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1 Lesson 28 Zeros of a function: - the inputs that make the function equal to zero (same values as the x- coordinates of the x-intercepts) o if ( ), ( ) when o zeros are 2 and - when a function is graphed, the zeros indicate the -intercepts (the points where the graph touch or cross the horizontal axis) o -intercepts of are ( ) and ( ) - Note: Zeros are numbers; x-intercepts are ordered pairs. Example 1: Given the graph of the function ( ) ( )( ), a. find the zeros of the function. (Hint: Let ( ) and solve for x and/or look at the graph of the function f below.) zeros are and. b. list the intervals of inputs that are formed by the zeros in the graph below. c. determine when the function is positive or above the x-axis (express in interval notation). d. determine when the function is negative or below the x-axis (express in interval notation)

2 How can we determine where a polynomial function is positive or negative without using its graph? If ( ) ( )( ), where is ( ) (above x-axis) and ( ) (below x- axis)? We will answer this question in example 2 using a sign chart. Example 2: Solve the inequality ( )( ), and express the solution in interval notation. (Note: According to the inequality above, we want the negative results from our sign chart, including the zeros. Note: In the sign chart below, I placed solid dots to indicate including the zeros.) Zeros are 3 and -1. ( ) ( ) ( ) Result Solving nonlinear inequalities: 1. set one side of the inequality to zero 2. combine like terms, then factor the polynomial side 3. find the values that make each factor zero (including restrictions), the zeros 4. set-up a sign chart using the values from step 3 5. find the intervals that satisfy the inequality by using test values The method above used to solve nonlinear inequalities is different than what is shown in the book. This method is also different than what is shown in the video lessons and PowerPoint presentations on the course website. On homework, quizzes, and exams, you are free to use whatever you method you prefer. ***We use the Zero Factor Theorem (shown below) to find the zeros of a function ( ( ) ), but we cannot use it to find the intervals where a function is positive or negative ( ( ) ( ) )?*** if and only if or ***THIS IS NOT TRUE FOR INEQUALITIES. does not lead to the conclusion We must use the steps 1 5 above.*** 2

3 Example 3: Solve the inequality ( )( ), and express the solutions in interval notation. a) Factor and find the zeros: b) Determine intervals and make a sign chart. c) Solution: Do not assume that the signs always alternate from left to right in the sign chart. Example 4: Given the function ( ), find the following: a. ( ) b. ( ) 3

4 Example 5: Solve the inequality and express the solution in interval notation. Why can t we multiply both sides of the inequality above by the denominator to eliminate the fraction? (Answer: We would not know if we were multiplying by a positive value or a negative value; should we change the inequality symbol?) We must subtract the 2 from both sides and combine. Example 6: Solve the inequality and express the solution in interval notation. Why can t we divide both sides of the inequality above by to simplify the inequality? Answer: We would be eliminating a zero. NEVER divide both sides of an equation or inequality by a variable. Note: Write the inequality with zero on one side first (step 1). 4

5 Example 7: Solve the inequality and express the solution in interval notation. Why don t we cancel a factor of from the numerator and denominator to simplify the inequality? Answer: We would be eliminating a zero. Never cancel a variable in such an inequality. Example 8: A golf ball is projected upward from ground level with an initial velocity of 112 feet per second. Its height above the ground seconds after it is hit is ( ). When is the ball more than 160 feet above the ground? Height > 160 feet 5

6 Example 9: The average speed of a driver during a round trip between Indianapolis and Louisville is given by the equation, where is the average speed going (Indy to Louisville) and is the average speed returning (Louisville to Indy). For what speeds would be greater than? (In other words, when would the return speed be greater than the going speed?) Round your answer to the nearest tenth of a mile per hour. 6