Practical Algebra. A Step-by-step Approach. Brought to you by Softmath, producers of Algebrator Software

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1 Practical Algebra A Step-by-step Approach Brought to you by Softmath, producers of Algebrator Software

2 2 Algebra e-book Table of Contents Chapter 1 Algebraic expressions 5 1 Collecting... like terms 5 Chapter 2 Fractions 11 1 Adding... and subtracting fractions 11 2 Reducing... fractions 15 Chapter 3 Complex numbers 22 1 Introduction Operating Chapter 4 Rational expressions 28 1 Adding... and subtracting 28 Chapter 5 Radicals 33 1 Adding... and subtracting 33 2 Multiplying... radicals 39 3 Rationalizing... denominators 40 4 Combining... like radical terms 43 5 Combining... operations 44 Chapter 6 Equations 47 1 Linear... equations 47 Graphing linear... equations 51 2 Absolute... value equations 54 Chapter 7 Systems of equations 57 1 Solving... systems by elimination 57 2 Solving... systems by substitution 58 Chapter 8 Inequalities 62 1 Linear... inequalities 62 Graphing linear... inequalities 64 2 Simultaneous... inequalities 69 Chapter 9 Polynomials 73 1 Adding... and subtracting 73 2 Factoring... 76

3 Contents 3 3 Evaluating Higher... degree polynomials 81 5 Completing... the square 82 Chapter 10 Functions 86 1 Adding... and subtracting 86 2 Linear... functions 88 Graphing linear... functions 89 3 Quadratic... functions 91 Graphing quadratic... functions 92 4 Composition Exponential... functions 96 Chapter 11 Trigonometry Angles... and degree measures Trigonometric... functions 102 3

4 Chapter 1 Algebraic expressions

5 Algebraic expressions 1 Algebraic expressions 1.1 Collecting like terms 5 Like terms in an algebraic expression are terms with identical symbolic or variable parts. Thus 3x and 7x are like terms because both contain the symbolic part x 5x 2 yz and 13x 2 yz are like terms because both contain the symbolic part x 2 yz 4x 2 and 7x are not like terms because even though the symbol present in both is x, the symbolic part x 2 is not identical to the symbolic part x. Algebraic expressions that contain like terms can be simplified by combining each group of like terms into a single term. The reason why this is possible and valid is quite easy to see. For instance, consider the expression 3x + 7x which is the sum of two like terms, representing the accumulation of three x s and another seven x s. Clearly, the end result is a total of ten x s. In notation 3x + 7x = (3 + 7)x = 10x This process of combining (or collecting ) like terms can be performed for each group of like terms that appear in an expression. The net effect will be that the original expression can now be written with fewer terms, yet which are entirely equivalent to the terms in the original expression. Example: Simplify: 5x x + 4x 2 + 8x + 7. solution: This expression has six terms altogether. However, we notice that two of the terms have the literal part x 2 and so are like terms we can replace 5x 2 + 4x 2 by (5 + 4)x 2 = 9x 2 two of the terms have the same literal part x and so are also like terms. We can replace -3x + 8x by (-3 + 8)x = 5x two of the terms are just constants, and so obviously can be combined arithmetically: = 16.

6 6 Algebra e-book So 5x x + 4x 2 + 8x + 7 = 5x 2 + 4x 2 + (-3x) + 8x = (5 + 4)x 2 + (-3 + 8)x + 16 = 9x 2 + 5x Thus, in simplest form, the original six term expression can be rewritten as 9x 2 + 5x + 16 consisting of just three terms. Algebrator can easily solve these type of problems:

7 Algebraic expressions Clicking on the "Solve Once" button will show the first step of the solution process: 7

8 8 Algebra e-book An explanation can be obtained by clicking on the "Explain" button:

9 Algebraic expressions 9

10 Chapter 2 Fractions

11 Fractions 2 Fractions 2.1 Adding and subtracting fractions 11 Addition and subtraction of algebraic fractions works basically the same way as does addition and subtraction of simple numerical fractions: (i.) rewrite all of the fractions as mathematically equivalent fractions, but having the same denominator (ideally, the simplest such common denominator). (ii.) combine the numerators of these fractions as indicated (adding and subtracting), retaining the common denominator. When we work with algebraic fractions, there is then often a third very important step: (iii.) simplify the resulting fraction as much as possible. It is always necessary to make sure that once the addition/subtraction step is completed, you make a reasonable effort to simplify the result. Usually this simplification step will be made very much easier if you take care in step (i.) to make sure that you have used the simplest possible common denominator. We will illustrate the process for determining that simplest possible denominator when the fractions to be added/subtracted are algebraic fractions. You will see that the strategy is essentially the same as the strategy used to determine the least common denominator when adding or subtracting numerical fractions, except that now symbolic factors may occur as well as numerical factors. Example 1: Combine and simplify: solution: Here, the two denominators share some factors in common and so the simplest common denominator for these fractions is not simply the product of the two denominators. Noting that 15x = (3)(5)(x) and 12x 2 = (2 2)(3)(x 2) we conclude that the simplest common denominator here is (2 2)(3)(5)(x 2) = 60x 2. So

12 12 Algebra e-book The numerator is a difference of two terms, and so we try multiplying to remove the brackets followed by collection of like terms in hope of being able to achieve some simplification. When the like terms were combined, we noticed that the two terms remaining shared a common factor of 2x, which has been factored out in the last line. Thus, our original difference of two fractions becomes as the final answer. Since what is left in the numerator cannot be factored further, this is as simple as we can get for the final answer. Example 2: Simplify solution: As mentioned before, the instruction to simplify here means combine the two fractions into a single fraction, and reduce that resulting single fraction to as simple a form as possible. We ve now done several examples of this type, so here we ll more or less just show the work without including a lot of discussion. This would be a good example for you to try as a practice exercise before you look at our solution given below.

13 Fractions 13 The two denominators factor as 6xy 2 = (2)(3)(x)(y 2) and 15x 2 y = (3)(5)(x 2)(y) Thus, the simplest common denominator is (2)(3)(5)(x 2)(y 2) = 30x 2 y 2 So, going back to the original problem There is no obvious way to factor this numerator, so it does not appear that any further simplification of this fraction is possible. So, the last form above must be the requested final answer. Example 3: Simplify solution: Some people experience an overwhelming temptation here to simply cancel the x s: confusing this addition problem with the much simpler multiplication problem

14 14 Algebra e-book where the cancellation to get the simple result 1 is justified. However, no such quick cancellations can be done in the addition problem we have here because the two x s aren t even part of the same fraction! In fact, the first term here is not even a fraction. Perhaps the clearest way to proceed is to turn the first term into a fraction with denominator equal to 1, and then apply the methods of the previous examples for adding two fractions. We get This is the final answer, since the numerator cannot be factored further and so there is no additional simplification possible. Example 4: Combine and simplify: solution: Again, we start by turning this problem into a problem with two fractions, and then we just apply the methods already illustrated in the previous examples.

15 Fractions 15 The simple form of this last fraction is obtained from the previous step by collecting like terms in the numerator. The terms -5y + 5y add up to zero y s, and the two numbers sum to -28. The numerator of the final fraction shown above, y 2 28, cannot be factored further to give a form that leads to cancellations with the denominator, so this must be the required final answer. 2.2 Reducing fractions To reduce a numerical fraction to simplest form means to rewrite it as an equivalent fraction which has the smallest possible denominator. People refer to this operation as reducing a fraction to lowest terms. Recall that starting with a fraction, we can get an equivalent fraction by either: (i) multiplying the numerator and denominator by the same nonzero value or (ii) dividing the numerator and denominator by the same nonzero value. Obviously if the goal of simplification is to find an equivalent fraction with a smaller denominator, we will have to use the second principle: dividing the numerator and denominator by the same nonzero value. The strategy for finding the values to divide into the numerator and denominator is quite systematic: Step 1: Factor both the numerator and denominator into a product of prime factors using the method described in the previous note in this series. Step 2: If the numerator and denominator both have a prime factor which is the same, then divide the current numerator and denominator by that value. The result will be the numerator and

16 16 Algebra e-book denominator of an equivalent fraction, but with a smaller denominator. Repeat step 2 as often as possible. When the numerator and denominator have no further prime factors in common, they form the desired equivalent fraction which has the smallest denominator. We have then obtained the simplest form of the original fraction. Example: Reduce the fraction to simplest form. solution: For this example, we ll go through the process step-by-step in some detail. Then we ll illustrate the familiar shortcuts in a couple of examples. The prime factorizations of the numerator and denominator here are easily obtained: 42 = 2 21 = = 2 35 = So, the numerator and denominator of the fraction we are given both contain the prime factor 2. Thus The new fraction, 21 / 35, is simpler than the old fraction, 42 / 70, because its denominator, 35, is smaller than the original denominator of 70. Still, 21 / 35 is equivalent to 42 / 70 because it was obtained by dividing both the numerator and denominator of 42 / 70 by the same value, 2. But 21 / 35 is still not in simplest form because the factorization of its numerator and denominator (shown in brackets above) indicates that they both still contain a common prime factor of 7. So Thus, we have These two fractions, the original 42 / 70and the final 3 / 5, are equivalent because we got 3 / 5 by dividing the numerator and denominator of 42 / 70 by the same values (2 and 7 in turn, or

17 Fractions 17 effectively, 14, if you think of doing it in one step). Furthermore, the numerator and denominator of 3 / 5clearly do not share any further common prime factors, and so this simplification process cannot be carried further. Therefore, 3 / 5 is the simplest form of 42 / 70. Example: Reduce the fraction to simplest form. solution: Yes! This is the same problem as the first one. What we want to do here is show the shortcut form of the strategy for simplifying fractions. We begin as before by rewriting both the numerator and the denominator as a product of prime factors: Now, if you study the steps of the previous example, you will see that dividing the numerator and denominator by 2 results in those two factors disappearing from each. We indicate this by drawing slashes through them: Dividing the numerator and denominator in the resulting factored equation by 7 again just results in those two factors of 7 disappearing from each. So again, Now there are no common factors left in the numerator and denominator, so the process ends, and we conclude that 3 / 5 is the simplest form of 42 / 70. In practice, this whole process is typically done in a single step, crossing out pairs of factors without rewriting intermediate forms:

18 18 Algebra e-book Algebrator can easily solve these type of problems:

19 Fractions 19 Clicking on the "solve step" button once shows the first step of the solution process. An explanation can be obtained by clicking on the "Explain" button.

20 20 Algebra e-book Algebrator shows an explanation for any step you want.

21 Chapter 3 Complex numbers

22 22 Algebra e-book 3 Complex numbers 3.1 Introduction In this section we define a set of numbers that has the real numbers as a subset. Definition The equation 2x = 1 has no solution in the set of integers, but in the set of rational numbers, 2x = 1 has a solution. The situation is similar for the equation x2 = -4. It has no solution in the set of real numbers because the square of every real number is nonnegative. However, in the set of complex numbers x2 = -4 has two solutions. The complex numbers were developed so that equations such as x2?4 would have solutions. The complex numbers are based on the symbol. In the real number system this symbol has no meaning. In the set of complex numbers this symbol is given meaning. We call it i. We make the definition that Complex Numbers The set of complex numbers is the set of all numbers of the form a + bi, where a and b are real numbers,, and i2 = -1. In the complex number a + bi, a is called the real part and b is called the imaginary part. If b? 0, the number a + bi is called an imaginary number. In dealing with complex numbers, we treat a + bi as if it were a binomial, with i being a variable. Thus we would write 2 + (-3)i as 2-3i. We agree that 2 + i3, 3i + 2, and i3 + 2 are just different ways of writing 2 + 3i (the standard form). Some examples of complex numbers are For simplicity we write only 7i for 0 + 7i. The complex number 9 + 0i is the real number 9, and 0 + 0i is the real number 0. Any complex number with b = 0 is a real number. For any real number a, a + 0i a. The set of real numbers is a subset of the set of complex numbers. Take a look at the figure below.

23 Complex numbers Operating A complex number is an expression of the form x + iy where x and y are real numbers. Either x or y may be equal to zero: real numbers and imaginary numbers are special cases of complex numbers. Given a complex number z = x + iy the real number x is known as the real part of z and the real number y is known as the imaginary part of z. The notation Re is used for real part and Im is used for imaginary part. Note that the imaginary part of z is a real number y (NOT iy). Adding and Subtracting Complex Numbers To add or subtract complex numbers, we add or subtract their real and imaginary parts separately: Examples: 1) (2 + 3i) + (4 + i) = 6 + 4i 2) (3-5i) - 7i = 3-12i 3) (5-4i) + (3 + 2i) - (8 + i) = -3i 4) 12-4i + (3 + 4i) = 15

24 24 Algebra e-book Multiplying complex numbers To multiply complex numbers, we use the usual rules and the identity i2 = -1: Examples: 1) (2 + 3i)(4 + 5i) = (2 4) + (3i 4) + (2 5i) + (3i 5i) = 8 12i + 10i - 15 = i 2) (2 + 3i)3 = (2 + 3i)(2 + 3i)(2 + 3i) = ( i)(2 + 3i) = i The complex conjugate Given a complex number z = x + iy, the complex conjugate is given by z* = x - iy (pronounced z star ): To find the complex conjugate of any expression, replace i by i. Examples: 1) (3 + 4i)* = 3-4i 2) (5-2i)* = 5 + 2i 3) (6i)* = -6i 4) [(2 + 4i)*]* = [2-4i]* = 2 + 4i

25 Complex numbers 25 The last expression demonstrates that A complex number and its complex conjugate have the property that: Examples: 1) (3 + 4i) + (3 + 4i)* = 3 + 4i + 3-4i = 6 2) (3 + 4i) - (3 + 4i)* = 3 + 4i i = 8i 3) 5i + (5i)* = 5i - 5i = 0 4) 5i - (5i)* = 10i Complex conjugate of a product The modulus of a complex number A complex number z = x + iy multiplied by its complex conjugate is a real number: The modulus of a complex number z = x + iy is written z and is equal to the positive square root of the sum of the squares of its real and imaginary parts: Examples:

26 26 Algebra e-book The modulus is a real number The quotient of complex numbers In order to simplify an expression with a complex number in the denominator, we multiply both the numerator and denominator by the complex conjugate: In this way, we get a real number in the denominator (the modulus squared of a + bi). Examples: In general:

27 Chapter 4 Rational expressions

28 28 Algebra e-book 4 Rational expressions 4.1 Adding and subtracting Definition A rational expression is a ratio (fraction) of two polynomials. All rational expressions are valid for all real numbers except those values that make the denominator zero. are rational expressions whereas are not. When adding (or subtracting) rational expressions with a common denominator, the process is straightforward: Simply add (or subtract) the numerators, and put the result in a rational expression with the same denominator; i.e., where a, b and c are polynomials. Example 1 Example 2 Note that there is no restriction on x because, as will be shown in another paper. When performing any operation with rational expressions of a single variable, we must make sure our final answer is simplified: a. All polynomials are in descending order; i.e., the largest power of the variable appears on the left and as we proceed to the right, the powers of the variable decrease. b. All leading terms (highest degree terms) in the denominator have positive coefficients.

29 Rational expressions 29 c. All like (similar) terms have been combined. d. The greatest common factor of the numerator and denominator is 1; i.e., all common factors have been divided out. e. Unless told otherwise, the numerator is to multiplied out and simplified, while the denominator may be left in factored form. The expressions,, are considered to be simplified. The expressions, are not considered to be simplified. The following shows how to convert the latter pair of expressions into simplified forms. [Factor out -1 from denominator] [Put in descending order] [Factor out -1 from (-x + 1)] [Eliminate the leading neg sign in denom.] [Simplify] Example 3 Substract: Solution: The factorizations of the denominators are Hence the LCD is:

30 30 Algebra e-book LCD = (x - 2)(x + 2)(x + 1) Use these factorizations to convert the rational expressions to rational expressions that have the LCD for their denominators, perform the subtraction, and simplify the result. [Acceptable as being simplified] [Preferrable] Note that the above is valid for all into result. For example, let x = 3. Then we obtain and. What this means is that if we substitute any value of, and evaluate the expressions, we obtain the same

31 Rational expressions 31

32 Chapter 5 Radicals

33 Radicals 5 Radicals 5.1 Adding and subtracting 33 The procedure for adding and subtracting square roots which may contain algebraic expressions is very similar to the one we usewhen adding and subtracting numerical square roots: simplify the square root in each term of the expression combine terms whose square roots are identical Read the following examples: Example 1: Simplify solution: We have Thus as the final answer. The method should be fairly obvious by now. Use the remaining three examples as practice problems. Try to do them yourself before looking at our solutions. Example 2: Simplify solution:

34 34 Algebra e-book First, we check to simplify each of the two square roots individually: Thus as the final answer. Algebrator can easily solve these type of problems:

35 Radicals Clicking on the "solve step" button once shows the first step of the solution process. 35

36 36 Algebra e-book An explanation can be obtained by clicking on the "Explain" button.

37 Radicals 37

38 38 Algebra e-book Example 3: Simplify solution: So as the final answer. Example 4: Simplify solution: Each square root must first be simplified: and Therefore

39 Radicals 39 This is as simple as we can get the expression. 5.2 Multiplying radicals The product rule for radicals, index, such as allows multiplication of radicals with the same Caution The product rule does not allow multiplication of radicals that have different indices. We cannot use the product rule to multiply Example 1 Multiplying radicals with the same index Multiply and simplify the following expressions. Assume the variables represent positive numbers. Solution Product rule for radicals

40 40 Algebra e-book Product rule for radicals Simplify. Simplify. Product rule for radicals Simplify Rationalize the denominator. We find a product such as by using the distributive property as we do when multiplying a monomial and a binomial. A product such as found by using FOIL as we do for the product of two binomials. 5.3 Rationalizing denominators Examples with Solutions Example 1: Simplify solution: can be

41 Radicals 41 There is no factorization of the expression in the square root possible here, so the only simplification necessary is to rationalize the denominator of the fraction. Since the square root in the denominator is already as simple as possible, we need to just multiply the numerator and denominator by that square root to achieve the required result. Although there is a t in both the numerator and the denominator, it is not a factor in the denominator, and therefore we cannot cancel the t s between the numerator and denominator of this last form. In fact, there is obviously no common factors to cancel between the numerator and denominator of this last form, so this last expression is as simple as possible and so it must be the required final answer. NOTE: Sometimes as they focus on simplifying fractions such as the final result of Example 1 above, people note that the numerator and denominator contain a power of the same expression in this case of (s + t). Without thinking they do the following: or Superficially, it may appear that one or both of these lines represents a useful simplification, but in fact, both lines really just amount to reversing the simplification previously done. The second line obviously just gives back the original problem, and the first line does as well, though in a somewhat less recognizable form. So, while it is very useful to break the simplification of complicated algebraic expressions down into a series of relatively simple independent steps, it is also important to review your entire solution to make sure that you haven t lost track of what the problem actually requires you to accomplish, and make sure that that is indeed what your final answer has accomplished. Example 2: Rationalize the denominator in and simplify the result as much as possible. solution: If we simply follow the steps illustrated in the first and second examples above, we get

42 42 Algebra e-book which has a rational denominator, but clearly can be simplified further. First 8x 3 = x 2 x so Thus, as the final simplified result. Alternatively, we could have noted that the square root in the denominator of the original fraction is not in simplest form, and so we could have started by simplifying both the square root and the fraction itself first: Now the denominator of this fraction can be rationalized by multiplying the numerator and denominator by : giving the same final simplified result as we obtained earlier. Example 3:

43 Radicals 43 Simplify solution: No common factors can be found to cancel between the numerator and denominator here. Also, the square roots which appear here obviously cannot be simplified further. So, all that s left to do to simplify this expression is to rationalize the denominator. This is accomplished by multiplying the numerator and denominator by, giving: as the final result (since there are clearly no common factors that can be cancelled between the numerator and denominator of this final expression). Notice that we used brackets explicitly to ensure that the entire numerator was multiplied by 5.4 in the first step here. Combining like radical terms Before we can combine like radical terms, we sometimes need to simplify the individual radicals to obtain like terms. Example 1 Find: First simplify The radicand, 24, has a perfect square factor, 4. Replace with Combine the like radical terms, Simplify.

44 44 Algebra e-book So, Example 2 Find: Solution First simplify The radicand, 32, has a perfect cube factor, 8. Replace with Multiply Combine the like radical terms, Also, combine the like terms -5 and -2. Simplify. So, 5.5 Combining operations Now we will simplify some radical expressions that involve more than one operation. Example 1 Simplify: Solution Factor each radicand. Write each radical as the product of radicals.

45 Radicals Simplify square roots of perfect squares. For each term, multiply the radicals. Combine like radical terms. Therefore, Example 2 Simplify: Solution To remove the parentheses, distribute Multiply radicand. by Then factor the Simplify the second term: Combine like radical terms. Thus, 45

46 Chapter 6 Equations

47 Equations 6 Equations 6.1 Linear equations 47 Equations that can be written in the form ax + b = 0 where a and b are real numbers, with a 0, are linear equations. Examples oflinear equations include 5y + 9 = 16 and 8x = 4 The following properties are used to solve linear equations. PROPERTIES OF EQUALITY For all real numbers a, b, and c: 1. If a = b then a + c = b + c Addition property of equality (The same number may be addedto both sides of an equation.) 2. If a = b then ac = bc Multiplication property of equality (Both sides of an equation may be multiplied by the same number.) Solving Linear Equations EXAMPLE (a) If x -2 = 3 then x = = 5 Addition property of equality (b) If x/2=3 then x = 2 3 = 6 Multiplication property of equality The following example shows how these properties are used to solve lineare quations. Of course, the solutions should always be checked by substitution inthe original equation. EXAMPLE Solve 2x = 3x + 2(2-3x) Solution 2x = 3x + 4-6x Distributive property 2x + 3 = -3x + 4 Combine like terms 5x + 3 = 4 Add 3x to both sides 5x = 1 Add -3 to both sides Multiply both sides by.

48 48 Algebra e-book Check by substituting in the original equation. The left side becomes 2(1/5)-5+8 and the right side becomes 3(1/5)+2(2-3(1/5)). Verify that both of these expressions simplify to 17/5. Algebrator can easily solve these type of problems: Clicking on the "solve step" button once shows the first step of the solution process.

49 Equations An explanation can be obtained by clicking on the "Explain" button. 49

50 50 Algebra e-book

51 Equations Graphing linear equations Using the Slope-Intercept Method Objectives 1. Determine the slope of a line from the graph. 2. Determine the slope of a line from two points on the line. 3. Determine the slope of a line from its linear equation in 2 variables. 4. Graph a linear equation in 2 variables using the slope-intercept method. The slope of a line is the ratio of the amount of rise to the amount of run, Slope is positive if the line rises. Slope is negative if the line falls. Slope is zero if the line is horizontal. Slope is undefined if the line is vertical. Slope of a line through 2 given points: Given 2 ordered pairs, (x1, y1) and (x2, y2) Slope intercept form of a linear equation: y = mx + b m is the slope b is the y-intercept Note that the coefficient of y must be 1!! To Graph when you know a point and the slope: 51

52 52 Algebra e-book 1. Plot the known point. 2. Count the rise and run from the known point. For a positive rise, count upward. For a negative rise, count downward. For a positive run, count to the right. For a negative run, count to the left. ( hint: if the slope is a whole number, make it a fraction over one) 3. Place the second point where the next ordered pair is located based on your counting above. 4. Draw a line (not a segment!) connecting the 2 points. Examples: Graph a line that contains a) the point (1, -3) with b) the point (2, 5) with m = -4 To graph a line using the slope-intercept method: 1. Put the equation in slope-intercept form (solve for y ) 2. Determine the slope and y -intercept from the equation.

53 Equations 3. Plot the y -intercept. 4. Locate the next point in the line by counting the slope. 5. Draw a line (not a segment) connecting the 2 points. Examples: c) y = -2 x + 4 d) 6 x + 3 y = -9 e) 4y = 14 53

54 Algebra e-book Absolute value equations Solving an Equation of the Form z = w Here are several examples of solving equations of the form z = w. Example 1 Solve: 4x - 34 = 6x + 14 Solution 4x - 34 = 6x + 14 Replace 4x - 34 with z and 6x + 14 with w: z = w So: z = w or z = -w Substitute 4x - 34 for z and 6x + 14 for w. Now, solve for x. 4x - 34 = 6x + 14 or 4x = 6x + 48 or -2x = 48 or or So: x = -24 or 4x - 34 = Check x = -24 4x - 34 = 6x + 14 Check x = -3 4x - 34 = 6x x x - 34 = -6x x = 20 10x = 2 x= Let's check the solutions: -(6x + 14)

55 Equations Is 4(-24) - 34 = 6(-24) + 14? Is 4(2) - 34 = 6(2) + 14 Is -130 = -130? Is -26 = 26? Is 130 = 130? Yes 55 Is 26 = 26? Yes So, the solutions are x = -24 and x = 2. Example 2 Solve: 3x - 4 = 3x + 16 Solution 3x - 4 = 3x + 16 Replace 3x - 4 with z and 3x + 16 with w: z = w So: z = w or z = -w Substitute 3x - 4 for z and 3x + 16 for w. Now, solve for x. 3x + 4 = 3x + 16 or 3x - 4 = -(3x + 16) 3x = 3x + 20 or 3x - 4 = -3x - 16 or 6x = = 20 x = -2 Since 0 = 20 is a contradiction, the left equation does not lead to a solution. Check x = -2 3x - 4 = 3x + 16 Is 3(-2) - 4 = 3(-2) + 16? Is -10 = 10? Is 10 = 10? Yes Thus, -2 is the only solution.

56 Chapter 7 Systems of equations

57 Systems of equations 7 Systems of equations 7.1 Solving systems by elimination Example Use elimination to find the solution of this system. x - 3y = -17 First equation -x + 8y = 52 Second equation Solution Add the two equations. Simplify. The x-terms have been eliminated. To solve for y, divide both sides by 5. 3 x - y = -x + 8 = y 5 2 0x + 5 = y 3 5 5y = 35 y =7 To find the value of x, substitute 7 for y in either of the original equations. Then solve for x. We will use the first equation. Substitute 7 for y. Multiply. Add 21 to both sides. The solution of the system is (4, 7). To check the solution, substitute 4 for x and 7 for y into each original equation. Then simplify. In each case, the result will be a true statement. The details of the check are left to you. 1 7 x - 3y = -17 x - 3(7) = -17 x - 21 = -17 x =4 57

58 58 Algebra e-book In the two original equations in the previous example, the coefficients of x were opposites. Thus, when the equations were added, the x-terms were eliminated. 3 1x - y = x + 8 = y = y 3 5 When the coefficients of neither variable are opposites, we choose a variable. Then we multiply both sides of one (or both) equations by an appropriate number (or numbers) to make the coefficients of that variable opposites. Note: The Multiplication Principle of Equality enables us to multiply both sides of an equation by the same nonzero number without changing the solutions of the equation. 7.2 Solving systems by substitution Graphing is not always the best way to find the solution of a system of equations. It may be difficult to read the coordinates of the point of intersection. This is especially true when the coordinates are not integers. Instead we can use algebraic methods to solve the system. One algebraic method for finding the solution of a linear system is the substitution method. Procedure To Solve a Linear System By Substitution Step 1 Solve one equation for one of the variables in terms of the other variable. Step 2 Substitute the expression found in Step 1 into the other equation. Then, solve for the variable. Step 3 Substitute the value obtained in Step 2 into one of the equations containing both variables. Then, solve for the remaining variable. Step 4 To check the solution, substitute it into each original equation. Then simplify. Example Use substitution to find the solution of this system.

59 Systems of equations 59 2x + y = 4 First equation 3x + y = 7 Second equation Solution Step 1 Solve one equation for one of the variables in terms of the other variable. Either equation may be solved for either variable. For instance, let s solve the first equation for y. 2x + y = 4 Subtract 2x from both sides. y = -2x + 4 The equation y = -2x + 4 means that y and the expression -2x + 4 are equivalent. Step 2 Substitute the expression found in Step 1 into the other equation. Then, solve for the variable. 3x + y = 7 Substitute -2x + 4 for y in the second equation. Combine like terms. Subtract 4 from both sides. Now we know x = 3. 3x + (-2x + = 7 4) =7 x+ 4 =3 x Next, we will find y. Step 3 Substitute the value obtained in Step 2 into one of the equations containing both variables. Then, solve for the remaining variable. We will use the equation from Step 1. y = -2x + 4 Substitute 3 for x. y = -2(3) + 4 y = -2 Simplify. The solution of the system is x = 3 and y = -2. The solution may also be written as (3, -2). Step 4 To check the solution, substitute it into each original equation. Then simplify. Substitute x = 3 and y = -2 into both of the original equations: First equation Second equation 2x + y =4 3x + y =7

60 60 Algebra e-book Is 2(3) + (-2) = 4? Is 6 - Is 3(3) + 2 =4? Is 4 = 4? Yes Is 9 - (- = 7? 2) =7? 2 = 7? Yes 7 Since (3, -2) satisfies both equations, it is the solution of the system. Note: If we graphed the system, the lines would intersect at the point (3, -2).

61 Chapter 8 Inequalities

62 62 Algebra e-book 8 Inequalities 8.1 Linear inequalities To write that one number is greater than or less than another number, we use the following symbols. INEQUALITY SYMBOLS < means is less than means is less than or equal to > means is greater than means is greater than or equal to An equation states that two expressions are equal; an inequality states that they are unequal. A linear inequality is an inequality that can be simplified to the form ax < b (Properties introduced in this section are given only for <, but they are equally valid for >, or.) Linear inequalities are solved with the following properties. PROPERTIES OF INEQUALITY For all real numbers a, b, and c: 1. If a < b then a + c < b + c 2. If a < b and if c > 0 then ac < bc 3. If a < b and if c < 0 then ac > bc Pay careful attention to property 3; it says that if both sides of an inequality are multiplied by a negative number, the direction of the inequality symbol must be reversed. Solving Linear Inequalities EXAMPLE Solve 4-3y 7 + 2y. Solution Use the properties of inequality. 4-3y + (-4) -3y 7 + 2y + (-4) Add -4 to both sides y Remember that adding the same number to both sides never changes the direction of the inequality symbol.

63 Inequalities -3y + (-2y) -5y y + (-2y) Add -2y to both sides. 3 Multiply both sides by -1/5. Since -1/5 is negative, change the direction of the inequality symbol. CAUTION It is a common error to forget to reverse the direction of the inequality sign when multiplying or dividing by a negative number. For example, to solve -4x 12 we must multiply by -1/4 on both sides and reverse the inequality symbol to get x -3. The solution y -3/5 in the previous example represents an interval on the number line. Interval notation often is used for writing intervals. With interval notation, y -3/5 is written as. This is an example of a half-open interval, since one endpoint, -3/5, is included. The open interval (2, 5) corresponds to 2 < x < 5, with neither endpoint included. The closed interval [2, 5] includes both endpoints and corresponds to 2 x 5. The graph of an interval shows all points on a number line that correspond to the numbers in the interval. To graph the interval, for example, use a solid circle at -3/5 since -3/5 is part of the solution. To show that the solution includes all real numbers greater than or equal to -3/5 draw a heavy arrowpointing to the right (the positive direction). See Figure 1. Graphing Linear Inequalities EXAMPLE Solve -2 < 5 + 3m < 20. Graph the solution. Solution The inequality -2 < 5 + 3m < 20 says that 5 + 3m is between -2 and 20. Solve this inequality with an extension of the properties given above.work as follows, first adding -5 to each part (-5) < 5 + 3m + (-5) < 20+ (-5) -7 < 3m < 15

64 64 Algebra e-book Now multiply each part by 1/3. A graph of the solution is given in Figure 2; here open circles are used to show that -7/3 and 5 are not part of the graph. (Some textbooks use brackets in place of solid circles for the graph of a closed interval, and parentheses in place of open circles for the graph of an open interval.) Graphing linear inequalities You have probably already studied linear equations. We now turn our attention to linear inequalities. Definition A linear inequality is a linear equation with the equal sign replaced by an inequality symbol. Linear Inequality If A, B, and C are real numbers with A and B not both zero, then Ax + By = C is called a linear inequality. In place of =, we can also use =, <, or >. Graphing Linear Inequalities Consider the inequality -x + y > 1. If we solve the inequality for y, we get y > x + 1. Which points in the xy-plane satisfy this inequality? We want the points where the y-coordinate is larger than the x-coordinate plus 1. If we locate a point on the line y = x + 1, say (2, 3), then the ycoordinate is equal to the x-coordinate plus 1. If we move upward from that point, to say (2, 4), the y-coordinate is larger than the x-coordinate plus 1. Because this argument can be made at every point on the line, all points above the line satisfy y > x + 1. Likewise, points below the line satisfy y < x + 1. The solution sets, or graphs, for the inequality y > x + 1 and the inequality y < x + 1 are the shaded regions shown in the figures below. In each case the line y = x + 1 is dashed to indicate that points on the line do not satisfy the inequality and so are not in the solution set. If the inequality symbol is = or =, then points on the boundary line also satisfy the inequality, and the line is drawn solid.

65 Inequalities 65 Every nonvertical line divides the xy-plane into two regions. One region is above the line, and the other is below the line. A vertical line also divides the plane into two regions, but one is on the left side of the line and the other is on the right side of the line. An inequality involving only x has a vertical boundary line, and its graph is one of those regions. Graphing a Linear Inequality 1. Solve the inequality for y, then graph y = mx + b. y > mx + b is satisfied above the line. y = mx + b is satisfied on the line itself. y < mx + b is satisfied below the line. 2. If the inequality involves x and not y, then graph the vertical line x = k. x > k is satisfied to the right of the line. x = k is satisfied on the line itself. x < k is satisfied to the left of the line. Example 1 Graphing linear inequalities Graph each inequality. a) b) y - 2x + 1

66 66 Algebra e-book c) 3x - 2y < 6 Solution a) The set of points satisfying this inequality is the region below the line. To show this region, we first graph the boundary line. The slope of the line is and the y-intercept is (0, -1). Start at (0, -1) on the y-axis, then rise 1 and run 2 to get a second point of the line. We draw the line dashed because points on the line do not satisfy this inequality. The solution set to the inequality is the shaded region shown in the figure below. b) Because the inequality symbol is, every point on or above the line satisfies this inequality. To show that the line y = - 2x + 1 is included, we make it a solid line. See the figure below. c) First solve for y: 3x <6 2y -2y < -3x + 6

67 Inequalities y 67 Divide by -2 and reverse the inequality. To graph this inequality, use a dashed line for the boundary the line. See the figure below for the graph. and shade the region above Caution In Example 1(c) we solved the inequality for y before graphing the line. We did that because < corresponds to the region below the line and > corresponds to the region above the line only when the inequality is solved for y. Algebrator can easily solve these type of problems:

68 68 Algebra e-book Clicking on the "Graph All" button shows the graphical representation.

69 Inequalities 69 Graphing multiple inequalities together is one of the many features of Algebrator. 8.2 Simultaneous inequalities If we join to simple inequalities with the connective "and" or the connective "or", we get a compound inequality. A compound inequality using the connective "and" is true if and only if both simple inequalities are true. Example 1

70 70 Algebra e-book Compound inequalities using the connective "and" Determine whether each compound inequality is true. a) 3 > 2 and 3 < 5 b) 6 > 2 and 6 < 5 Solution a) The compound inequality is true because 3 > 2 is true and 3 < 5 is true. b) The compound inequality is false because 6 < 5 is false. A compound inequality using the connective "or" is true if one or the other or both of the simple inequalities are true. It is false only if both simple inequalities are false. Example 2 Compound inequalities using the connective "or" Determine whether each compound inequality is true. a) 2 < 3 or 2 > 7 b) 4 < 3 or 4 = 7 Solution a) The compound inequality is true because 2 < 3 is true. b) The compound inequality is false because both 4 < 3 and 4 = 7 are false. If a compound inequality involves a variable, then we are interested in the solution set to the inequality. The solution set to an "and" inequality consists of all numbers that satisfy both simple inequalities, whereas the solution set to an "or" inequality consists of all numbers that satisfy at least one of the simple inequalities. Example 3 Solutions of compound inequalities

71 Inequalities Determine whether 5 satisfies each compound inequality. a) x < 6 and x < 9 b) 2x - 9 = 5 or -4x = -12 Solution a) Because 5 < 6 and 5 < 9 are both true, 5 satisfies the compound inequality. b) Because = 5 is true, it does not matter that -4 5 = -12 is false. So 5 satisfies the compound inequality. 71

72 Chapter 9 Polynomials

73 Polynomials 9 Polynomials 9.1 Adding and subtracting 73 Objective Learn how to add and subtract polynomials. This lesson is not difficult conceptually, but it is important that you get sufficient practice in addition and subtraction of polynomials. Adding Polynomials Let's begin with an example in order to see how to add polynomials. Example 1 Add x 3 + x + 1 and 3x 3 + x 2 + 2x. Solution First, write the polynomials side by side. ( x 3 + x + 1 ) + ( 3x 3 + x 2 + 2x ) The monomials x 3 and x each occur twice, so group the terms together and combine them. x 3 + x x 3 + x 2 = ( x 3 + 3x 3 ) + x 2 + ( x + 2x ) + 2x +1 = 4x 3 + x 2 + 3x + 1 This polynomial is simpler than the original two polynomials written side by side. Like Terms To be explicit about the steps taken to add polynomials, let's talk a little bit about like terms. Definition of Like Terms When we are given two polynomials, we say the monomials in the polynomials are like terms if they contain exactly the same number of occurrences of each variable. Example 2

74 74 Algebra e-book Name the like terms in x 4 + 4x 3 + 6x x + 1 and x 3 + x Solution There are several pairs of like terms, shown in the diagram below. The pairs of like terms are connected by arrows. So, the like terms are 4x 3 and x 3, 6x 2 and x 2, 1 and 1. Example 3 Name the like terms in ab + 2a + 2b and ab - a + 1. Solution In this case, there are two pairs of like terms, ab and ab, 2a and -a, shown in the diagram below. Whenever there are like terms, collect them and add them together to get a single term. In Example 2, add 4x 3 and x 3 together to get 5x 3. In the same way, add 6x 2 and x 2 together to get 7x 2, and finally, = 2. When we do this, we get a simpler polynomial. ( x 4 + 4x 3 + 6x x + 1 ) + ( x 3 + x ) = x 4 + 5x 3 + 7x 2 + 4x + 2 In Example 3, collect the ab and a terms to get ( ab + 2a + 2b ) + ( ab - a + 1) = 2ab + a + 2b + 1. Key Idea Sums of polynomials can be simplified by adding together like terms. In the same way, differences of polynomials can be simplified by collecting together all pairs of like terms.

75 Polynomials 75 Subtracting Polynomials Example 4 Simplify ( x 3-3x x - 1) - ( x 2 + 2x + 1). Solution First write this as an addition expression by adding the additive inverse. ( x 3-3x x - 1) - ( x 2 + 2x + 1) = ( x 3-3x x - 1) + ( -x 2-2x - 1) To simplify the difference, collect all pairs of like terms. These are shown in the diagram below. Collect the terms to get the following polynomial. x 3 + ( - 3-1) x 2 + (3-2) x + ( - 1-1) = x 3-4x 2 + x - 2 There is another method of adding and subtracting polynomials that is similar to adding and subtracting numbers. Write the polynomials one over the other, with like terms lined up in columns. To find the difference, either subtract the coefficients in each of the columns, or add the additive inverse. Example 5 Subtract 4x 3 + 5x 2 + 6x + 7 and 2x 3 + 3x + 4. Solution Write the polynomials with the like terms in columns. Notice that a zero coefficient is added for each missing term. Now add the coefficients in each column.

76 Algebra e-book Factoring After studying this lesson, you will be able to: Factor various types of problems. Steps of Factoring: 1. Factor out the GCF 2. Look at the number of terms: 2 Terms: Look for the Difference of 2 Squares 3 Terms: Factor the Trinomial 4 Terms: Factor by Grouping 3. Factor Completely 4. Check by Multiplying This section is a review of the types of factoring we've covered so far. Follow the steps listed above to factor the problems. Example 1 Factor 3x st : Look for a GCF...the GCF is 3 so we factor out 3: 3( x 2-9) 2 nd : Look at the number of terms in the parenthesis. There are 2 terms and it is the difference of 2 squares. We factor the difference of 2 squares (keeping the 3). 3(x + 3) ( x - 3) 3 rd : Now, make sure the problem is factored completely. It is. 4 th : Check by multiplying. Example 2 Factor 9y 2-42y + 49

77 Polynomials 77 1 st : Look for a GCF...the GCF is 1 so we don't have to worry about that. 2 nd : Look at the number of terms. There are 3 terms so we factor the trinomial. -make 2 parentheses -using the sign rules, we know the signs will be the same because the constant term is positive - we also know they will be negative because the inside/outside combination must equal -58y -find the factors of the 1 st term: 1y, 9y and 3y, 3y. Let's try 3y, 3y -find the factors of the constant term: 1, 49 and 7, 7. Let's try 7, 7 (3y - 7) (3y - 7) -check the inside/outside combination: inside we have -21y and outside we have -21y which adds up to -42y 3 rd : Now, make sure the problem is factored completely. It is. 4 th : Check by multiplying. Algebrator can easily solve these type of problems:

78 78 Algebra e-book Clicking on the "solve step" button once shows the first step of the solution process. An explanation can be obtained by clicking on the "Explain" button.

79 Polynomials Clicking on the "Solve All" button will show all steps in the solution process. 79

80 80 Algebra e-book Example 3 Factor x 3-5x 2-9x st : Look for a GCF...the GCF is 1 so we don't have to worry about that. 2 nd : Look at the number of terms. There are 4 terms so we factor by grouping. Group the terms (x 3-5x 2 ) + (- 9x + 45 ) Take the GCF of the each group: x 2 (x - 5 )(- 9(x - 5 )) Take the GCF of the entire problem: (x - 5 )(x 2-9) 3 rd : Now, make sure the problem is factored completely. It isn't. We can factor the second parenthesis. (x - 5 )(x + 3)(x - 3) 4 th : Check by multiplying. 9.3 Evaluating A polynomial is an algebraic expression. Like other algebraic expressions involving variables, a polynomial has no specific value unless the variables are replaced by numbers. A polynomial can be evaluated without using funcion notation. Value of a Polynomial a) Find the value of -3x4 - x3 + 20x + 3 when x = 1. b) Find the value of -3x4 - x3 + 20x + 3 when x = -2. Solution a) Replace x by 1 in the polynomial: -3x4 - x3 + 20x + 3 = -3(1)4 - (1)3 + 20(1) + 3 =

81 Polynomials 81 = 19 So the value of the polynomial is 19 when x = 1. b) Replace x by -2 in the polynomial: -3x4 - x3 + 20x + 3 = -3(-2)4 - (-2)3 + 20(-2) + 3 = -3(16) - (-8) = = -77 So the value of the polynomial is -77 when x = Higher degree polynomials It is not necessary always to use substitution to factor polynomials with higher degrees or variable exponents. In the next example we use trial and error to factor two polynomials of higher degree and one with variable exponents. Remember that if there is a common factor to all terms, factor it out first. Example 1 Higher-degree and variable exponent trinomials Factor each polynomial completely. Variables used as exponents represent positive integers. a) x8-2x4-15 b) -18y7 + 21y4 + 15y c) 2u2m - 5um - 3 Solution a) To factor by trial and error, notice that x8 = x4 x4. Now 15 is 3 5 or Using 1 and 15 will not give the required -2 for the coefficient of the middle term. So choose 3 and -5 to get the -2 in the middle term:

82 82 Algebra e-book x8-2x4-15 = (x4-5)(x4 + 3) b) -18y7 + 21y4 + 15y = 3y(6y6-7y3-5) Factor out the common factor -3y first. = 3y(2y3 + 1)(3y3-5) Factor the trinomial by trial and error. c) Notice that 2u2m = 2um um and 3 = 3 1. Using trial and error, we get 2u2m - 5um - 3 = (2um + 1)(um - 3). 9.5 Completing the square The essential part of completing the square is to recognize a perfect square trinomial when given its first two terms. For example, if we are given x2 + 6x, how do we recognize that these are the first two terms of the perfect square trinomial x2 + 6x + 9? To answer this question, recall that x2 + 6x + 9 is a perfect square trinomial because it is the square of the binomial x + 3: (x + 3)2 = x x + 32 = x2 + 6x + 9 Notice that the 6 comes frommultiplying 3 by 2 and the 9 comes from squaring the 3. So to find the missing 9 in x2 + 6x, divide 6 by 2 to get 3, then square 3 to get 9. This procedure can be used to find the last term in any perfect square trinomial in which the coefficient of x2 is 1. Rule for Finding the Last Term The last term of a perfect square trinomial is the square of one-half of the coefficient of the middle term. In symbols, the perfect square trinomial whose first two terms are. Helpful hint Review the rule for squaring a binomial: square the first term, find twice the product of the two terms, then square the last term. If you are still using FOIL to find the square of a binomial, it is time to learn the proper rule. Example 1

83 Polynomials 83 Finding the last term Find the perfect square trinomial whose first two terms are given. Solution a) One-half of 8 is 4, and 4 squared is 16. So the perfect square trinomial is x2 + 8x b) One-half of -5 is c) One-half of d) One-half of is, and, and is squared is squared is, and. So the perfect square trinomial is.. So the perfect square trinomial is. So the perfect square trinomial is Another essential step in completing the square is to write the perfect square trinomial as the square of a binomial. Recall that a2 + 2ab + b2 = (a + b)2 and a2-2ab + b2 = (a - b)2. Example 2 Factoring perfect square trinomials Factor each trinomial. Solution a) The trinomial x2 + 12x + 36 is of the form a2 + 2ab + b2 with a = x and b = 6. So x2 + 12x + 36 = (x + 6)2. Check by squaring x + 6.

84 84 Algebra e-book is of the form a2-2ab + b2 with a = y and b) The trinomial Check by squaring c) The trinomial. So. is of the form a2-2ab + b2 with a = z and. So

85 Chapter 10 Functions

86 86 Algebra e-book 10 Functions 10.1 Adding and subtracting Recall that a function is a rule that assigns exactly one output number to each input number. We can combine functions in several ways in order to make new functions. Definion Sum and Difference of Two Functions Given two functions, f(x) and g(x): The sum of f and g, written (f + g)(x), is defined as (f + g)(x) = f(x) + g(x). The difference of f and g, written (f - g)(x), is defined as (f - g)(x) = f(x) - g(x). The domains of (f + g)(x) and (f - g)(x) consist of all real numbers that are in the domain of both f(x) and g(x). Example 1 Given f(x) = 0.5x + 4 and g(x) = 0.25x + 1, find the sum (f + g)(x). Solution Use the definition for the sum of functions. (f + g)(x) = f(x) + g(x) Substitute for f(x) and g(x). = (0.5x + 4) + (0.25x + 1) Combine like terms. = 0.75x + 5 So, (f + g)(x) = 0.75x + 5.

87 Functions 87 Example 2 Given f(x) = 5x2 + 6x - 12 and g(x) = 8x - 15, find the difference (f - g)(x). Solution Use the definition for the difference of functions. (f - g)(x) = f(x) - g(x) Substitute for f(x) and g(x). = (5x2 + 6x - 12) - (8x 15) Remove parentheses. = 5x2 + 6x x + 15 Combine like terms. = 5x2-2x + 3 So, (f - g)(x) = 5x2-2x + 3. You have already learned to evaluate a function for a specific value of the input. For example, if f(x) = 2x + 1, then we can find f(3) by substituting 3 for x in the f(x) = 2x + 1 function rule. f(3) = 2(3) + 1 =7 Likewise, we can evaluate the sum or difference of two functions for a given number. There are two methods that are typically used. Procedure To Evaluate the Sum or Difference of Functions Step 1 Find (f + g)(x) or (f - g)(x).

88 88 Algebra e-book Step 2 Use x = a to find (f + g)(a) or (f - g)(a) Linear functions You have learned that a function is a rule that assigns to each input number, x, exactly one output number, y. A linear function is a rule that can be written in the form: y = Ax + B or f(x) = Ax + B For example, the following are linear functions: y= x or f(x) = x y = -2x + 6 or f(x) = -2x + 6 The following functions are not linear functions: y = x2-7 or f(x) = x2-7 (Not linear since x is squared.) (Not linear since x is in the denominator.) For any linear function y = Ax + B: The domain is all real numbers, which is the interval (-8, +8 ). This is because we can multiply any input x, by any real number A, and then add any real number B. If A? 0, then the range is all real numbers, (-8, +8 ). If A = 0, then the graph is a horizontal line. In that case, the range is the single number B. In applications of linear functions, however, the domain and range may be restricted to only a portion of the real numbers.

89 Functions Graphing linear functions We can graph a linear function by calculating two ordered pairs, plotting the corresponding points on a Cartesian coordinate system, and then drawing a line through the points. We typically calculate and plot a third point as a check. To find ordered pairs, we select values for x and then calculate the corresponding values for y. Thus, the output value y depends on our choice of the input value x. For this reason, the variable y is frequently called the dependent variable and the variable x is called the independent variable. Example 1 Make a table of at least three ordered pairs that satisfy the function f(x) = 3x - 1. Then, use your table to graph the function. Solution To make a table, select 3 values for x. We ll let x = -2, 0, and 2. Substitute the values of x into the function and simplify. x f(x) = 3x f(-2) = 3(-2) - 1 = -6-1 = f(0) = 3(0) - 1 = 0-1 = -1 f(2) = 3(2) - 1 = 6-1 = 5 (x, y) (-2, -7) (0, -1) (2, 5)

90 90 Algebra e-book Now, plot the points (-2, -7), (0, -1), and (2, 5) and then draw a line through them. Two important characteristics of the graph of a linear function are its y-intercept and its slope. The y-intercept is the point where the line crosses the y-axis. The slope measures the steepness or tilt of the line. Slope is defined as the ratio of the rise to the run of the line. When moving from one point to another on the line, the rise is the vertical change and the run is the horizontal change. The linear function, f(x) = Ax + B, is another way of writing the familiar slope-intercept form for the equation of a line, y = mx + b. In f(x) = Ax + B the slope of the line is given by A and the y-intercept is given by B. f(x) = Ax + B is equivalent to y = mx + b This means that the graphs of all linear functions are straight lines. That is why such functions are called linear.

91 Functions 91 Example 2 Graph the function: Solution This has the form of a linear function. Thus, the slope is and the y-intercept is b = -3. To graph the line, first plot the y-intercept; that is, plot the point (0, -3). From this point rise in the y-direction 5 units (the numerator of the slope) and run in the x-direction 4 units (the denominator of the slope). The new location (4, 2) is a second point on the line. Finally, connect the plotted points Quadratic functions A Quadratic (or Second-Degree) Function is a function of the form f(x) = ax2 + bx + c, where a, b and c are real numbers and a? 0. This form is called the General Form of a Quadratic Function. The graph of a quadratic function is a parabola with a "turning point" called the vertex. The domain of a quadratic function is all real numbers.

92 92 Algebra e-book f(x) = x2 is concave up while f(x) = -x2 + 4x + 21 concave down. So, when looking at the general form for a quadratic function, f(x) = ax2 + bx + c, what does a tell us about the graph? The sign of a tells us whether the parabola is concave up or concave down. There is another form of a quadratic function, called the Standard Form of a Quadratic Function y = a(x - h)2 + k is a quadratic function with vertex (h, k) and axis of symmetry x = h. It opens up if a > 0 and down if a < 0. It is wider than f(x) = x2 if -1 < a < 1, and narrower if a > 1 or if a < -1. Now if the function is in General Form, f(x) = ax2 + bx + c, the vertex is not as obvious. There is a helpful formula, however: Given a quadratic function f(x) = ax2 + bx + c, the vertex is the point: When you graph a Quadratic Function (parabola), you should show: Vertex Axis of Symmetry x-intercepts (or ) if it has any zeros y-intercept Graphing quadratic functions The Graph of f(x) = Ax2 + Bx + C To graph a quadratic function, first calculate several ordered pairs. Then, plot the corresponding points on a Cartesian coordinate system. Finally, connect the points with a smooth line. Example 1 Make a table of five ordered pairs that satisfy the function f(x) = x2. Then, use the table to graph the function. Solution To make a table, select 5 values for x. We will let x = -2, -1, 0, 1, and 2. Substitute those values of

93 Functions 93 x into the function and simplify. x f(x) = x2 (x,y) -2 f(-2) = (-2)2 = 4 (-2, 4) -1 f(-1) = (-1)2 = 1 (-1, 1) 0 f(0) = (0)2 = 0 (0, 0) 1 f(1) = (1)2 = 1 (1, 1) 2 f(2) = (2)2 = 4 (2, 4) Now, plot the points and connect them with a smooth curve. The graph of a quadratic function, f(x) = Ax2 + Bx + C, has a distinctive shape called a parabola. The sign of the coefficient of x2, A, determines whether the graph opens up or down. When A is positive, the parabola opens up. When A is negative, the parabola opens down. Graph the functions and state the domain and range of each.

94 94 Algebra e-book a. f(x) = x2 + 3 b. f(x) = x2-5 Solution a. Graph The graph of f(x) = x2 + 3 is related to the graph of y = x2: For each input value x, the output of f(x) = x2 + 3 is 3 more than the output of f(x) = x2. Thus, the graphs have the same shape, but the graph of f(x) = x2 + 3 is shifted up 3 units. Domain Since we can square any real number, the domain is all real numbers or (-8, +8 ). Range We can see from the graph that the smallest value of y is 3. Therefore, the range is y = 3 or [3, +8 ). b. Graph To graph f(x) = x2-5, we note that for each input value x, the output of f(x) = x2-5 is 5 less than the output of f(x) = x2. Thus, the graph is shifted down 5 units. Domain Since we can square any real number, the domain is all real numbers or (-8, +8 ). Range We can see from the graph that the smallest value of y is -5. Therefore, the range is y = -5 or [-5, +8 ).

95 Functions Composition Another way to combine two functions is called the composition of functions. This is where the output of one function becomes the input of a second function. Definition Composition of Functions Let f(x) and g(x) represent two functions. The composition of f and g, written (f? g)(x), is defined as (f? g)(x) = f[g(x)] Here, g(x) must be in the domain of f(x). If it is not, then f[g(x)] will be undefined. Note: In the composition (f? g)(x), the output from g(x) is the input for f(x). Example 1 Given f(x) = 3x - 5 and g(x) = x2-7, find: a. (f? g)(x)

96 96 Algebra e-book b. (g? f)(x) Solution a. Use g(x) as the input for f(x). (f? g)(x) = f[g(x)] Replace g(x) with x2-7. = f[x2-7] In f(x), replace x with x2-7. = 3(x2-7) - 5 Remove parentheses. = 3x Subtract. = 3x2-26 So, (f? g) = 3x2-26. b Use f(x) as the input for g(x).. Replace f(x) with 3x - 5. (g? f)(x) = g[f(x)] = g[3x - 5] In g(x), replace x with 3x - 5. = (3x - 5)2-7 Square the binomial. = 9x2-15x - 15x Combine like terms. = 9x2-30x + 18 So, (g? f)(x) = 9x2-30x Notes: In (f? g)(x) = f[g(x)], we substitute the expression for g(x) into f(x). In (g? f)(x) = g[f(x)], we substitute the expression for f(x) into g(x). In this example: (f? g)(x) = 3x2-19 (g? f)(x) = 9x2-30x In most cases, (f? g)(x) and (g? f)(x) do NOT yield the same result Exponential functions Definition of an Exponential Function

97 Functions 97 Previously, you have studied functions that have terms where the base is a variable and the exponent is a constant. For example, In this lesson, you will study exponential functions. Exponential functions have terms where the base is a constant and the exponent contains a variable. Definition Exponential Function An exponential function is a function that has the form: f(x) = bx where b and x are real numbers, b > 0, and b? 1. The domain is all real numbers. The range is all positive real numbers. Note: We restrict the values of the base, b, so that the function f(x) = bx is a one to one function.? 1. A multiple of an exponential function is also an exponential function. This includes the following forms: General form Example f(x) = A bx f(x) = 5 2x f(x) = C + A bx f(x) = x

98 98 Algebra e-book Note the similarities and differences between the graphs of linear, quadratic, and exponential functions: Linear: f(x) = 2x (straight line) Quadratic: f(x) = x2 (parabola) Exponential: f(x) = 2x (a curve)

99 Chapter 11 Trigonometry

100 100 Algebra e-book 11 Trigonometry 11.1 Angles and degree measures An angle has three parts: an initial ray, a terminal ray, and a vertex (the point of intersection of the two rays), as shown in the figure below. An angle is in standard position if its initial ray coincides with the positive x-axis and its vertex is at the origin. We assume that you are familiar with the degree measure of an angle. It is common practice to use θ (the Greek lowercase theta) to represent both an angle and its measure. Angles between 0º and 90º are acute, and angles between 90º and 180º are obtuse. Positive angles are measured counterclockwise, and negative angles are measured clockwise. For instance, the figure below shows an angle whose measure is -45º. You cannot assign a measure to an angle by simply knowing where its initial and terminal rays are located. To measure an angle, you must also know how the terminal ray was revolved. For example, this figure shows that the angle measuring -45º has the same terminal ray as the angle measuring 315º. Such angles are coterminal. In general, if θ is any angle, then θ + n(360 ), n is a nonzero integer is coterminal with θ.

101 Trigonometry 101 An angle that is larger than 360º is one whose terminal ray has been revolved more than one full revolution counterclockwise, as shown in the following figure. NOTE It is common to use the simbol θ to refer to both an angle and its measure. For instance, in the previous figure, you can write the measure of the smaller angle as θ = 45. Radian Measure To assign a radian measure to an angle θ, consider θ to be a central angle of a circle of radius 1, as shown in the figure below. The radian measure of θ is then defined to be the length of the arc of the sector. Because the circumference of a circle is 2πr, the circumference of a unit circle (of radius 1) is 2π. This implies that the radian measure of an angle measuring is 360. In other words, 360 = 2π radians. Using radian measure for θ, the length s of a circular arc of radius r is s = rθ, as shown in the figure below.

102 102 Algebra e-book You should know the conversions of the common angles shown in the next figure. For other angles, use the fact that is equal to radians Trigonometric functions There are two common approaches to the study of trigonometry. In one, the trigonometric functions are defined as ratios of two sides of a right triangle. In the other, these functions are defined in terms of a point on the terminal side of an angle in standard position. We define the six trigonometric functions, sine, cosine, tangent, cotangent, secant, and cosecant (abbreviated as sin, cos, etc.), from both viewpoints. Definition of the Six Trigonometric Functions Right triangle definitions, where (see the figure below)

103 Trigonometry 103 Circular function defiinitions, where θ is any angle (see the figure below). The following trigonometric identities are direct consequences of the definitions ( is the Greek letter phi). Trigonometric Identities (Note that sin2θ is used to represent (sin θ)2.)

104 104 Algebra e-book