1. 2 - 1 Section 2.1 Properties of Functions

4. Definition of a Function

8. Examples

10. Practice 1 Find the domain and range for the function Solution: The domain includes only those values of x satisfying since the denominator cannot be zero. Using the methods for solving a quadratic inequality produces the domain Because the numerator can never be zero, the denominator can take on any positive real number except for 0, allowing y to take on any positive value except for 0, so the range is

11. Practice 2 Given the function find each of the following. (a) (b) All values of x such that (a) Solution: Replace x with the expression x + h and simplify. (b) Solution: Set f (x) equal to − 5 and then add 5 to both sides to make one side equal to 0. Continued

12. Practice 2 continued This equation does factor as Set each factor equal to 0 and solve for x.

13. Piecewise-defined Functions: Example:

14. Symmetry: Even and Odd Functions A function f is even if for every number x in its domain the number -x is also in its domain and f(-x) = f(x) A function f is odd if for every number x in its domain the number -x is also in its domain and f(-x) = - f(x)

17. 2 - 17 Section 2.2 Quadratic Functions; Translation and Reflection

23. Translation and Reflection

24. Vertex location Example Where: h= k= Rewrite in this form

25. Practice 1 For the function (a) complete the square, (b) find the y-intercept, (c) find the x intercepts, (d) find the vertex, and (e) sketch the graph. Solution (a): To begin, factor 2 from the x-terms so the coefficient of x 2 is 1: Next, we make the expression inside the parentheses a perfect square by adding the square of one-half of the coefficient of x, Continued

26. Practice 1 continued Solution (b):The y-intercept (where x = 0) is − 1. Solution (c): To find the x-intercepts, solve Use the quadratic formula to verify that the x-intercepts are at Solution (d): The function is now in the form

27. Translation and Reflection

29. 2 - 29 Section 2.3 Polynomial and Rational Functions

31. Examples

32. Practice 1 Graph Solution: Using the principles of translation and reflection, we recognize that this is similar to the graph of but reflected vertically (because of the negative in front of x 6 ) and 64 units up.

34. Examples

35. Practice: Identify degree and leading coefficient for each below

36. Rational Functions

37. Example: f(x)=1/x

39. Examples of graphs of rational functions: Notice the domain! Asymptotes

40. 2 - 40 Section 2.4 Exponential Functions

41. Examples:

42. More examples of exponentials: Most common base for exponential is: e = 2.718281828 This number is called: Euler number. Occurs in natural phenomena and is used for financial modeling.

43. Your Turn 1 Solve Solution: Since the bases must be the same, write 25 as 5 2 and 125 as 5 3.

44. Application: Interests If m is made larger and larger, ultimately we get to:

45. Your Turn 2 Find the interest earned on $4400 at 3.25% interest compounded quarterly for 5 years. Solution: Use the formula for compound interest with P = 4400, r = 0.0325, m = 4, and t = 5. The investment plus the interest is $5172.97. The interest amounts to $5172.97 − $4400 = $772.97.

46. Your Turn 3 Find the amount after 4 years if $800 is invested in an account earning 3.15% compounded continuously. Solution: In the formula for continuous compounding, let P = 800, t = 4 and r = 0.0315 to get or $907.43.

47. 2 - 47 Section 2.5 Logarithmic Functions

49. Example 1

52. Example 2 If all the following variable expressions represent positive numbers, then for a > 0, a ≠ 1, the statements in (a)–(c) are true.

53. Your Turn 1 Write the expression as a sum, difference, or product of simpler logarithms. Solution: Using the properties of logarithms,

55. Your Turn 2 Evaluate Solution: Using the change-of-base theorem for logarithms with x = 50 and a = 3 gives

56. Your Turn 3 Solve for x: Solution: This leads to two solutions: x = − 4 and x = 2. But notice that x = − 4 is not a valid value for x in the original equation, since the logarithm of a negative number is undefined. The only solution is, therefore, x = 2.

57. Your Turn 4 Solve for x: Solution: Taking natural logarithms on both sides gives

59. 2 - 59 Section 2.6 Applications: Growth and Decay; Mathematics of Finance

61. Your Turn 1 Yeast in a sugar solution is growing at a rate such that 5 g grows exponentially to 18 g after 16 hours. Find the growth function, assuming exponential growth. Solution: The values of y 0 and k in the exponential growth function y = y 0 e kt must be found. Since y 0 is the amount present at time t = 0, y 0 = 5. To find k, substitute y = 18, t = 16, and y 0 = 5 into the equation y = y 0 e kt. Now take natural logarithms on both sides and use the power rule for logarithms and the fact that Continued

62. Your Turn 1 continued The exponential growth function is where y is the number of grams of yeast present after t hours.

63. Recap: Graphs of Basic Functions