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Aim: What are the higher degree function and equation? Do Now: a) Graph f(x) = x 3 + x 2 – x – 1 on the calculator b) How many times does the graph intersect.

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Presentation on theme: "Aim: What are the higher degree function and equation? Do Now: a) Graph f(x) = x 3 + x 2 – x – 1 on the calculator b) How many times does the graph intersect."— Presentation transcript:

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2 Aim: What are the higher degree function and equation? Do Now: a) Graph f(x) = x 3 + x 2 – x – 1 on the calculator b) How many times does the graph intersect the x-axis? HW: Worksheet

3 The intersecting points of the graph and the x-axis are called the zeros of the function, because at those points the value of y is 0. f(x) = x 3 + x 2 – x – 1

4 The zeros of the function f(x) = x 3 + x 2 – x –1 is just like the root of the equation x 3 + x 2 – x – 1 = 0 Why are there only two zeros for a third degree function? We can tell the zeros are 1 and –1 from the graph The graph is tangent to the x-axis at x = – 1, therefore x = –1 is a repeated zeros (double roots)

5 To find the zeros (roots) of a function, we can use either graphical or algebraic methods. Graphically: find the x-values of the intersecting points of the graph of the function and the x-axis. Algebraically: Let f(x) = 0, then solve the variable

6 Ex: x 3 + x 2 – x – 1 = 0 Regrouping: (x 3 + x 2 ) – (x + 1) = 0 Factor: x 2 (x + 1) – (x + 1) = 0 Factor GCF: (x + 1)(x 2 – 1) = 0 x + 1 = 0 x = –1

7 a. Graph: f(x) = x 4 – 3x 2 + 2 b. Solve: x 4 – 3x 2 + 2 = 0 f(x) = x 4 – 3x 2 + 2

8 We can treat this equation as a quadratic equation Factor: Set each binomial equals zero x 2 – 2 = 0 x 2 – 1 = 0 Solve for x:

9 x 2 (x 2  4x  4)  0 Factor out x 2. x 2 (x  2) 2  0 Factor completely. x 2  or (x  2) 2  0 Set each factor equal to zero. x  0 x  2 Solve for x. x 4  4x 3  4x 2  0 Multiply both sides by  1. (optional step) Solve for x:  x 4  4x 3  4x 2 = 0

10 Solve equation by grouping Solve for x: x 4 – x 3 + x – 1 = 0 (x 4 – x 3 ) + (x – 1) = 0 Grouping as two binomials x 3 (x – 1) + (x – 1) = 0 Factor x 3 (x – 1)(x 3 + 1) = 0 Factor (x – 1) x – 1 = 0 x 3 + 1 = 0 Set each binomial equals zero x = 1x = -1 Solve for x

11 Solve for x: 1. x 3 – 2x 2 – 3x = 0 2. x 4 + 5x 2 – 36 = 0 3. 5x 3 + 30x 2 + 45x = 0 4. 32x 3 – 16x 2 – 18x + 9 = 0

12 Find the solution to polynomial equations of higher degree that can be solved using factoring and/or the quadratic formula and / graphically Objective

13 Students will be able to 1. define the degree of a polynomial equation 2. factor polynomial expressions of degree > 3 3. identify polynomials that are written in ‘quadratic form’ 4. state and apply the quadratic formula 5. express irrational solutions in simplest radical form 6. graphically identify (estimate) x-intercepts as solution of a polynomial equation 7. graphically identify (estimate) the x-coordinate of the point of intersection of a system of polynomial equations as a solution of that system

14 Choose an effective approach to solve a problem from a variety of strategies (numeric, graphic, algebraic) Use multiple representations to represent and explain problem situations Use a variety of strategies to extend solution methods to other problems Determine information required to solve the problem, choose methods for obtaining the information, and define parameters for acceptable solutions Evaluate the relative efficiency of different representations and solution methods of a problem Performance Standard


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