Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 1 Chapter 6 Polynomial Functions

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide Using Factoring to Solve Polynomial Equations

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 3 Zero Factor Property Let A and B be real numbers. If AB = 0, then A = 0 or B = 0. In words, if the product of two numbers is zero, then at least one of the numbers must be zero.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 4 Example: Solving a Quadratic Equation Solve (x – 5)(x + 2) = 0.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 5 Solution

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 6 Solution Check that both 5 and –2 satisfy the original equation: So, the solutions are 5 and –2.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 7 Example: Solving a Quadratic Equation Solve w 2 – 2w – 8 = 0.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 8 Solution

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 9 Check that both –2 and 4 satisfy the original equation: So, the solutions are –2 and 4. Solution

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 10 Example: Finding x-Intercepts of the Graph of a Quadratic Function Find the x-intercepts of the graph of f(x) = x 2 – 7x + 10

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 11 Solution To find the x-intercepts, substitute 0 for f(x) and solve for x: So, the x-intercepts are (2, 0) and (5, 0).

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 12 Solution Use “zero” on a graphing calculator to verify our work.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 13 Connection Between x-Intercepts and Solutions Let f be a function. If k is a real-number solution of the equation f(x) = 0, then (k, 0) is an x-intercept of the graph of the function f. Also, if (k, 0) is an x-intercept of the graph of f, then k is a solution of f(x) = 0.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 14 Example: Solving a Quadratic Equation Solve 2x 2 – 8x = 5x – 20.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 15 Solution

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 16 Solution To verify that these are solutions to the equation, use a graphing calculator table to check that for each input, the outputs for y = 2x 2 – 8x is equal to the output for y = 5x – 20.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 17 Example: Solving a Quadratic Equation Solve (x + 2)(x – 4) = 7.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 18 Solution Although the left-hand side of the equation is factored, the right-hand side is not zero. So first, find the product of the left-hand side:

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 19 Solution Therefore, the solutions are –3 and 5. To verify the work, enter the equations y = (x + 2)(x – 4) and y = 7.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 20 Solution Use “intersect” to find the intersection points (–3, 7) and (5, 7). The x-coordinates of these points are the solutions of the original equation.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 21 Example: Solving a Quadratic Equation That Contains Fractions Solve

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 22 Solution To clear the equation of fractions, multiply both sides by the LCD, 6:

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 23 Solution

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 24 Example: Finding an Input and an Output of a Quadratic Function Let f(x) = x 2 – 3x – Find f(5).2. Find x when f(x) = 5.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 25 Solution 1. f(5) = 5 2 – 3(5) – 23 = 25 – 15 – 23 = –13 2. Substitute 5 for f(x) in the equation: Let f(x) = x 2 – 3x – 23.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 26 Solution 2. Verify that f(7) = 5 and f(–4) = 5 by hand. Or, verify the work in Problems 1 and 2 using a graphing calculator table.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 27 Solving Quadratic or Cubic Equations by Factoring If an equation can be solved by factoring, we solve it by the following steps: 1. Write the equation so one side of the equation is equal to zero. 2. Factor the nonzero side of the equation. 3. Apply the zero factor property. 4. Solve each equation that results from applying the zero factor property.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 28 Example: Finding x-Intercepts of the Graph of a Cubic Function Find the x-intercepts of the graph of f(x) = x 3 – 5x 2 – 4x + 20

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 29 Solution Substitute 0 for f(x) and solve for x: So, the x-intercepts are (–2, 0), (2, 0), and (5, 0).

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 30 Solution Use “zero” on a graphing calculator to verify our work.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 31 Zero Factor Property Warning The expression x 2 (x – 5) – 4(x – 5) = 0 is not factored, because it is a difference, not a product. Only after we factor the left side of the equation can we apply the zero factor property

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 32 Cubic Functions The solutions set of a cubic equation in one variable may contain one, two, or three real numbers (one, two, or three x-intercepts).

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 33 Example: Using Graphing to Solve an Equation in One Variable Use graphing to solve x 2 – x – 7 = –x 2.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 34 Solution Use “intersect” on a graphing calculator to find the solutions of the system y = x 2 – x – 7 y = –x 2

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 35 Solution The approximate solutions of the system are (–1.64, –2.68) and (2.14, –4.57). The x-coordinates of these ordered pairs are the approximate solutions of the equation.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 36 Quadratic Model Definition A quadratic model is a quadratic function, or its graph, that describes the relationship between two quantities in an authentic situation.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 37 Example: Modeling with a Quadratic Function Annual revenues of restaurants are shown in the table for various years. Let f(t) be the annual revenue (in billions of dollars) of restaurants at t years since A model of the situation is

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 38 Example: Modeling with a Quadratic Function 1. Use a graphing calculator to draw the graph of the model and, in the same viewing window, the scattergram of the data. Does the model fit the data well? 2. Predict the revenue in Predict when the annual revenue was $84 billion.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 39 Solution 1. The graph of the model and the scattergram of the data show that the model appears to fit the data quite well.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 40 Solution 2. To predict the revenue in 2018, find f(48): The revenue in 2018 will be $754.8 billion, according to the model.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 41 Solution 3. To estimate when the annual revenue was $84 billion, substitute 84 for f(t) and solve for t:

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 42 Solution 3. The inputs –30 and 5 represent the years 1940 and 1975, respectively. The estimate of 1940 is an example of model breakdown, as a little research would show the revenue in 1940 was much less than $84 billion. Therefore, we estimate that it was 1975 when the revenue was $84 billion.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 43 Solution Use a graphing calculator table to verify our work in Problems 2 and 3.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 44 Area of Rectangular Objects The area A of a rectangle is given by the formula A = LW, where L is the length of the rectangle and W is the width.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 45 Example: Solving an Area Problem A person has a rectangular garden with a width of 9 feet and a length of 12 feet. She plans to place mulch outside of the garden to form a border of uniform width. If she has just enough mulch to cover 100 square feet of land, determine the width of the border.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 46 Solution Step 1: Define each variable. Let x be the width (in feet) of the mulch border.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 47 Solution Step 2: Write an equation in one variable. For the outer rectangle (garden and border), then length is 2x + 12 and the width is 2x + 9. So the area is:

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 48 Solution Step 2: The area of the garden is 12 ∙ 9 = 108 square feet, and the area of the border is given as 100 square feet. The area of the garden plus the area of the border is equal to the area of the outer rectangle:

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 49 Solution Step 3: Solve the equation.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 50 Solution Step 3:

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 51 Solution Step 4: Describe each result. For the result the border width is negative. Model breakdown has occurred, because a width must be positive. The border width is 2 feet.

Copyright © 2015, 2008, 2011 Pearson Education, Inc. Section 6.7, Slide 52 Solution Step 5: Check. If the border width is 2 feet, then the outer rectangle has a width of 13 feet and a length of 16 feet. Therefore, the total area of the outer rectangle is 13 ∙16 = 208 square feet. This checks with our calculation near the beginning of our work.