UNIT POLYNOMIAL FUNCTIONS Date Lesson Text TOPIC Homework Sept. 0.0.0 Factoring Review WS.0 Sept. 1.1.5 Dividing Polnomials Pg. 168 #,, 4, (5 10)doso, 11, 1 Sept...6 Factoring Polnomials Factor and Remainder Theorems Pg. 176 # (1 7)doso, 9, 10, 1, 14 Sept. 6..7 Factoring a Sum or Difference of Cubes Pg. 18 # 1-5 Sept. 7.4.1 Exploring Polnomial Functions Investigation FINISH INVESTIGATION Sept. 8.4-II Work Period for Lesson.4 WS.4 Pg. 17 # 1 5, 7 Sept. 9.5. Characteristics of Polnomial Functions Part I Odd/Even, Leading Coefficient, End Behaviour, Turning Points Pg. 16 # 1 6 Sept. 0.5 (II). Characteristics of Polnomial Functions Part II continued Pg. 17 # 7 16 Oct..6. Characteristics of Polnomial Functions in Factored Form Families of Polnomial Functions Pg. 146 # 1 6, 8, 9, 1, 1 Oct. 4.7.4 Transformations of Cubic and Quartic Functions Pg. 155 # 1 5, 6doso, 7, 10 Oct. 5.8 Review for Unit Test Pg. 184 # 1, (4, 5)a, 6, 7, 8be, 9bd, 10bd, 1d, 1, 14bc, 15ac, 17, 18 Oct. 6.9 UNIT TEST
MHF 4U Lesson.0 Review of Factoring Ex. Factor each of the following completel. a) 4 4p 8p 5p b) x x 10 c) 50 8b d) a 7ab 18b e) x x xz z f) x 7x 15 g) 8x 10x h) 4a 8ab 49b i) x 4 1x 6 j) 4( a b) c 4 k) 16x 10 5 WS.0
MHF 4U Lesson.1 Dividing Polnomials Dividing polnomials can be done in more than one wa. It is important to use the most efficient wa in order to solve problems in the simplest manner. I. Dividing b Factoring. x 11x 6 x 6 Whenever it is possible to divide polnomials b factoring, it is the simplest and most efficient wa to solve the problem. Failing to recognize this will cause ou to waste time and effort on a more inefficient method of solving the problem. II. Dividing using Long Division Ex. 1 a) Divide x x 8x 1 b x 1 and express our answer in quotient form. b) State an restrictions on the variable. c) Write a corresponding statement that can be used to check the division. d) Verif our answer.
Ex. a) Divide 4x 9x 1 b x 1 and express our answer in quotient form. b) Write a corresponding statement that can be used to check the division. Ex. The volume, V, in cubic centimeters, of a rectangular box is given b V ( x) x 7x 14x 8. Determine expressions for possible dimensions of the box if the height, h, in centimeters is given b x.
Ex. Divide each of the following using snthetic division. a) ( x ) ( x ) x b) (x 5x 9) ( x ) c) (8 4x 6x ) (x 1) x d) ( x 1) ( x 1) Pg. 168 #,, 4, (5 10)doso, 11, 1
MHF 4U Lesson. The Factor and Remainder Theorems The Remainder Theorem The remainder theorem states: When a polnomial f(x), is divided b x a, the remainder is equal to f(a). Ex. 1 a) Given: f(x) = x x x evaluate each of the following. (i) f() (ii) f(-) (iii) f(1) b) Divide b each of the following. (i) x (ii) x (iii) x 1 The Factor Theorem From the remainder theorem, we have seen that the remainder can be found b determining the value of f(a). B extrapolating, we can determine that if the remainder is zero, the function is evenl divisible b the divisor. The factor theorem states: x a is a factor of f(x) if and onl if (iff) f(a) = 0. Ex. Determine whether or not x is a factor of f ( x) x x x 4.
Ex. Factor completel. 4 a) x x 5x 6 b) x x 7x 7 x 18 c) x x 7x
Ex. 4 When x mx nx is divided b x 1the remainder is -1 and x is a factor. Determine the values of m and n. Ex. 5 If when x 4x kx 5 is divided b x the remainder is 7, what is the value of k? Pg. 176 # (1 7)doso, 9, 10, 1, 14
MHF 4U Lesson. Sum and Difference of Cubes A sum or difference of cubes is in the form a b or a b. Ex. 1 Factor a b using the factor theorem. If we use the factor theorem on a b, we can see that a b = ( a b)( a ab b ). Ex. Factor each of the following completel. a) x 64 b) x 81 c) 8x 1 d) 64x 1 7
e) 8 15 6 x 64 f) 15 x 16 15 6 g) ( x 1) 16 h) ( x ) ( ) Pg. 18 # 1-5
MHF 4U INV.4 Exploring Polnomial Functions Investigation n n 1 n Polnomial functions are functions in the form f ( x) ax bx cx..., where a, b, c, are real numbers and each exponent is a WHOLE number. 0 18 16 Equation in expanded form: 14 1 10 8 6 4 5 4 1 1 4 5 x 4 6 Degree: Sign of Leading Coefficient (SOLC): Number of real roots: Starts in Q, Ends in Q. Number of Turning Points: 8 10 1 14 16 18 0 10 9 8 7 6 5 4 1 Equation in expanded form: Degree: Sign of Leading Coefficient (SOLC): Number of real roots: 5 4 1 1 1 4 5 x 4 Starts in Q, Ends in Q. Number of Turning Points: 5 6 7 8 9 10
10 9 8 Equation in expanded form: 7 6 5 4 1 5 4 1 1 1 4 5 x Degree: Sign of Leading Coefficient (SOLC): Number of real roots: Starts in Q, Ends in Q. Number of Turning Points: 4 5 6 7 8 9 10 10 9 8 7 6 5 4 1 Equation in expanded form: Degree: Sign of Leading Coefficient (SOLC): Number of real roots: 5 4 1 1 1 4 5 x 4 Starts in Q, Ends in Q. Number of Turning Points: 5 6 7 8 9 10
10 9 8 Equation in expanded form: 7 6 5 4 1 5 4 1 1 1 4 5 x Degree: Sign of Leading Coefficient (SOLC): Number of real roots: Starts in Q, Ends in Q. Number of Turning Points: 4 5 6 7 8 9 10 60 54 48 4 6 0 4 18 1 6 Equation in expanded form: Degree: Sign of Leading Coefficient (SOLC): Number of real roots: 5 4 1 1 4 5 x 6 1 18 4 Starts in Q, Ends in Q. Number of Turning Points: 0 6 4 48 54 60
10 9 8 Equation in expanded form: 7 6 5 4 1 5 4 1 1 1 4 5 x Degree: Sign of Leading Coefficient (SOLC): Number of real roots: Starts in Q, Ends in Q. Number of Turning Points: 4 5 6 7 8 9 10 10 9 8 7 6 5 4 1 Equation in expanded form: Degree: Sign of Leading Coefficient (SOLC): Number of real roots: 5 4 1 1 1 4 5 x 4 Starts in Q, Ends in Q. Number of Turning Points: 5 6 7 8 9 10
0 18 16 Equation in expanded form: 14 1 10 8 6 4 5 4 1 1 4 5 x 4 6 Degree: Sign of Leading Coefficient (SOLC): Number of real roots: Starts in Q, Ends in Q. Number of Turning Points: 8 10 1 14 16 18 0 150 15 10 105 90 75 60 45 0 15 Equation in expanded form: Degree: Sign of Leading Coefficient (SOLC): Number of real roots: 5 4 1 1 4 5 x 15 0 45 60 Starts in Q, Ends in Q. Number of Turning Points: 75 90 105 10 15 150
50 45 40 Equation in expanded form: 5 0 5 0 15 10 5 5 4 1 1 4 5 x 5 10 15 Degree: Sign of Leading Coefficient (SOLC): Number of real roots: Starts in Q, Ends in Q. Number of Turning Points: 0 5 0 5 40 45 50 50 45 40 5 0 5 0 15 10 5 Equation in expanded form: Degree: Sign of Leading Coefficient (SOLC): Number of real roots: 5 4 1 1 4 5 x 5 10 15 0 Starts in Q, Ends in Q. Number of Turning Points: 5 0 5 40 45 50
10 9 8 7 6 5 4 1 Equation in expanded form: Degree: Sign of Leading Coefficient (SOLC): Number of real roots: 5 4 1 1 1 4 5 x 4 Starts in Q, Ends in Q. Number of Turning Points: 5 6 7 8 9 10 HW: FINISH INVESTIGATION
MHF 4U INV.5 (Part I) Graphs of Polnomial Functions 1. Sketch each of the following.
. What do ou notice about all graphs that have: a) Odd degree and negative leading coefficient b) Odd degree and positive leading coefficient c) Even degree and negative leading coefficient d) Even degree and positive leading coefficient. Predict that general characteristics of the graph of a function that has: a) a degree of 5 and a negative leading coefficient b) a degree of 6 and a positive leading coefficient c) check our predictions b sketching one possible graph from parts a) and b). MULTIPLICITY Pg. 16 # 1-6
MHF 4U Inv..5 (Part II) Graphs of Polnomial Functions 1. This time tr to complete the table before sketching the curves. Polnomial Odd or Even Degree Sign of Leading Coefficient Number of Turning Points Number of real zeros End Behaviour x x e) ( x 1)( x 4) f) ( x 1) g) ( x ) ( x ) h) x( x )( x 4) i) ( x 1)( x ) j) ( x 1) ( x ) k) ( x )( x ) l) x( x 1)( x ) m) ( x )( x 4) n) ( x ) ( x 4) o) ( x ) ( x 4) p) x( x 1)
. Now use the information above to sketch the above functions. Pa close attention to the end behaviours and to the zeros (x-intercepts). Use the graphing calculators to check our graphs onl.
Now we will examine the multiplicit and behavior of each zero in the previous functions. Polnomial x int Multiplicit & Behaviour x int Multiplicit & Behaviour x int Multiplicit & Behaviour a) x b) x 4 c) ( x 1) d) ( x )( x ) e) ( x 1)( x 4) f) ( x 1) g) ( x ) ( x ) h) x( x )( x 4) i) ( x 1)( x ) j) ( x 1) ( x ) k) ( x )( x ) l) m) n) o) x( x 1)( x ) ( x )( x 4) ( x ) ( x 4) ( x ) ( x 4) p) x( x 1) 5. Given the polnomial = (x + 1) (x )(x )(x 9), determine with the help of the tables above, but without the use of graphing technolog: a) the quadrants in which the graph originates/terminates. b) the zeros of the function. c) the x-intercepts of the function. d) the -intercept of the function. e) the behaviour of the graph at each of the x -intercepts. f) Sketch the graph of the function. Pg 17 # 7-16
SUMMARY Graphs of Polnomial Functions Functions with an odd degree When the leading coefficient is positive, the graph When the leading coefficient is negative, the graph extends from the rd quadrant to the 1 st quadrant. extends from the nd quadrant to the 4 th quadrant. Opens up to the right Opens down to the right * All cubic functions are smmetrical about a point. Functions with an even degree When the leading coefficient is positive, the graph When the leading coefficient is negative, the graph extends from the nd quadrant to the 1 st quadrant. extends from the rd quadrant to the 4 th quadrant. Opens up Opens down * All quadratic functions are smmetrical about a line.
MHF 4U Investigation/Lesson.6 More Polnomial Functions in Factored Form 1. Draw a sketch of each graph using the properties of polnomial functions, clearl identifing all the intercepts. Check our sketches with a TI-8. a) f (x)= (x - 4)(x + ) b) f (x) = -(x 1)(x + 4)(x + 1) c) f (x) = (x - 1)(x + 1) x x x c) f (x) = x(x -) d) f (x) = - (x - ) (x + ) f) f (x) = x(x - )(x + 1)(x+) x x x g) f (x) = x (x-4) h) f (x) = (x +) (x - ) i) f (x) = x(x +)(x -1)(x-)(x+ 4) x x x
Ex. Sketch a possible graph of the function f(x) = (x + )(x 4) (x 1). 50 5 00 175 150 15 100 75 50 5 1 1 11 10 9 8 7 6 5 4 1 5 1 4 5 6 7 8 9 10 11 1 1 x 50 75 100 15 150 175 00 5 50 Finding Equations of Polnomial Functions Ex. A quadratic function passes through the points (1, 0), (-, 0), and (, -1). Algebraicall determine the equations of this function. Ex. Each member of a famil of cubic functions has zeros of -,, and 5. a) Write the equation of the famil of curves.
b) Determine the equation of the member of the famil that has a -intercept of 6. Ex. Determine the equation of the following functions. a) A quartic function has zeros at 1, 0,, and and passes through the point (, 9). b) Pg. 146 # 1 6, 8, 9, 1, 1
MHF4U INV.7 Transformations of Cubic and Quartic Functions HW: p. 155 # 1-5, 6doso, 7, 10 Parent Function: x. Point: (, 8) Review: Pg. 184 # 1, (4, 5)a, 6, 7, 8be, 9bd, 10bd, 1d, 1, 14bc, 15ac, 17, 18 Transformation 1 New Pt. Transformation New Pt. Transformation New Pt. Transformation 4 New Pt. x 1 x 1 x 4 6 8 x Parent Function: 4 x. Point: (, 16) x 4 Transformation 1 New Pt. Transformation New Pt. Transformation New Pt. Transformation 4 New Pt. x 1 4 x 4 4 1 1 1 x 4 10