomial f (x). 3. (a, 0) is an x-intercept of the graph of y = f (x). A turning point of a graph of a function is a point at which the graph changes from increasing to decreasing or vice versa. A polynomial function of degree n has at most n – 1 turning points and at most n zeros. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 10

Repeated Zeros For any factor (x – a) k of f (x) a is a zero of the function and k is called the multiplicity. 1. If k is odd the graph of f (x) crosses the x-axis at (a, 0). 2. If k is even the graph of f (x) touches, but does not cross through, the x-axis at (a, 0). Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 11

Repeated Zeros For any factor (x – a) k of f (x) a is a zero of the function and k is called the multiplicity. 1. If k is odd the graph of f (x) crosses the x-axis at (a, 0). 2. If k is even the graph of f (x) touches, but does not cross through, the x-axis at (a, 0). Ex 3: Determine end behavior and the multiplicity of the zeros, the graph: f (x) = (x – 2)3(x +1)4. As Zero 2 – 1 , and as Multiplicity Behavior crosses x-axis 3 odd at (2, 0) touches x-axis 4 even at (– 1, 0) Copyright © by Houghton Mifflin Company, Inc. All rights reserved. , y x 12

Now graph this business Copyright © by Houghton Mifflin Company, Inc. All rights reserved.

Ex 4: Find the end behavior, all the real zeros and turning points of the graph of f (x) = x 4 – x 3 – 2 x 2. Factor completely: f (x) = x 4 – x 3 – 2 x 2 = x 2(x + 1)(x – 2). The real zeros are x = – 1, x = 0, and x = 2. y These correspond to the x-intercepts (– 1, 0), (0, 0) and (2, 0). The graph shows that there are three turning points. Since the degree is four, this is Turning point the maximum number possible. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. Turning point x Turning point f (x) = x 4 – x 3 – 2 x 2 14

Ex 5: Sketch the graph of f (x) = 4 x 2 – x 4. 1. Write the polynomial function in standard form: f (x) = –x 4 + 4 x 2 The leading coefficient is negative and the degree is even. as , 2. Find the zeros of the polynomial by factoring. f (x) = –x 4 + 4 x 2 = –x 2(x 2 – 4) = – x 2(x + 2)(x – 2) Zeros: x = – 2, 2 multiplicity 1 x = 0 multiplicity 2 x-intercepts: (– 2, 0), (2, 0) crosses through (0, 0) touches only y (– 2, 0) (2, 0) x (0, 0) Example continued Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 15

Ex 5 continued: Sketch the graph of f (x) = 4 x 2 – x 4. 3. Since f (–x) = 4(–x)2 – (–x)4 = 4 x 2 – x 4 = f (x), the graph is symmetrical about the y-axis. 4. Plot additional points and their reflections in the y-axis: (1. 5, 3. 9) and (– 1. 5, 3. 9 ), ( 0. 5, 0. 94 ) and (– 0. 5, 0. 94) y 5. Draw the graph. (– 1. 5, 3. 9 ) (– 0. 5, 0. 94 ) Copyright © by Houghton Mifflin Company, Inc. All rights reserved. (1. 5, 3. 9) (0. 5, 0. 94) x 16

Ex 6: Find the zeros and determine their multiplicity g(t) = 5 t – 3 6 t + 9 t g(t) = t 5 – 6 t 3 + 9 t g(t) = t(t 4 – 6 t 2 + 9) g(t) = t(t 2 – 3)2

The Intermediate Value Theorem • Let a and b be real numbers such that a < b. • If f is a polynomial function such that f(a) ≠ f(b) then in the interval [a, b], f takes on every value between f(a) and f(b).

The Intermediate Value Theorem • Think of an elevator: – If it reaches the 2 nd floor and the 30 th floor, it must reach every floor in between the 2 nd and 30 th.

The Intermediate Value Theorem

Why do we care? Since f(1)= -6 f(3) = 4 We know that for 1< x < 3 There is an x such that f(x)=0

1 -8 A Matching 1. 2. 3. 4. A 5. 6. 7. 8. f(x) = -2 x + 3 f(x) = x 2 – 4 x f(x) = -2 x 2 – 5 x f(x) = 2 x 3 – 3 x + 1 f(x) = -¼ x 4 + 3 x 2 f(x) = -1/3 x 3 + x 2 – 4/3 f(x) = x 4 + 2 x 3 f(x) = 1/5 x 5 -2 x 3 + 9/5 x

1 -8 B Matching 1. 2. 3. 4. 5. 6. 7. B 8. f(x) = -2 x + 3 f(x) = x 2 – 4 x f(x) = -2 x 2 – 5 x f(x) = 2 x 3 – 3 x + 1 A f(x) = -1/3 x 3 + x 2 – 4/3 f(x) = x 4 + 2 x 3 f(x) = 1/5 x 5 -2 x 3 + 9/5 x

1 -8 C Matching C 1. 2. 3. 4. 5. 6. 7. 8. f(x) = -2 x + 3 f(x) = x 2 – 4 x f(x) = -2 x 2 – 5 x f(x) = 2 x 3 – 3 x + 1 A f(x) = -1/3 x 3 + x 2 – 4/3 f(x) = x 4 + 2 x 3 B

1 -8 D Matching 1. 2. 3. 4. 5. 6. D 7. 8. C f(x) = x 2 – 4 x f(x) = -2 x 2 – 5 x f(x) = 2 x 3 – 3 x + 1 A f(x) = -1/3 x 3 + x 2 – 4/3 f(x) = x 4 + 2 x 3 B

1 -8 E Matching 1. 2. 3. 4. 5. E 6. 7. 8. C f(x) = x 2 – 4 x f(x) = -2 x 2 – 5 x f(x) = 2 x 3 – 3 x + 1 A f(x) = -1/3 x 3 + x 2 – 4/3 D B

1 -8 F Matching 1. 2. 3. F 4. 5. 6. 7. 8. C f(x) = x 2 – 4 x f(x) = -2 x 2 – 5 x f(x) = 2 x 3 – 3 x + 1 A E D B

1 -8 G Matching 1. G 2. 3. 4. 5. 6. 7. 8. C f(x) = x 2 – 4 x f(x) = -2 x 2 – 5 x F A E D B

1 -8 H Matching 1. 2. H 3. 4. 5. 6. 7. 8. C G f(x) = -2 x 2 – 5 x F A E D B

H Dub • 2. 2 Page 148 #1 -8 all, 13 -21 ODD, 2741 ODD (part c by hand – no calculators!!!), 65, 66, 92