Functions, Equations, and Inequalities

2. Functions, Equations,and Inequalities Chapter 2

3. 2.1Linear Equations, Functions, and Models Solve linear equations. Solve applied problems using linear models. Find zeros of linear functions. Solve a formula for a given variable.

4. Linear Equations An equation is a statement that two expressions are equal. One Variable mx + b = 0, where m and b are real numbers and m 0. Examples: 4x + 6 = 18 5(x2) = 7x + 8

5. Equivalent Equations Equations that have the same solution set. Examples: 4x + 8 = 16 and x = 2 are equivalent because 2 is the solution of each equation. 5y + 6 = 21 and y = 4 are not equivalent because y is equal to 3 in the first equation.

6. Addition If a = b is true, then a + c = b + c is true. Example: You can also use the intersect feature on a graphing calculator to solve equations. y1 = 9x 7, y2 = 2 Equation-Solving Principles

7. Multiplication If a = b is true, then ac = bc is true. Example: We graph y1 = 3(5 + 2x) and y2 = 4 2(x + 3) Equation-Solving Principles

8. Applications Using Linear Models • Mathematical techniques to answer questions in real-world situations. • Often modeled by linear equations and functions.

9. Five Steps for Problem Solving 1. Familiarize yourself with the problem situation. Make a drawing Write a list Assign variables Organize into a chart or table Find further information Guess or estimate the answer 2. Translate to mathematical language or symbolism. 3. Carry out some type of mathematical manipulation. 4. Check to see whether your possible solution actually fits the problem situation. 5. State the answer clearly using a complete sentence.

10. The Motion Formula • The distance d traveled by an object moving at rate r in time t is given by d = rt. • Example: On a bicycle tour, Dave rides his bicycle 10 mph faster than his father, Bill. In the same time that Dave travels 90 miles, his father travels 60 miles. Find their speeds.

11. Distance Rate Time Dave 90 r + 10 t Bill 60 r t Example 1. Familiarize. We can organize the information in a table as follows. 2. Translate. Using the formula , we get two expressions for t. and Since they occur in the same time, we have the following equation: . 3. Carry out. We solve the equation.

12. Example 4. Check. If Bill travels at rate of 20 mph for a distance of 60 miles it will take him 3 hours. If Dave travels 90 miles at a rate of r + 10, or 30 mph, it will also take him 3 hours. Therefore, the answer checks. 5. State. Dave travels at a rate of 30 mph while his father, Bill, travels at a rate of 20 mph on their bicycles.

13. Simple-Interest Formula • I = Prt • I = the simple interest ($) • P = the principal ($) • r = the interest rate (%) • t = time (years)

14. Example The Robinson’s have two loans that total $15,000. One loan is at 4% simple interest and the other is at 6.5%. After 1 year, they owe $700 in interest. What is the amount of each loan? • Solution: 1.Familiarize. We will let x = the amount borrowed at 4% interest. Then the remainder is $15,000 – x, borrowed at 6.5% interest. Amount Borrowed Interest Rate Time Amount of Interest 4% Loan x 4% or 0.04 1 yr 0.04x 6.5% Loan 15,000 –x 6.5% or 0.065 1 yr 0.065(15,000 – x) Total 15,000 700

15. Example 2. Translate. The total amount of interest on the two loans is $700. Thus we write the following equation. 0.04x + 0.065(15,000 x) = 700 3.Carry out. We solve the equation. 0.04x + 0.065(15,000 x) = 700 0.04x + 975  0.065x = 700  0.025x + 975 = 700 0.025x = 275 x = 11,000 If x = 11,000, then 15,000  11,000 = 4000. 4.Check. The interest on $11,000 at 4% for 1 yr is $11,000(0.04)(1), or $440. The interest on $4000at 6.5% for 1 yr is $4000(0.065)(1) or $260. Since $440 + $260 = $700, the answer checks. 5.State. The Robinson’s borrowed $11,000 at 4% interest and $4000 at 6.5% interest.

16. Zeros of Linear Functions • An input c of a function f is called a zero of the function, if the output for c is 0. • f(c) = 0 • A linear function f(x) = mx + b, with m 0, has exactly one zero.

17. Find the zero of f(x) = 3x 18. Algebraic Solution: 3x 18 = 0 3x = 18 x = 6 Find the zero of f(x) = 3x – 18. Graphic Solution: The x-intercept of the graph is (6, 0). Thus, 6 is the zero of the function. Example

18. Formulas • A formulais an equation that can be used to model a situation. • Example: Solve y = ax + bx 2 for x. y = ax + bx 2 y + 2 = ax + bx y + 2 = x(a + b)

19. 2.2The Complex Numbers Perform computations involving complex numbers.

20. The Complex-Number System • The complex-number system is used to find zeros of functions that are not real numbers. • When looking at a graph of a function, if the graph does not cross the x-axis, it has no real-number zeros.

21. iis not under the radical The Number i • The number i is defined such that . Examples: Express each number in terms of i.

22. Complex Numbers A complex number is a number of the form a + bi, where a and b are real numbers. The number a is said to be the real part of a + bi and the number b is said to be the imaginary part of a + bi. Imaginary Numbera + bi, a 0, b  0 Pure Imaginary Numbera + bi, a= 0, b  0

23. Add: (9 + 5i) + (2 + 4i) (9 + 2) + (5i + 4i) 11 + (5 + 4)i 11 + 9i Subtract: (6 + 7i)  (4  3i) (6  4) + [7i  (3i)] 2 + 10i Addition and Subtraction Complex numbers obey the commutative, associative, and distributive laws.

24. Multiplication When and are real numbers, . Examples: Multiply and simplify. (1 + 5i)(1 + 7i) = 1 + 7i + 5i + 35i2 = 1 + 7i + 5i 35 = 34 + 12i (7 6i)2 = 72  2  7  6i + (6i)2 = 49  84i + 36i2 = 49  84i  36 = 13  84i

25. Simplifying Powers of i • Recall that 1 raised to an even power is 1, and 1 raised to an odd power is 1. • Examples: • i47 = i46  i = (i2)23  i = (1)23  i = 1  i = i • i72 = i71 i = (i2)36  i = (1)36  i = 1  i = i

26. Conjugates The conjugate of a complex number a + bi is a bi. The numbers a + bi and a  bi are complex conjugates. Examples: 6 + 7i and 6  7i 8  3i and 8 + 3i 14i and 14i • The product of a complex number and its conjugate is a real number.

27. Example: (4 + 9i)(4  9i) = 42  (9i)2 = 16 81i2 = 16  81(1) = 97 Example: (7i)(7i) =  49i2 = 49(1) = 49 Multiplying Conjugates

28. Dividing Conjugates • Example: Divide 4  3i by 1  8i.

29. 2.3Quadratic Equations, Functions, and Models Find zeros of quadratic functions and solve quadratic equations by using the principle of zero products, by using the principle of square roots, by completing the square, and by using the quadratic formula. Solve equations that are reducible to quadratic. Solve applied problems using quadratic equations.

30. Quadratic Equations A quadratic equation is an equation equivalent to ax2 + bx + c = 0, a 0, where a, b, and c are real numbers. A quadratic equation written in this form is said to be in standard form.

31. Quadratic Functions A quadratic functionf is a function that can be written in the form f(x) = ax2 + bx + c, a 0, where a, b, and c are real numbers. The zeros of a quadratic function f(x) = ax2 + bx + c are the solutions of the associated quadratic equation ax2 + bx + c = 0. Quadraticfunctions can have real-number or imaginary-number zeros and quadratic equations can have real-number or imaginary-number solutions.

32. Equation-Solving Principles • The Principle of Zero Products: If ab = 0 is true, then a = 0 or b = 0, and if a = 0 or b = 0, then ab = 0. • Example: Solve 3x2 5x = 2. • Solution: 3x2 5x = 2 3x2 5x  2 = 0 (3x + 1)(x  2) = 0 3x + 1 = 0 or x  2 = 0 3x = 1 or x = 2 x =

33. Check: 3(2)2 5(2) = 2 3(4)  10 = 2 12  10 = 2 2 = 2 The solutions are and 2. Checking the Solutions

34. Graphical Solution • The solutions of the equation y = 3x2 5x = 2, or the equivalent equation 3x2 5x  2 = 0, are the zeros of the function f(x) = 3x2 5x  2. They are also the first coordinates of the x-intercepts of the graph of f(x) = 3x2 5x  2.

35. Equation-Solving Principles The Principle of Square Roots: If x2 = k, then x = or x =  . Example: Solve 2x2 6 = 0. • Solution: 2x2 6 = 0 2x2= 6 x2= 3 x = or x =  • Check: 2x2 6 = 0 2( )2  6  0 2  3  6  0 6  6  0 0 = 0 Solutions are and  .

36. Completing the Square To solve a quadratic equation by completing the square: • Isolate the terms with variables on one side of the equation and arrange them in descending order. • Divide by the coefficient of the squared term if that coefficient is not 1. • Complete the square by taking half the coefficient of the first-degree term and adding its square on both sides of the equation. • Express one side of the equation as the square of a binomial. • Use the principle of square roots. • Solve for the variable.

37. Example Find the zeros of f(x) = x2 + 6x 16 by completing the square. Solution: x2 + 6x 16 = 0 x2 + 6x= 16 x2 + 6x + =16 + x2 + 6x + 9 = 25 (x + 3)2 = 25 x + 3 =  x + 3 =  5 x = 5  3 or x = 5  3 x = 2 or x = 8

38. Graphical Solution: x2 + 6x 16 = 0

39. Quadratic Formula • The solutions of ax2 + bx + c = 0, a 0, are given by . • This formula can be used to solve any quadratic equation.

40. Example Solve: 4x2 + 3x = 8 Solution: 4x2 + 3x 8 = 0 a = 4, b = 3, c = 8 The approximate solutions are 1.088 and 1.838. The exact solutions are:

41. Graphical Solution: 4x2 + 3x = 8 • Graph y1 = 4x2 + 3x and y2= 8 The approximate solutions are 1.088 and 1.838.

42. Discriminant • When you apply the quadratic formula to any quadratic equation, you find the value of b2 4ac, which can be positive, negative, or zero. This expression is called the discriminant. For ax2 + bx + c = 0: b2 4ac = 0 One real-number solution; b2 4ac > 0Two different real-number solutions; b2 4ac < 0Two different imaginary-number solutions, complex conjugates.

43. Equations Reducible to Quadratic • Some equations can be treated as quadratic, provided that we make a suitable substitution. • Example:x4 10x2 + 9 = 0 Knowing that x4= (x2)2, we can substitute u for x2 and the resulting equation is then u2  10u + 9 = 0. This equation can then be solved for u by factoring or using the quadratic formula. Then the substitution can be reversed by replacing u with x2, and solving for x. Equations like this are said to be reducible toquadratic, or quadratic in form.

44. Solving an Equation Reducible to Quadratic Solve: x4 10x2 + 9 = 0 u2  10u + 9 = 0 (substituting u for x2) (u  9)(u  1) = 0 u  9 = 0 or u  1 = 0 u = 9 or u = 1 x2 = 9 or x2 = 1 (substitute x2 for u and solve for x) x = ±3 or x = ±1 The solutions are 3, 3, 1, and 1.

45. Application Some applied problems can be translated to quadratic equations. Example:Free Fall. An acorn falls from the top of a tree that is 32 feet tall. How long will it take to reach the ground?

46. Solving an Application 1. Familiarize. The formula s = 16t2 is used to approximate the distance s, in feet, that an object falls freely from rest in t seconds. 2. Translate. Substitute 32 for s in the formula: 32 = 16t2. 3. Carry out. Use the principle of square roots. 32 = 16t2 4. Check. In 1.414 seconds, an acorn would travel a distance of 16(1.414)2, or about 32 ft. The answer checks. 5. State. It would take about 1.414 sec for an acorn to reach the ground from the top of a 32 ft tall tree.

47. Use the Intersect method, we replace t with x, graph y1 = 32 and y2 = 16x2, and find the first coordinates of the points of intersection. Time cannot be negative in this application, so we need to find only the point of intersection with a positive first coordinate. The window [-5, 5, -10, 40], with Xscl = 1 and Yscl = 5. Graphical Solution: 32 = 16t2

48. 2.4Analyzing Graphs of Quadratic Functions Find the vertex, the axis of symmetry, and the maximum or minimum value of a quadratic function using the method of completing the square. Graph quadratic functions. Solve applied problems involving maximum and minimum function values.

49. The graph of a quadratic function is called a parabola. The point (h, k) at which the graph turns is called the vertex. The maximum or minimum value of f(x) occurs at the vertex. Each graph has a line x = h that is called the axis of symmetry. Graphing Quadratic Functions of the Type f(x) = a(x h)2 + k

50. Example Find the vertex, the axis of symmetry, and the maximum or minimum value of f(x) = x2 8x + 12. Solution: Complete the square. Vertex: (4, 4) Axis of symmetry: x = 4 Minimum value of the function: 4