Applied Math Book, Sem 1, 2024-25 PDF

2 This work is licensed under a Creative Commons “Attribution-NonCommercial- ShareAlike 4.0 International” license. License : https://creativecommons.org/licenses/by-nc-sa/4.0/deed.en The content of this course material is compiled and organized by Saba Hasan, Mohammed Sarfaraz, Wardah Al Majraﬁ and Jithin Mathew of UTAS, Ibri, while updates and corrections are done by Dr.Naseer Ahmed and Shreemathi Vendantarajagopalan, UTAS, Nizwa. Majority of the content of this material is taken from https://openstax.org/. Resources from this material can be directly accessed for free from the following websites: 1. Precalculus at https://openstax.org/details/books/precalculus 2. Algebra and Trigonometry at https://openstax.org/details/books/algebra-and-trigonometry 3. Introductory Statistics at https://openstax.org/details/books/introductory-statistics 4. Applied Finite Mathematics at http://cnx.org/content/col10613/1.5 Correspondingly, each page of the material is provided with proper attribution, the link of which will direct to the original source. Contents Chapter 1 System of linear Equations and Inequalities 7 1.1 Solving Systems of Linear Equations with Two Variables 8 1.2 Solving Systems of Linear Inequalities 17 Chapter 2 Functions and their Graphs 25 2.1 Functions 26 2.2 Composition of Functions 39 2.3 Inverse Function 41 2.4 Quadratic Equations 43 2.5 Graphing Functions 46 Chapter 3 Exponential and Logarithmic Functions 57 3.1 Exponential and Logarithmic Functions: 58 3.2 Applications of Exponential and Logarithmic Functions 71 Chapter 4 Statistics 75 4.1 Measures of Central Tendency 77 4.2 Summarizing Data into Tables and Graphs 81 Chapter 5 Probability 89 5.1 Basic Concepts of Probability 90 5.2 Probability Using Tree Diagrams 97 5.3 Permutations and Combinations 99 5.4 Probability Using Permutations and Combinations 104 Chapter 6 Mathematics of Finance 109 6.1 Simple Interest and Discount 109 6.2 Compound Interest 114 3 4 Applied Mathematics Learning Outcomes A student who satisfactorily complete the course should be able to: (a) Solve two variables linear equations and inequalities and sketch their graph (b) Interpret a series of three simultaneous inequalities of two variables, display them graphically and deter- mine the solution set. (c) Demonstrate an understanding of the deﬁnition of a function and its graph. (d) Solve quadratic, exponential, logarithmic equations, and inequalities. (e) Solve simple real life problems involving linear, quadratic, and exponential functions graphically and algebraically. (f) Determine the zeros and the maximum or minimum of a quadratic function, and solve related problems, including those arising from real world applications. (g) Sketch the graphs of a quadratic, exponential, and logarithmic functions. (h) Compare simple and compound interest and relate compound interest to exponential growth. (i) Understand the inverse relationship between exponents and logarithms and use this relationship to solve related problems. (j) Understand basic concepts of descriptive statistics, mean,median,mode and summarize data into tables and simple graphs (bar charts, histogram, and pie chart). (k) Understand basic probability concepts and compute the probability of simple events using tree diagrams and formulas for permutations and combinations. 5 6 Chapter 1 System of linear Equations and Inequalities Contents 1.1 Solving Systems of Linear Equations with Two Variables 1.1.1 Solve a System of Linear Equations by Graphing 1.1.2 Solve a System of Equations by Substitution 1.1.3 Solve a System of Equations by Elimination 1.2 Solving Systems of Linear Inequalities 1.2.1 Solve a System of Linear Inequalities by Graphing Learning outcome covered: (a) Solve two variables linear equations and inequalities and sketch their graph. (b) Interpret a series of three simultaneous inequalities of two variables, display them graphically and determine the solution set. Learning Objectives By the end of this chapter, the students will be able to: à Determine whether an ordered pair is a solution of a system of equations à Solve a system of linear equations by graphing à Solve a system of equations by substitution à Solve a system of equations by elimination à Solve a system of linear inequalities 7 8 Introduction Climb into your car. Put on your seatbelt. Choose your destination and then relax. Thats right. You dont have to do anything else because you are in an autonomous car, or one that navigates its way to your destination! No cars are fully autonomous at the moment and so you theoretically still need to have your hands on the wheel. Self-driving cars may help ease traﬃc congestion, prevent accidents, and lower pollution. The technology is thanks to computer programmers who are developing software to control the navigation of the car. These programmers rely on their understanding of mathematics, including relationships between equations. In this chapter, you will learn how to solve systems of linear equations in diﬀerent ways and use them to analyze real-world situations. 1.1 Solving Systems of Linear Equations with Two Variables Determine whether an ordered pair is a solution of a system of linear inequalities In Solving Linear Equations, we learned how to solve linear equations with one variable. Now we will work with two or more linear equations grouped together, which is known as a system of linear equations. System of Linear Equations When two or more linear equations are grouped together, they form a system of linear equations. In this section, we will focus our work on systems of two linear equations in two unknowns. We will solve larger systems of equations later in this chapter. An example of a system of two linear equations is shown below. We use a brace to show the two equations are grouped together to form a system of equations. { 2x + y = 7 x − 2y = 6 A linear equation in two variables, such as 2x + y = 7, has an inﬁnite number of solutions. Its graph is a line. Remember, every point on the line is a solution to the equation and every solution to the equation is a point on the line. To solve a system of two linear equations, we want to ﬁnd the values of the variables that are solutions to both equations. In other words, we are looking for the ordered pairs (x, y) that make both equations true. These are called the solutions of a system of equations. Solutions of a System of Equations The solutions of a system of equations are the values of the variables that make all the equations true. A solution of a system of two linear equations is represented by an ordered pair (x, y). To determine if an ordered pair is a solution to a system of two equations, we substitute the values of the variables into each equation. If the ordered pair makes both equations true, it is a solution to the system. 9 Example 1 Determine whether the ordered pair is a solution to the system: { x − y = −1 2x − y = −5 (a) (−2, −1) (b) (−4, −3) Solution : { x − y = −1 Given 2x − y = −5 (a) Substitute x = −2 and y = −1 in both the equations. x − y = −1 2x − y = −5 ? ? −2 − (−1) = −1 2(−2) − (−1) = −3 −1 = −1 true −3 = −5 false The order pair (−2, −1) does not make true for the second equation. So it is not a solution. (b) Substitute x = −4 and y = −3 in both the above equations. x − y = −1 2x − y = −5 ? ? −4 − (−3) = −1 2(−4) − (−3) = −5 −1 = −1 true −5 = −5 true The order pair (−4, −3) makes the both equations true. So it is a solution. Example 2 Determine whether the ordered pair is a solution to the system: { 3x + y = 0 x + 2y = −5 (a) (1, −3) (b) (0, 0) Solution : { 3x + y = 0 Given x + 2y = −5 (a) Substitute x = 1 and y = −3 in both the equations. 3x + y = 0 x + 2y = −5 ? ? 3(1) − 3 = 0 1 + 2(−3) = −5 0=0 true −5 = −5 true The order pair (1, −3) makes the both equations true. So it is a solution. (b) Substitute x = 0 and y = 0 in both the above equations. Access for free at https://openstax.org/books/algebra- and- trigonometry/pages/1- introduction- to- prerequisites 10 3x + y = 0 x + 2y = −5 ? ? 3(0) + 0 = 0 0 + 2(0) = −5 0=0 true 0 ̸= −5 false The order pair (0, 0) does not make true for the second equation. So it is not a solution. 1.1.1 Solve a System of Linear Equations by Graphing In this section, we will use three methods to solve a system of linear equations. The ﬁrst method well use is graphing. The graph of a linear equation is a line. Each point on the line is a solution to the equation. For a system of two equations, we will graph two lines. Then we can see all the points that are solutions to each equation. And, by ﬁnding what the lines have in common, well ﬁnd the solution to the system. Most linear equations in one variable have one solution, but we saw that some equations, called contradictions, have no solutions and for other equations, called identities, all numbers are solutions. Similarly, when we solve a system of two linear equations represented by a graph of two lines in the same plane, there are three possible cases, as shown. How T o :: SOLVE A SYSTEM OF LINEAR EQUATIONS BY GRAPHING. Step 1. Graph the ﬁrst equation. Step 2. Graph the second equation on the same rectangular coordinate system. Step 3. Determine whether the lines intersect, are parallel, or are the same line. Step 4. Identify the solution to the system. How? If the lines intersect, identify the point of intersection. This is the solution to the system. If the lines are parallel, the system has no solution. If the lines are the same, the system has an inﬁnite number of solutions. Step 5. Check the solution in both equations. 11 In the next example, well ﬁrst re-write the equations into slopeintercept form as this will make it easy for us to quickly graph the lines. Example 3 { 2x + y = 7 Solve the system by graphing x − 2y = 6 Solution : Access for free at https://openstax.org/books/algebra- and- trigonometry/pages/1- introduction- to- prerequisites 12 In the next example, well ﬁrst re-write the equations into slopeintercept form as this will make it easy for us to quickly graph the lines. Example 4 Solve the system by graphing: { 3x + y = −1 2x + y = 0 Solution : Well solve both of these equations for y so that we can easily graph them using their slopes and y−intercepts. { 3x + y = −1 2x + y = 0 Solve the ﬁrst equation for y. y = −3x − 1 Find the slope and y−intercept. m = −3, b = −1 Solve the second equation for y. y = −2x Find the slope and y−intercept. m = −2, b = 0 Then graph the lines. Determine the point of intersection. The lines intersect at (−1, 2). Check the solution in both equations. 3x + y = −1 2x + y =0 ? ? 3(−1) + 2 = −1 2(−1) + 2 =0 −1 = −1, 0 = 0. 13 Solving systems of linear equations by graphing is a good way to visualize the types of solutions that may result. However, there are many cases where solving a system by graphing is inconvenient or imprecise. If the graphs extend beyond the small grid with x and y both between 10 and 10, graphing the lines may be cumbersome. And if the solutions to the system are not integers, it can be hard to read their values precisely from a graph. 1.1.2 Solve a System of Equations by Substitution We will now solve systems of linear equations by the substitution method. We will use the same system we used ﬁrst for graphing. { 2x + y = 7 x − 2y = 6 We will ﬁrst solve one of the equations for either x or y. We can choose either equation and solve for either variablebut well try to make a choice that will keep the work easy.Then we substitute that expression into the other equation. The result is an equation with just one variableand we know how to solve those! After we ﬁnd the value of one variable, we will substitute that value into one of the original equations and solve for the other variable. Finally, we check our solution and make sure it makes both equations true. How T o :: SOLVE A SYSTEM OF EQUATIONS BY SUBSTITUTION Step 1. Solve one of the equations for either variable. Step 2. Substitute the expression from Step 1 into the other equation. Step 3. Solve the resulting equation. How? Step 4. Substitute the solution in Step 3 into either of the original equations to ﬁnd the other variable. Step 5. Write the solution as an ordered pair. Step 6. Check that the ordered pair is a solution to both original equations. Example 5 { 2x + y = 7 Solve the system by substitution x − 2y = 6 Solution : Access for free at https://openstax.org/books/algebra- and- trigonometry/pages/1- introduction- to- prerequisites 14 1.1.3 Solve a System of Equations by Elimination We have solved systems of linear equations by graphing and by substitution. Graphing works well when the variable coeﬃcients are small and the solution has integer values. Substitution works well when we can easily solve one equation for one of the variables and not have too many fractions in the resulting expression. The third method of solving systems of linear equations is called the Elimination Method. When we solved a system by substitution, we started with two equations and two variables and reduced it to one equation with one variable. This is what well do with the elimination method, too, but well have a diﬀerent way to get there. The Elimination Method is based on the Addition Property of Equality. The Addition Property of Equality says that when you add the same quantity to both sides of an equation, you still have equality. We will extend the Addition Property of Equality to say that when you add equal quantities to both sides of an equation, the results are equal. For any expressions a, b, c, and d. if a = b and c = d then a + c = b + d. To solve a system of equations by elimination, we start with both equations in standard form. Then we decide which variable will be easiest to eliminate. How do we decide? We want to have the coeﬃcients of one variable be opposites, so that we can add the equations together and eliminate that variable. 15 Notice how that works when we add these two equations together:    3x + y = 5 2x − y = 0   5x = 5 The y ′ s add to zero and we have one equation with one variable. Lets try another one: { x + 4y = 2 2x + 5y = −2 This time we dont see a variable that can be immediately eliminated if we add the equations. But if we multiply the ﬁrst equation by −2, we will make the coeﬃcients of x opposites. We must multiply every term on both sides of the equation by −2. { −2(x + 4y) = −2(2) 2x + 5y = −2 Then rewrite the system of equations. { −2x − 8y = −4 2x + 5y = −2 Now we see that the coeﬃcients of the x terms are opposites, so x will be eliminated when we add these two equations. −2x − 8y = −4 2x + 5y = −2 −3y = −6 Once we get an equation with just one variable, we solve it. Then we substitute that value into one of the original equations to solve for the remaining variable. And, as always, we check our answer to make sure it is a solution to both of the original equations. Now well see how to use elimination to solve the same system of equations we solved by graphing and by substitution. How T o :: SOLVE A SYSTEM OF EQUATIONS BY ELIMINATION Step 1. Write both equations in standard form. If any coeﬃcients are fractions, clear them. Step 2. Make the coeﬃcients of one variable opposites. Decide which variable you will eliminate. Multiply one or both equations so that the coeﬃcients of that vari- able are opposites. How? Step 3. Add the equations resulting from Step 2 to eliminate one variable. Step 4. Solve for the remaining variable. Step 5. Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable. Step 6. Write the solution as an ordered pair. Step 7. Check that the ordered pair is a solution to both original equations. Access for free at https://openstax.org/books/algebra- and- trigonometry/pages/1- introduction- to- prerequisites 16 Example 6 { 2x + y = 7 Solve the system by elimination x − 2y = 6 Solution : 17 1.1 EXERCISES Determine Whether an Ordered Pair is a Solution of a System of Equations I. In the following exercises, determine if the following points are solutions to the given system of equations. { { { 2x − 6y = 0 −3x + y = 8 x+y =2 (1) (2) (3) 3x − 4y = 5 −x + 2y = −9 y = 43 x ( ) ( ) (a) (3, 1) (b) (−3, 4) (a) (−5, −7) (b) (−5, 7) (a) 87 , 67 (b) 1, 34 Solve a System of Linear Equations by Graphing II. In the following exercises, solve the following systems of equations by graphing. { { { 3x + y = −3 −x + y = 2 y =x+2 (1) (2) (3) 2x + 3y = 5 2x + y = −4 y = −2x + 2 { { { y =x−2 x + y = −4 2x − y = 4 (4) (5) (6) y = −3x + 2 −x + 2y = −2 2x + 3y = 12 Solve a System of Equations by Substitution III. In the following exercises, solve the systems of equations by substitution. { { { 2x + y = −4 2x + y = −2 x−3y = −9 (1) (2) (3) 3x−2y = −6 3x−y = 7 2x + 5y = 4 { { { 5x−2y = −6 x−2y = −5 −2x + 2y = 6 (4) (5) (6) y = 3x + 3 2x−3y = −4 y = −3x + 1 Solve a System of Equations by Elimination IV. In the following exercises, solve the systems of equations by elimination. { { { 5x + 2y = 2 6x−5y = −1 2x−5y = 7 (1) (2) (3) −3x−y = 0 2x + y = 13 3x−y = 17 { { { 5x−3y = −1 3x−5y = −9 4x−3y = 3 (4) (5) (6) 2x−y = 2 5x + 2y = 16 2x + 5y = −31 1.2 Solving Systems of Linear Inequalities Determine whether an ordered pair is a solution of a system of linear inequalities The deﬁnition of a system of linear inequalities is very similar to the deﬁnition of a system of linear equations. System of Linear Inequalities Two or more linear inequalities grouped together form a system of linear inequalities. A system of linear inequalities looks like a system of linear equations, but it has inequalities instead of equations. A system of two linear inequalities is shown here. { x + 4y ≥ 10 3x − 2y < 12 To solve a system of linear inequalities, we will ﬁnd values of the variables that are solutions to both inequal- ities. We solve the system by using the graphs of each inequality and show the solution as a graph. We will ﬁnd the region on the plane that contains all ordered pairs (x, y) that make both inequalities true. Access for free at https://openstax.org/books/algebra- and- trigonometry/pages/1- introduction- to- prerequisites 18 Solutions of a System of Linear Inequalities Solutions of a system of linear inequalities are the values of the variables that make all the inequalities true. The solution of a system of linear inequalities is shown as a shaded region in the x, y coordinate system that includes all the points whose ordered pairs make the inequalities true. To determine if an ordered pair is a solution to a system of two inequalities, we substitute the values of the variables into each inequality. If the ordered pair makes both inequalities true, it is a solution to the system. Example 1 Determine whether the ordered pair is a solution to the system: { x + 4y ≥ 10 3x − 2y < 12 (a) (−2, 4) (b) (3, 1) Solution : { x + 4y ≥ 10 Given 3x − 2y < 12 (a) Is the ordered pair (−2, 4) a solution? Substitute x = −2 and y = 4 in both the inequalities. x + 4y ≥ 10 3x − 2y < 12 ? ? −2 + 4(4) ≥ 10 3(−2) − 2(4) < 12 14 ≥ 10 true −14 < 12 true The ordered pair (−2, 4) made both inequalities true. Therefore (−2, 4) is a solution to this system. (b) Is the ordered pair (3, 1) a solution? Substitute x = 3 and y = 1 in both the inequalities. x + 4y ≥ 10 3x − 2y < 12 ? ? 3 + 4(1) ≥ 10 3(3) − 2(1) < 12 7 ≥ 10 false 7 < 12 true The ordered pair (3, 1) made one inequality true, but the other one false. Therefore (3, 1) is not a solution to this system. 19 1.2.1 Solve a System of Linear Inequalities by Graphing The solution to a single linear inequality is the region on one side of the boundary line that contains all the points that make the inequality true. The solution to a system of two linear inequalities is a region that contains the solutions to both inequalities. To ﬁnd this region, we will graph each inequality separately and then locate the region where they are both true. The solution is always shown as a graph. How T o :: SOLVE A SYSTEM OF LINEAR INEQUALITIES BY GRAPHING. Step 1. Graph the ﬁrst inequality. Graph the boundary line. Shade in the side of the boundary line where the inequality is true. Step 2. On the same grid, graph the second inequality. How? Graph the boundary line. Shade in the side of that boundary line where the inequality is true. Step 3. The solution is the region where the shading overlaps. Step 4. Check by choosing a test point. Example 2 Solve the system by graphing { y ≥ 2x − 1 y 3 1  y 3, by graphing x−y = 3 and testing a point. The intercepts are x = 3 and y = −3 and the boundary line will be dashed. Test (0, 0) which makes the inequality false so shade (red) the side that does not contain (0, 0). 1 1 Graph y < − x + 4 by graphing y = − x + 4 5 5 1 using the slope m = − and y−intercept b = 4. 5 The boundary line will be dashed Test (0, 0) which makes the inequality true, so shade (blue) the side that contains (0, 0). Choose a test point in the solution and verify that it is a solution to both inequalties. The point of intersection of the two lines is not included as both boundary lines were dashed. The solution is the area shaded twicewhich appears as the darkest shaded region. Example 4 Solve the system by graphing { x − 2y < 5 y > −4 Solution : Graph x−2y < 5, by graphing x−2y = 5 and testing a point. The intercepts are x = 5 and y = −2.5 The boundary line will be dashed. Test (0, 0) which makes the inequality true, so shade (red) the side that contains (0, 0). Access for free at https://openstax.org/books/algebra- and- trigonometry/pages/1- introduction- to- prerequisites 22 Graph y > −4, by graphing y = −4 and rec- ognizing that it is a horizontal line through y = −4. The boundary line will be dashed. Test (0, 0) which makes the inequality true so shade (blue) the side that contains (0, 0). The point (0, 0) is in the solution and we have already found it to be a solution of each inequality. The point of intersection of the two lines is not included as both boundary lines were dashed. The solution is the area shaded twice which appears as the darkest shaded region. Example 5 Solve the system by graphing    y ≥ −4 y 5 4x − y < 10 y > 32 x − 5 (1) (2) (3) 2x − y ≤ 10 −2x + 2y > −8 x + 12 y ≤ 4 (a) (3, −3) (b) (7, 1) (a) (5, −2) (b) (−1, 3) (a) (6, −4) (i) (3, 0) Solve a System of Linear Inequalities by Graphing II. In the following exercises, solve each system by graphing. { { { y ≤ 3x + 2 y < −2x + 2 y < 2x − 1 (1) (2) (3) y >x−1 y ≥ −x − 1 y ≤ − 21 x + 4 { { { y ≥ − 23 x + 2 x−y >1 x + 2y < 4 (4) (5) (6) y > 2x − 3 y < − 14 x + 3 y 10   y+4≥0 (7) x − 3y < 6 (8) 2x + y < 1 (9) 2x + 3y ≤ 1 1 1       2x − 3y > −1 x + 3y < −2 −3x + 2y ≤ 6 Access for free at https://openstax.org/books/algebra- and- trigonometry/pages/1- introduction- to- prerequisites 24 Chapter 2 Functions and their Graphs Contents 2.1 Functions 2.1.1 Types of functions 2.1.2 Vertical line test for a function 2.1.3 Horizontal line test for a one-to-one function 2.1.4 Find the domain of the function 2.2 Composition of Functions 2.3 Inverse Function 2.4 Quadratic Equations 2.5 Graphing Functions 2.5.1 Representing a Linear Function in Graphical Form 2.5.2 Representing a Quadratic Function in Graphical Form 2.5.3 Finding the Vertex of a Quadratic Function 2.5.4 Finding the Domain and Range of a Quadratic Function 2.5.5 Determining the Maximum and Minimum Values of Quadratic Functions 2.5.6 Finding the x−and y−Intercepts of a Quadratic Function 2.5.7 Sketch the Graph of the Quadratic Function Learning outcomes covered: (a) Demonstrate understanding of the definition of a function and its graph. (b) Solve quadratic equations using quadratic formula. Learning Objectives By the end of this chapter, the students will be able to: à Determine whether a relation represents a function. à Find the domain and range of the function. à Find the value of a function. 25 26 à Determine whether a function is one-to-one. à Use the vertical line test to identify functions à Use horizontal line test to identify one-to-one function. à Find the inverse function and composition of functions. à Solve quadratic equations using the quadratic Formula. à Sketch the graph of linear and quadratic functions. Introduction Toward the end of the twentieth century, the values of stocks of Internet and technology companies rose dramatically. As a result, the Standard and Poor’s stock market average rose as well. This ﬁgure tracks the value of that initial investment of just under $100 over the 40 years. It shows that an investment that was worth less than $500 until about 1995 skyrocketed up to about $1, 100 by the beginning of 2000. That ﬁve-year period became known as the "dot-com bubble" because so many Internet startups were formed. As bubbles tend to do, though, the dot-com bubble eventually burst. Many companies grew too fast and then suddenly went out of business. The result caused the sharp decline represented on the graph beginning at the end of 2000. Notice, as we consider this example, that there is a deﬁnite relationship between the year and stock market average. For any year we choose, we can determine the corresponding value of the stock market average. In this chapter, we will explore these kinds of relationships and their properties. 2.1 Functions A jetliner changes altitude as its distance from the starting point of a ﬂight increases. The weight of a growing child increases with time. In each case, one quantity depends on another. There is a relationship between the two quantities that we can describe, analyze, and use to make predictions. In this section, we will analyze such relationships. A relation is a set of ordered pairs. The set of the ﬁrst components of each ordered pair is called the domain of the relation and the set of the second components of each ordered pair is called the range of the relation. Consider the following set of ordered pairs. The ﬁrst numbers in each pair are the ﬁrst ﬁve natural numbers. The second number in each pair is twice the ﬁrst. {(1, 2), (2, 4), (3, 6), (4, 8), (5, 10)} The domain is {1, 2, 3, 4, 5} and the range is {2, 4, 6, 8, 10}. Note that the values in the domain are also known as an input values, or values of the independent variable, and are often labeled with the lowercase letter x. Values in the range are also known as an output 27 values, or values of the dependent variable, and are often labeled with the lowercase letter y. A function f is a relation that assigns a single value in the range to each value in the domain. In other words, no x-values are used more than once. For our example that relates the ﬁrst ﬁve natural numbers to numbers double their values, this relation is a function because each element in the domain, {1, 2, 3, 4, 5} is paired with exactly one element in the range, {2, 4, 6, 8, 10}. Now let’s consider the set of ordered pairs that relates the terms "even" and "odd" to the ﬁrst ﬁve natural numbers. It would appear as {(odd, 1), (even, 2), (odd, 3), (even, 4), (odd, 5)}. Notice that each element in the domain, {even, odd} is not paired with exactly one element in the range, {1, 2, 3, 4, 5}. For example, the term “odd" corresponds to three values from the domain, {1, 3, 5} and the term “even" corresponds to two values from the range,{2, 4}. This violates the deﬁnition of a function, so this relation is not a function. This image compares relations that are functions and not functions. (a) This relationship is a function because each input is associated with a single output. Note that input q and r both give output n. (b) This relationship is also a function. In this case, each input is associated with a single output. (c) This relationship is not a function because input q is associated with two diﬀerent outputs. Functions A function is a relation in which each possible input value leads to exactly one output value. We say “the output is a function of the input." The input values make up the domain, and the output values make up the range. Access for free at https://openstax.org/books/algebra- and- trigonometry/pages/1- introduction- to- prerequisites 28 Function Notation Once we determine that a relationship is a function, we need to display and deﬁne the functional relationships so that we can understand and use them, and sometimes also that we can program them into computers. There are various ways of representing functions. A standard function notation is one representation that makes it easier to work with functions. To represent “height is a function of age," we start by identifying the descriptive variables h for height and a for age. The letters f, g and h are often used to represent functions just as we use x, y and z to represent numbers and A, B and C to represent sets. h is f of a. We name the function f ; height is a function of age. h = f (a). We use parentheses to indicate the function input. f (a). We name the function f ; the expression is read as ‘‘f of a”. Remember, we can use any letter to name the function; we can use the notation h(a) to show that h depends on a. The input value a must be put into the function h to get an output value. The parentheses indicate that age is input into the function; they do not indicate multiplication. We can also give an algebraic expression as the input to a function. For example f (a + b) means ﬁrst add a and b , and the result is the input for the function. We must perform the operations in this order to obtain the correct result. Dependent and Independent Variable Function Notation The notation y = f (x) deﬁnes a function named f. This is read as “y is a function of x". The letter x represents the input value, or the independent variable. The letter y , or f (x) , represents the output value, or the dependent variable. Functions deﬁned by equations It is relatively easy to determine whether an equation is a function by solving for y. When you are given an equation and a speciﬁc value for x, there should only be one corresponding y-value for that x-value. 29 Example 1 Determine the following equations represent the function or not. (a) y = x + 1 (b) y = x2 − 1 (c) y 2 = x + 5 Solution : (a) y = x + 1 is a function because y will always be one greater than x. Equations with exponents can also be functions. (b) y = x2 − 1 is a function; although x-values of 1 and -1 give the same y-value (0), that is the only possible y-value for each of those x-values. (c) y 2 = x + 5 is not a function; if you assume that x = 4, then y 2 = 4 + 5 = 9. y 2 = 9 has two possible answers 3 and -3. Functions deﬁned by arrow diagrams Example 2 Let X = {1, 2, 3} and Y = {A, B, C, D}. If the arrow diagram on the right deﬁnes a function f then (a) What is the domain of f ? (b) What is the co-domain of f ? (c) What is f (3)? (d) What is the range of f ? (e) What is the pre-image of D? (f) What is the pre-image of A? Solution : (a) The domain of f is X = {1, 2, 3} (b) The co-domain of f is Y = {A, B, C, D} (c) f (3) = C (d) Range of f is {C, D} (e) The pre-image of D is 1. (f) A has no pre-image. Access for free at https://openstax.org/books/algebra- and- trigonometry/pages/1- introduction- to- prerequisites 30 2.1.1 Types of functions One-to-One Function (Injective Function) An injective function or a one-to-one function is a function f : A → B with the following property. For every element b in the codomain B there is maximum one element a in the domain A such that f (a) = b. For an injective function, f (a) = f (b) ⇒ a = b. Onto Function (Surjective Function) A surjective function or an onto function is a function f : A → B with the following property. For every element b in the codomain B there is at least one element a in the domain A such that f (a) = b. For a surjective function, the range and codomain are the same. Bijection (Bijective Function) A bijective function is a function f : A → B that is both injective and surjective. For every element b in the codomain B there is exactly one element a in the domain A such that f (a) = b. 31 2.1.2 Vertical line test for a function The vertical line test can be used to determine whether a graph represents a function. If we can draw any vertical line that intersects a graph more than once, then the graph does not deﬁne a function because a function has only one output value for each input value. 2.1.3 Horizontal line test for a one-to-one function Once we have determined that a graph deﬁnes a function, an easy way to determine if it is a one-to-one function is to use the horizontal line test. Draw horizontal lines through the graph. If any horizontal line intersects the graph more than once, then the graph does not represent a one-to-one function. 2.1.4 Find the domain of the function Polynomial Function Let n be a non-negative integer. A polynomial function is a function that can be written in the form f (x) = an xn + an−1 nxn−1 +... + a2 x2 + a1 x + a0 This is called the general form of a polynomial function. Each ai is a coeﬃcient and can be any real, but an cannot be 0. Each expression ai xi is a term of a polynomial function. The highest power of the variable that occurs in the polynomial function is called the degree of a polynomial function. Access for free at https://openstax.org/books/algebra- and- trigonometry/pages/1- introduction- to- prerequisites 32 How T o :: Find the domain of the function if function written in an equation form Step 1. Identify the input values. Step 2. Identify any restrictions on the input and exclude those values from the How? domain. Step 3. Write the domain in interval form, if possible. The domain of the polynomial function is the set of all real numbers denoted by (−∞, ∞) or R. Example 3 Find the domain of the function f (x) = x2 − 1 Solution : The input value, shown by the variable x in the equation, is squared and then the result is lowered by one. Any real number may be squared and then be lowered by one, so there are no restrictions on the domain of this function. The domain is the set of real numbers. In interval form, the domain of f is (−∞, ∞) or R (Set of real numbers). Example 4 Find the domain of the function f (t) = 2t7 + 2t4 − 3t3 + 7t2 − 5 Solution : The input value, shown by the variable t in the equation, is having diﬀerent positive powers multiplied with integers and then the result is lowered by ﬁve. Any real number may be powered by positive numbers multiplied with integers and then be lowered by ﬁve, so there are no restrictions on the domain of this function. The domain is the set of real numbers. In interval form, the domain of f is (−∞, ∞) or R (Set of real numbers). Rational Function A rational function is a function that can be written as the quotient of two polynomial functions P (x) and Q(x). P (x) ap xp + ap−1 xp−1 +... + a1 x + a0 f (x) = = ; Q(x) ̸= 0 Q(x) bq xq + bq−1 xq−1 +... + b1 x + b0 where P (x) and Q(x) are polynomial functions is called rational function. 33 Find the domain of the function if function written in an equation form How T o :: that includes a fraction Step 1. Identify the input values. Step 2. Identify any restrictions on the input. If there is a denominator in the function’s formula, set the denominator equal to zero and solve for x. How? If the function’s formula contains an even root, set the radicand greater than or equal to 0, and then solve. Step 3. Write the domain in interval form, making sure to exclude any restricted values from the domain. Example 5 x+1 Find the domain of the function f (x) = 2−x Solution : When there is a denominator, we want to include only values of the input that do not force the denominator to be zero. So, we will set the denominator equal to 0 and solve for x. 2−x=0 −x = −2 x=2 Now, we will exclude 2 from the domain. The answers are all real numbers where x < 2 or x > 2. In interval notation, we write the solution as: (−∞, 2) ∪ (2, ∞) or R − {2}, where R is the set of real numbers. Example 6 3x − 7 Find the domain of the function f (x) = 16 + x2 Solution : When there is a denominator, we want to include only values of the input that do not force the denominator to be zero. So, we will set the denominator equal to 0 and solve for x. 16 + x2 = 0 Now, we will not exclude any number from the domain because 16 + x2 is never zero for any real value of x. The answers are all real numbers. In interval form, the domain of f is (−∞, ∞) or R (Set of real numbers). Access for free at https://openstax.org/books/algebra- and- trigonometry/pages/1- introduction- to- prerequisites 34 Example 7 2x3 + 3x2 Find the domain of the function f (x) = 4 − x2 Solution : When there is a denominator, we want to include only values of the input that do not force the denominator to be zero. So, we will set the denominator equal to 0 and solve for x. 4 − x2 = 0 (2 − x)(2 + x) = 0 x = −2 or x = 2 Now, we will exclude −2 and 2 from the domain. The answers are all real numbers where x < −2 or −2 < x < 2 or x > 2. In interval notation, we write the solution as: (−∞, 2) ∪ (−2, 2) ∪ (2, ∞) or R − {−2, 2}, where R is the set of real numbers. Radical Function A function in the variable x equivalent to the form √ f (x) = n p(x) where index n is the whole number and p(x) is the non-negative polynomial function is called radical function Example 8 √ 3 Find the domain of the function p(r) = r+8 Solution : If the index of the radical function is odd number then any real number may be with no restrictions on the domain of this function. So, the domain is the set of real numbers. In interval form, the domain of p is (−∞, ∞) or R (Set of real numbers). Find the domain of the function if function written in an equation form How T o :: including an even root Step 1. Identify the input values. Step 2. Since there is an even root, exclude any real numbers that result in a negative number in the radicand. Set the radicand greater than or equal How? to zero and solve for x. Step 3. The solution(s) are the domain of the function. If possible, write the answer in interval form. 35 Example 9 √ Find the domain of the function g(t) = 8 + t2 Solution : If the index of the radical function is even number and 8 + t2 is positive for all real t then any real number may be with no restrictions on the domain of this function. So, the domain is the set of real numbers. In interval form, the domain of g is (−∞, ∞) or R (Set of real numbers). Example 10 √ Find the domain of the function f (x) = 7−x Solution : Set the radicand greater than or equal to zero and solve for x. 7−x≥0 −x ≥ −7 x≤7 Now, we will exclude any number greater than 7 from the domain. The answer is the set of all real numbers less than or equal to 7 or (−∞, 7] Example 11 √ Find the domain of the function f (x) = 9 − x2 Solution : Set the radicand greater than or equal to zero and solve for x. 9 − x2 ≥ 0 (3 − x)(3 + x) ≥ 0 x ≥ −3 and x ≤ 3 Now, we will exclude any number greater than 3 and less than −3 from the domain. The answer is the set of all real numbers less than or equal to 3 and greater than or equal to −3, or [−3, 3] Evaluating function at a point Example 12 Let f (t) = 4 − 5t and g(x) = 3m2 + 2m − 5. Find the following values of the indicated functions. (a) f (−2) (b) g(3) (c) g(−3) + f (−1) Access for free at https://openstax.org/books/algebra- and- trigonometry/pages/1- introduction- to- prerequisites 36 Solution : (a) f (−2) = 4 − 5 × (−2) = 14 g(−3) + f (−1) (c) = (3 × (−3)2 + 2 × (−3) − 5) + 4 − 5 × (−1) (b) g(3) = 3 × 32 + 2 × 3 − 5 = 28 = 27 − 6 − 5 + 4 + 5 = 25 Diﬀerence Quotient: The diﬀerence quotient is usually the name for the expression as below: f (x + h) − f (x) h Example 13 f (−3 + h) − f (−3) If f (x) = 7 − 4x then ﬁnd. h Solution : Given f (x) = 7 − 4x. Then f (−3) = 7 − 4 × (−3) = 19 f (−3 + h) = 7 − 4 × (−3 + h) = 7 + 12 − 4h f (−3 + h) = 19 − 4h Substitute the values of f (−3 + h) and f (−3) in the diﬀerence quotient, f (−3 + h) − f (−3) f (−3 + h) − f (−3) = h h (19 − 4h) − (19) = h −4h = h f (−3 + h) − f (−3) = −4 h 2.1 EXERCISES Answer the following questions: I. Let A = {a, b, c} and B = {1, 2, 3}. If the arrow diagram on the right deﬁnes a function f then (1) What is the domain of f ? (2) What is the co-domain of f ? (3) What is f (a)? (4) What is the range of f ? (5) What is the pre-image of 1? (6) What is the pre-image of 3? 37 II. Which of the following are functions? If it is a function, write the domain, co-domain and range of the functions. (1) (2) (3) (4) (5) (6) III. Which of the following are functions? If it is a function, write the domain and range for each relation. (1) {(3, 6), (4, 9), (5, 12), (6,13)} (2) {(2, -2), (2, -1), (2, 0), (2, 1) } (3) {(2, 1), (3, 1), (4, 1), (5, 1)} (4) {(10, 5), (11, 8), (12, 5), (13, 12)} (5) {(a, -1), (b, 1), (c, 3), (d, 5)} (6) {(2, 10), (3, 10), (4, 20), (5, 30)} IV. Determine which of the following equations deﬁnes a function. (1) x + y = 7 (2) 3x + 2y = −5 (3) x2 + 2y = 4 (4) x2 + y 2 = 9 √ (5) y = 1 − x2 (6) x + y 3 = −1 V. Identify the graph that represents a function, by using the vertical line test. (1) (2) (3) VI. Identify type of the functions in the following relations. (1) (2) (3) Access for free at https://openstax.org/books/algebra- and- trigonometry/pages/1- introduction- to- prerequisites 38 (4) (5) (6) VII. Identify the graph that represents a one-to-one function, by using the Horizontal line test. (1) (2) (3) VIII. Find the domain of the following indicated functions √ (1) f (x) = 2x2 + 5x − 3 (2) f (t) = 5 − t + t3 (3) f (t) = t−5 √ √ √ (4) f (x) = 5 + 2x (5) g(x) = 3 2 − x (6) h(u) = 3 u + 6 √ 3m2 + 1 5m (7) f (x) = −2 + x (8) F (m) = (9) f (x) = m−5 m+7 2x 5 − 2m (10) f (x) = (11) G(m) = 2 x2 − 25 m +9 IX. Let g(t) = 5 − 3t, f (x) = 3x + 7, h(m) = 3m2 + 2m − 5 and k(u) = u2 + u − 1. Find the following values of the indicated functions. (1) g(−2) (2) f (3) (3) g(3) + f (−1) h(2) − f (1) (4) 2h(−2) + k(5) (5) f (2) − k(−3) (6) g(2) X. Answer the following: F (2 + h) − F (2) (1) If F (s) = 2s + 3 then ﬁnd h f (3 + h) − f (3) (2) If f (x) = 2 − 5x then ﬁnd h f (−2 + h) − f (−2) (3) If f (x) = 10 − 3x then ﬁnd h f (x + h) − f (x) f (x) − f (a) XI. Find and simplify i) ii) for the following functions h x−a (1) f (x) = 4x + 1 (2) f (x) = 3 − 7x (3) f (x) = −2x2 + 3x − 1 39 2.2 Composition of Functions Performing algebraic operations on functions combines them into a new function, but we can also create functions by composing functions. When we wanted to compute a heating cost from a day of the year, we created a new function that takes a day as input and yields a cost as output. The process of combining functions so that the output of one function becomes the input of another is known as a composition of functions. The resulting function is known as a composite function. We represent this combination by the following notation: (f o g) (x) = f (g (x)) We read the left-hand side as "f composed with g at x" and the right-hand side as “f of g of x". The two sides of the equation have the same mathematical meaning and are equal. The open circle symbol ◦ is called the composition operator. We use this operator mainly when we wish to emphasize the relationship between the functions themselves without referring to any particular input value. Composition is a binary operation that takes two functions and forms a new function, much as addition or multiplication takes two numbers and gives a new number. However, it is important not to confuse function composition with multiplication because, as we learned above, in most cases f (g (x)) ̸= f (x)g(x). It is also important to understand the order of oper- ations in evaluating a composite function. We follow the usual convention with parentheses by starting with the innermost parentheses ﬁrst, and then working to the outside. In the equation above, the function g takes the input x ﬁrst and yields an output g(x). Then the function f takes g(x) as an input and yields an output f (g(x)). In general, f ◦ g and g ◦ f are diﬀerent functions. In other words, in many cases f (g (x)) ̸= g (f (x)) for all x. We will also see that sometimes two functions can be composed only in one speciﬁc order. Example 1 If f (x) = x2 and g (x) = x + 2, then ﬁnd f (g (x)) and g (f (x)). Solution : Given f (x) = x2 and g(x) = x + 2, ( ) f (g(x)) = f (x + 2) g(f (x)) = g x2 = (x + 2)2 = x2 + 2 = x2 + 4x + 4 These expressions are not equal for all values of x, so the two functions are not equal. It is irrelevant that 1 the expressions happen to be equal for the single input value x = −. 2 Access for free at https://openstax.org/books/algebra- and- trigonometry/pages/1- introduction- to- prerequisites 40 Example 2 If f (x) = 2x + 1 and g(x) = 3 − x, then ﬁnd (f ◦ g)(x) and (g ◦ f )(x). Solution : Given f (x) = 2x + 1 and g(x) = 3 − x, (f ◦ g)(x) = f (g(x)) (g ◦ f )(x) = g(f (x)) = f (3 − x) = g(2x + 1) = 2(3 − x) + 1 = 3 − (2x + 1) = 6 − 2x + 1 = 3 − 2x − 1 = 7 − 2x = 2 − 2x 2.2 EXERCISES I. Find f (g (x)) and g (f (x)) for the following functions. (1) f (x) = 3x + 2 and g (x) = x + 1 (2) f (x) = x + 2 and g (x) = x2 √ (3) f (x) = x2 + 1 and g (x) = x + 2 √ (4) f (x) = x + 2 and g (x) = x2 + 3 (5) f (x) = |x| and g (x) = 5x + 1 II. Evaluate each composite value. (1) f (x) = 2x + 5 and g (x) = x − 2, ﬁnd (f ◦ g) (2) √ (2) f (x) = x + 1 and g (x) = x − 7, ﬁnd (f ◦ g) (16) 1 (3) f (x) = x + 5 and g (x) = , ﬁnd (f o g) (2) and (g ◦ f ) (1) x−3 x (4) f (x) = x + 3 and g (x) = , ﬁnd (f ◦ g) (6) and (g ◦ f ) (2) x−4 1−x 1 (5) f (x) = and g (x) = , ﬁnd (g ◦ f ) (2) x 1 + x2 41 2.3 Inverse Function An inverse function reverses the operation done by a particular function. In other words, whatever a function does, the inverse function undoes it. In this section, we deﬁne an inverse function formally and state the necessary conditions for an inverse function to exist. We examine how to ﬁnd an inverse function. Existence of an Inverse Function We begin with an example. Given a function f and an output y = f (x), we are often interested in ﬁnding what value or values of x were mapped to y by f. For example, consider the function f (x) = x3 + 4. Since √ any output is f (x) = x3 + 4, we can solve this equation for x to ﬁnd that the input is x = 3 y − 4. This equation deﬁnes x as a function of y. Denoting this function as f −1 , and writing x = f −1 (y), we see that for any x in the domain of f , f −1 (f (x)) = f −1 (x3 + 4) = x. A function with this property is called the inverse function of the original function. Inverse Function Given a function f with domain D and range R, its inverse function (if it exists) is the function f −1 with domain R and range D such that f −1 (y) = x if f (x) = y. In other words, for a function f and its inverse f −1 , ( ) f −1 (f (x)) = x, for all x in D and f f −1 (y) = y, for all y in R Recall that a function has exactly one output for each input. Therefore, to deﬁne an inverse function, we need to map each input to exactly one output. For example, let’s try to ﬁnd the inverse function for f (x) = x2. Solving the equation f (x) = x2 , √ we arrive at the equation x = ± y. This equation does not describe x as a function of y because there are two solutions to this equation for every y > 0. The problem with trying to ﬁnd an inverse function for f (x) = x2 is that two inputs are sent to the same output for each output y > 0. The function f (x) = x3 + 4 discussed earlier did not have this problem. For that function, each input was sent to a diﬀerent output. A function that sends each input to a diﬀerent output is called a one-to-one function. How T o :: Find a Function’s Inverse Step 1. Make sure f is one-to-one function. If f is not one-to-one, then f −1 does not exist. Step 2. Solve for x and write x = f −1 (y). How? Step 3. Interchange x and y. Step 4. Find the domain of f −1.The domain of f −1 must be the same as the range of f. Access for free at https://openstax.org/books/algebra- and- trigonometry/pages/1- introduction- to- prerequisites 42 Example 1 Find the inverse of the function f (x) = 3x − 4. Solution : Let y = f (x), y = 3x − 4 Set up an equation. y+4 = 3x Add 4 both sides. y+4 x = divide both sides by 3 3 y+4 f −1 (y) = Replace x = f −1 (y) 3 x+4 f −1 (x) = Replace y with x 3 Since the domain of f is (−∞, ∞), the range of f −1 is (−∞, ∞). Since the range of f is (−∞, ∞), the domain of f −1 is (−∞, ∞). Example 2 2 Find the inverse of the function f (x) = + 4. x−3 Solution : Let y = f (x), 2 y = +4 Set up an equation. x−3 2 y−4 = Subtract 4 from both sides. x−3 2 x−3 = Multiply both sides by x − 3 and divide by y − 4 y−4 2 x = +3 Add 3 to both sides. y−4 2 f −1 (y) = +3 Replace x = f −1 (y) y−4 2 f −1 (x) = +3 Replace y with x x−4 Since the domain of f is R − {3}, the range of f −1 is R − {3}. Since the range of f is R − {4}, the domain of f −1 is R − {4}. 43 2.3 EXERCISES I. Find the inverse function f −1 if exist. (1) f = {(3, 6), (4, 9), (5, 12), (6, 13)} (2) f = {(2, −2), (4, −1), (6, 0), (8, 1)} (3) f = {(2, 1), (3, 1), (4, 1), (5, 1)} (4) f = {(10, 5), (11, 8), (12, 5), (13, 12)} (5) f = {(a, −1), (b, 1), (c, 3), (d, 5)} (6) f = {(2, 10), (3, 10), (4, 20), (5, 30)} II. Find the inverse function f −1 (x). (1) f (x) = x + 3 (2) f (x) = 2 − x, also ﬁnd f −1 (5) 1 (3) f (x) = 5x − 1 (4) f (x) = x + 5 7 x−4 2x − 3 (5) f (x) = (6) f (x) = − 3 5 5 2x + 3 (7) f (x) = −6 (8) f (x) = , also ﬁnd f −1 (−1) 8+x 5x + 4 √ (9) f (x) = 2 + x − 4 2.4 Quadratic Equations In this section we discuss the method of solving a quadratic equation by using the quadratic formula, a formula that will solve all quadratic equations. Although the quadratic formula works on any quadratic equation in standard form, it is easy to make errors in substituting the values into the formula. Pay close attention when substituting, and use parentheses when inserting a negative number. Quadratic Formula The roots of a quadratic equation of the form ax2 + bx + c = 0 where a ̸= 0 are given by the formula: √ −b ± b2 − 4ac x= 2a To use the Quadratic Formula, we substitute the values of a, b, and c from the standard form into the ex- pression on the right side of the formula. Then we simplify the expression. The result is the pair of roots of the quadratic equation. How T o :: Solve a quadratic equation using the quadratic formula Step 1. Write the quadratic equation in standard form, ax2 +bx+c = 0. Identify the values of a, b, and c. How? Step 2. Write the Quadratic Formula. Then substitute the values of a, b, and c. Step 3. Simplify. Step 4. Check the roots. Example 1 Solve by using the Quadratic Formula: 2x2 + 9x − 5 = 0 Access for free at https://openstax.org/books/algebra- and- trigonometry/pages/1- introduction- to- prerequisites 44 Solution : Given 2x2 + 9x − 5 = 0. Here a = 2, b = 9, and c = −5 √ −b ± b2 − 4ac x= √2a −9 ± (9)2 − 4(2)(−5) = Substitute the values of a, b, c 2(2) √ −9 ± 81 + 40 = √4 −9 ± 121 = 4 −9 ± 11 x= 4 −9 + 11 −9 − 11 x= , x= 4 4 2 −20 x= , x= 4 4 1 x = , x = −5 2 Example 2 Solve by using the Quadratic Formula: x2 − 6x = −5 Solution : Given x2 − 6x = −5. We can rewrite as x2 − 6x + 5 = 0. Here a = 1, b = −6, and c = 5 √ −b ± b2 − 4ac x= 2a √ −(−6) ± (−6)2 − 4(1)(5) x= 2(1) √ 6 ± 36 − 20 x= √2 6 ± 16 x= 2 6+4 6−4 x= , x= 2 2 10 2 x= , x= 2 2 x = 5, x = 1 Example 3 1 2 2 1 Solve by using the Quadratic Formula: u + u= 2 3 3 45 Solution : 1 2 2 1 Given u + u =. Multiply both the sides by the LCD 6, 2 3 3 ( ) ( ) 1 2 2 1 6 u + u =6 2 3 3 3u2 + 4u = 2 3u2 + 4u − 2 = 0 Here a = 3, b = 4, and c = −2 √ −b ± b2 − 4ac u= 2a √ −4 ± (4)2 − 4(3)(−2) = 2(3) √ −4 ± 40 = 6 √ −4 ± 2 10 = 6√ √ −2 + 10 −2 − 10 u= , u= 3 3 Example 4 Solve by using the Quadratic Formula: 3z 2 + 2z + 9 = 0 Solution : Given 3z 2 + 2z + 9 = 0. Here a = 3, b = 2, and c = 9 √ −b ± b2 − 4ac z= √ 2a −2 ± (2)2 − 4(3)(9) = 2(3) √ −2 ± 4 − 108 = √6 −2 ± −104 = 6√ −2 ± i 104 = 6 √ −2 ± 2i 26 = 6 √ 2(−1 ± i 26) z= 6√ √ −1 + i 26 −1 − i 26 z= , z= 3 3 2.4 EXERCISES I. Solve the following quadratic equations using quadratic formula. Access for free at https://openstax.org/books/algebra- and- trigonometry/pages/1- introduction- to- prerequisites 46 (1) 4z 2 + 2z − 6 = 0 (2) 2x2 + 10x + 11 = 0 (3) 3x2 − 5x + 2 = 0 (4) y 2 − 24 = −10y 3 2 1 3 (5) y + y= (6) 2y 2 − 2y = −5 4 2 8 (7) x2 − 2x + 5 = 0 (8) 4z 2 − 2z + 8 = 0 (9) x (x + 2) + 5 = 0 (10) x (x − 6) = −10 2.5 Graphing Functions 2.5.1 Representing a Linear Function in Graphical Form In Linear Functions, we saw that that the graph of a linear function is a straight line. We were also able to see the points of the function as well as the initial value from a graph. By graphing two functions, then, we can more easily compare their characteristics. The function describing the train’s motion is a linear function, which is deﬁned as a function with a constant rate of change, that is, a polynomial of degree 1. There are several ways to represent a linear function, in- cluding word form, function notation, tabular form, and graphical form. We will describe the train’s motion as a function using each method. Linear Function A linear function is a function whose graph is a line. Linear functions can be written in the slope- intercept form of a line f (x) = mx + b where b is the initial or starting value of the function (when input, x = 0 ), and m is the constant rate of change, or slope of the function. The y-intercept is at (0, b). 47 Finding the x−intercept of a Line: To ﬁnd the x-intercept, set a function equal to zero and solve for the value of x. f (x) = 0 mx + b = 0 mx = −b −b x= m b ∴ x − intercept = − m Example 1 1 Find the slope and intercepts, and then sketch the graph of the linear function deﬁned by f (x) = x−3. 2 Solution : Given 1 f (x) = x−3 2 f (x) = mx + b 1 slope (m) = and b = −3 2 −b 3 x − intercept = = =6 m 1/ 2 2.5.2 Representing a Quadratic Function in Graphical Form The graph of a quadratic function is a U-shaped curve called a parabola. One important feature of the graph is that it has an extreme point, called the vertex. If the parabola opens up, the vertex repre- sents the lowest point on the graph, or the minimum value of the quadratic function. If the parabola opens down, the vertex represents the highest point on the graph, or the maximum value. In either case, the vertex is a turning point on the graph. The graph is also symmetric with a vertical line drawn through the vertex, called the axis of symmetry. The y-intercept is the point at which the parabola crosses the y-axis. The x-intercepts are the points at which the parabola crosses the x-axis. If they exist, the x-intercepts represent the zeros, or roots, of the quadratic function, the values of x at which y = 0. Access for free at https://openstax.org/books/algebra- and- trigonometry/pages/1- introduction- to- prerequisites 48 Quadratic Functions A quadratic function is a function of degree two. The graph of a quadratic function is a parabola. The general form of a quadratic function is f (x) = ax2 +bx+c where a, b and c are real numbers and a ̸= 0. 2 The standard form of a quadratic function is f (x) = a(x − h) + k. b The vertex (h, k) is located as h = − and k = f (h) 2a Properties: 1. If a > 0 the parabola opens up and if a < 0 the parabola opens down. 2. Determine whether a is positive or negative. If a is positive, the parabola has a minimum. If a is negative, the parabola has a maximum. 3. Determine the maximum or minimum value of the parabola,k. 4. If the parabola has a minimum, the range is given by f (x) ≥ k or [k, ∞). If the parabola has a maximum, the range is given by f (x) ≤ k or (−∞, k]. 5. If a > 0, then the graph decreases in the interval (−∞, h] and increases in the interval [h, ∞). 6. If a < 0, then the graph increases in the interval (−∞, h] and decreases in the interval [h, ∞). 2.5.3 Finding the Vertex of a Quadratic Function How T o :: Find the vertex of the parabola for a quadratic function in general form, Step 1. Identify a, b, and c. Step 2. Find h, the x−coordinate of the vertex, by substituting a and b into −b How? h=. 2a Step 3. Find k, the y−coordinate of the vertex, by evaluating ( ) −b k = f (h) = f 2a Example 2 Find the vertex of the quadratic function f (x) = 2x2 −6x + 7. Rewrite the quadratic in standard form (vertex form). Solution : The horizontal coordinate of the vertex The vertical coordinate of the vertex b k = f (h) h=− ( ) 2a 3 −6 =f =− 2 2(2) ( )2 ( ) 6 3 3 = =2 −6 +7 4 2 2 3 5 = = 2 2 49 Rewriting into standard form, the stretch factor will be the same as the a in the original quadratic. First, ﬁnd the horizontal coordinate of the vertex. Then ﬁnd the vertical coordinate of the vertex. Substitute the values into standard form, using the a from the general form. f (x) = ax2 + bx + c f (x) = 2x2 −6x + 7 The standard form of a quadratic function prior to writing the function then becomes the following: ( )2 3 5 f (x) = 2 x − + 2 2 2.5.4 Finding the Domain and Range of a Quadratic Function Any number can be the input value of a quadratic function. Therefore, the domain of any quadratic function is the set of all real numbers. Because parabolas have a maximum or a minimum point, the range is restricted. Since the vertex of a parabola will be either a maximum or a minimum, the range will consist of all y−values greater than or equal to the y−coordinate at the turning point or less than or equal to the y−coordinate at the turning point, depending on whether the parabola opens up or down. How T o :: Find the domain and range Step 1. Identify the domain of any quadratic function is always (−∞, ∞) Step 2. Determine whether a is positive or negative. If a is positive, the parabola has a minimum. If a is negative, the parabola has a maximum. How? Step 3. Determine the maximum or minimum value of the parabola, k. Step 4. If the parabola has a minimum, the range is given by f (x) ≥ k, or [k, ∞). If the parabola has a maximum, the range is given by f (x) ≤ k, or (−∞, k]. Example 3 Find the domain and range of f (x) = −5x2 + 9x−1. Solution : As with any quadratic function, the domain is all real numbers. Here, a = −5, b = 9, c = −1 Because a is negative, the parabola opens downward and has a maximum value. We need to determine the maximum value. We can begin by ﬁnding the x−value of the vertex. b h=− 2a 9 =− 2(−5) 9 = 10 The maximum value is given by f (h). Access for free at https://openstax.org/books/algebra- and- trigonometry/pages/1- introduction- to- prerequisites 50 ( ) ( )2 ( ) 9 9 9 f = −5 +9 −1 10 10 10 61 = 20 ( ] 61 61 The range is f (x) ≤ , or −∞, 20 20 2.5.5 Determining the Maximum and Minimum Values of Quadratic Functions The output of the quadratic function at the vertex is the maximum or minimum value of the function, depending on the orientation of the parabola

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