1. ALGEBRA 2 MILEPOST PROBLEMSNotes to Students(and their Teachers)Different students master different skills at different speeds. No two students learn exactly the sameway at the same time. However, at some point you will be expected to perform certain skills withgood accuracy. In Algebra 1, your teacher did not want you to grab for your calculator if you neededto calculate -16 + 9 or 81. The same thing is true of your Algebra 2 teacher. There are certainalgebra skills that you need to be able to do on your own and with good accuracy. Most of theMilepost problems are skills that you should have been developing in Algebra 1 and Geometry. Ifyou have not mastered these skills by now it does not necessarily mean that you will not be successfulin this class. However, it does mean that to be successful you are going to need to do some extra workoutside of class. You need to get caught up on the algebra skills that this year's teacher and possiblynext year's pre-calculus teacher expect.Starting in Unit 2 and finishing in Unit 9, there are twenty one problems designated as Milepostproblems. Each one has an icon like the one above. After you do each of the Milepost problems,check your answers. If you have not done them correctly, this is your reminder that you need to putin some extra practice on that skill. The following practice sets are keyed to each of the Milepostproblems in the textbook. Each of the Milepost practice sets has the topic clearly labeled, followed bysome completed examples. Next, the solution to the Milepost problem from the book is given.Following that are more problems to practice with answers included.Warning! Looking is not the same as doing. You will never become good at any sport just bywatching it. In the same way, reading through the worked out examples and understanding the stepsare not the same as being able to do the problems yourself. An athlete only gets good with practice.The same thing is true of your algebra skills. If you did not get the Milepost problem correct you needthe extra practice. How many of the extra practice problems do you need to try? That is really up toyou. Remember that your goal is to be able to do that skill on your own confidently and accurately.Another warning! You should not expect your teacher to spend time in class going over the solutionsto the Milepost practice sets. After reading the examples and trying the problems, if you still are notsuccessful, talk to your teacher about getting a tutor or extra help outside of class time.Two other sources for help with the Milepost problems and the other new topics in Algebra 2 are theParent's Guide with Review to Math 1 (Algebra 1) and the Parent's Guide with Review to Math 3(Algebra 2). Information about ordering these supplements can be found inside the front page of thestudent text. These resources are also available free from the internet at www.cpm.org.Milepost Topics1. x- and y-intercepts of a quadratic equation 12. Graphing linear inequalities2. Graphing lines using slope and y-intercept 13. Multiplication and division of rational expressions3. Simplifying expressions with exponents 14. Solving rational equations4. Solving systems of linear equations 15. Addition and subtraction of rational expressions5. Multiplying polynomials 16. Integral and rational exponents6. Factoring quadratic expressions 17. Completing the square; locator points for parabolas7. Solving multi-variable equations 18. Absolute value equations and inequalities8. Slope of the line and distance 19. Writing and solving exponential equations9. Function notation; domain and range 20. Finding the equation for the inverse of a function10. Writing equations of lines given two points 21. Solving a system of equations in three variables11. Slopes of parallel and perpendicular linesSkill Building Resources 495

2. Milepost Number 1BB-62Finding the X- and Y-Intercepts of a Quadratic Equation.The y-intercept of a equation is the location where the graph crosses the y-axis. To find they-intercept of an equation, substitute x = 0 into the equation and solve for y. For example:Find the y-intercept for the equation y = x2 + 4x - 12.If x = 0, then y = (0)2 + 4(0) - 12 = -12. The y-intercept is (0, -12).The x-intercept of a equation is the location where the graph crosses the x-axis. To find thex-intercept of an equation, substitute y = 0 into the equation and solve for x by factoring or using thequadratic formula. Here are two examples:Find the x-intercept for the equation y = x2 + 4x - 12.If y = 0, thenBy factoring and using the zero productproperty0 = x2 + 4x - 120 = (x + 6)(x - 2)x = -6 or x = 2The x-intercepts are (-6, 0) and (2, 0)Find the x-intercept for the equation y = 2x2 - 3x - 3.If y = 0, thenSince we cannot factor the trinomial weuse the Quadratic Formula to solve forx. If ax2 + bx + c = 0, thenx =-b!! b2!-!4ac2a .substitute for a, b, and csimplify and addfind !!! valuesimplify the fractions0 = 2x2 - 3x - 3x =-(-3)!! (-3)2!-!4(2)(-3)2(2)=3!! 9!+!244 =3!! 334"3!!5.7454 or3!+!5.7454 and3!-!5.7454and the x-intercepts are approximately(2.19, 0) and (-0.69, 0).496 MILEPOSTS 3. Now we can go back and try the original question. Find the x- and y-intercepts for the graph of:y = x2 + 4x - 17.To find the y-intercept let x = 0 so y = (0)2 + 4(0) - 17 = -17 .To find the x-intercept let y = 0 so 0 = x2 + 4x - 17.Since we cannot factor we use the Quadratic Equation with a = 1, b = 4, and c = -17x =-(4)!! (4)2!-!4(1)(-17)2(1) =-4!! 16!+!682 =-4!! 842 =-4!!2 212 = -2 21 ,The answers are (0, -17), (-2 21 , 0) or (2.58, 0), (-6.58, 0)Here are some more to try. Find the x- and y-intercepts for each equation,1) y = 2x2 - 9x - 35 = 0 2) y = 2x2 - 11x + 53) 3x2 + 2 + 7x = y 4) 8x2 + 10x + 3 = y5) y + 2 = x2 - 5x 6) (x - 3)(x + 4) - 7x = y7) -4x2 + 8x + 3 = y 8) 0.09x2 - 0.86x + 2 = y9) y = 2x3 - 50x 10) y = 3x2 + 4xAnswers:Note that the coordinates of the intercepts are (x, 0) and (0, y).1. x = 7, -5/2 y = -35 2. x = 5, 1/2 y = 53. x = -1/3 , -2 y = 2 4. x = -3/4, -1/2 y = 35. x = (5 #33)/2, " 5.37, -0.37 y = -2 6. x = (6 #84)/2 " 7.58, -1.58 y = -127. x = (-8 #112)/-8 " -0.32, 2.32 y = 3 8. x = 5.56, 4 y = 29. x = 0, 5, -5 y = 0 10. x = 0, -4/3 y = 0Skill Building Resources 497 4. Milepost Number 2BB-66Graphing a Line using Slope and Y-Intercept2323If an equation of a line is written in the form y = mx + b, then the y-intercept is the point (0, b). Theslope of the line is the coefficient of x, represented in the general form of the equation as m. So in theequation y = x + 7, the slope is and the y-intercept is (0, 7).Let's first see how to use the information in slope-intercept equations to graph a line.23Without making a table, graph each line. Start with the y-intercept, then use the slope.a) y = x - 2 b) y = 4 - 2x23In part (a), we start by identifying the slope and y-intercept. Theslope is and the y-intercept is (0, -2). To graph the line weplot the y-intercept. (Before continuing, imagine what the linewill look like.) The fact that the slope is positive tells us thedirection of the line is upward left to right. Then, knowing thatthe slope is 23, we can find another point on the line by startingat the y-intercept, moving our pencil up vertically two units andthen horizontally (to the right) three units. Just remember thatthe slope is positive! After moving vertically 2 units andhorizontally 3 units, we arrive at another point on the line.xy32In part (b), y = 4 2x, don't let the form of the equation foolyou. The slope is -2 or - 21and the y-intercept is (0, 4). Theslope is always the coefficient of x and the y-intercept is alwaysthe constant. Rearranging their order doesn't change theirmeaning.xyIf the equation is not already in slope/intercept (y-form) then the equation must first be solved for y.If the equation to graph is 2x + 5y = 10, then after solving we get y = 10-2x5 or y =-25 x + 2 so they-intercept is (0, 2) and the slope is-25 .498 MILEPOSTS 5. Two other special cases to remember are vertical and horizontal lines.y = 2 is a horizontal line (slope equal to zero) passing throughthe y-axis at (0, 2).x = 3 is a vertical line (undefined slope) passing through thex-axis at (3, 0).xyy = 2x = 3Now we can go back to the original question.Graph each line and find the intersection.x + y = 5y = 13x + 1The first equation needs to be solved for y: y = -x +5 so they-intercept is (0, 5) and the slope is -1 or-11 . The secondequation has a y-intercept of (0, 1) and a slope of 13. Aftergraphing the two lines you see that they intersect at the point(3, 2).xyHere are some more to try. Use the slope and y-intercept to graph each line and tell the point ofintersection.1) y = -x + 8y = x - 22) y = -x + 3y = x + 33) y = -3xy = -4x +24) y = -x + 5y = 12x + 25) y = -2x - 1y = -4x + 36) 3x + 3y = 4 + x4 - 2x = 3y7) y = 22x + y = 48) x = 32x + 3y = 09) 2x + 3y = 02x - 3y = 010) 3x - 2y = 42y = 3x - 6Answers:1. (5, 3) 2. (0, 3) 3. (2, -6) 4. (2, 3) 5. (2, -5)6. same line 7. (1, 2) 8. (3, -2) 9. (0, 0) 10. no solution-parallel linesSkill Building Resources 499 6. Milepost Number 3BB-107Simplifying Expressions with Positive ExponentsThere are three basic patterns for expressions with positive exponents. They are summarized belowwith some examples.1) xa . xb = xa + b examples: x3 . x4 = x3 + 4 = x7; 27 24 = 2112)xaxb = xa - b examples: x10 x4 = x10 - 4 = x6;2427= 2-33) (xa)b = xab examples: (x4)3 = x4 3 = x12; (2x3)4 = 24x12 = 16x12Now we can go back and try the original problem. Simplify each expression.(-6x)2 = -3x2y336x2 = - y3a) (2x2y)4 = 24x8y4 = 16x8y4 b) -3x2y3123xy5 = 16x8y43xy5 = 16x7c) (2x2y)43yd) 5(5xy)2(x3y) = 5(25x2y2)(x3y)=125x5y3Here are some more to try. Use the properties of exponents to write each of the followingexpressions in a simpler form.1. 3x2 x2.n12n33. (x3)24. (-2x2)(-2x)5.-8x6y2-4xy6. (2x3)37. (103)4 8. 32 35 9. 105 10310. x2y3 x3y4 11. (x3)412.6x2y32xy13. -3x2 4x3 14. (2x2)3 15. (x3y)2(2x)316.m16y31m12y17 17. (6x3z)318. (3x2 )2 (6x4 )19. (5x)2(3y)320. (3x11z5)2 21.3 23. (6x)2 (24x3) 24.(2b)5 (3k2)222. 3x26x56x2y32xy500 MILEPOSTS 7. Answers:1. 3x3 2. n9 3. x6 4. 4x3 5. 2x5y 6. 8x97. 1012 8. 37 9. 102 10. x5y7 11. x12 12. 3xy213. -12x5 14. 8x6 15. 8x9y2 16. m4y14 17. 216x9z3 18. 3219. 675x2y3 20. 9x22y10 21. 288b5k4 22. 18x9 23. 32x 24. 3xy2Milepost Number 4FX-36Solving Systems of Linear Equations in Two VariablesYou can solve systems of equations with a variety of methods. You can graph, use the SubstitutionMethod, or the Elimination Method. Each method works best with certain forms of equations. Hereare some examples and then we can return to the original problem.For each system below, determine which method would be best (easiest) to use. Then solvethe system to find the point of intersection.a) x = 4y - 73x - 2y = 1b) y = 34x - 1y = - 13x - 1c) x + 2y = 16x - y = 2d) x + 3y = 43x - y = 2Although the method that is easiest for one person may not be the easiest for another, the mostcommon methods are shown on the next two pages. You should use the method you are comfortablewith and with which you are most successful.Skill Building Resources 501 8. a) x = 4y - 73x - 2y = 1Substitution: Substitute 4y - 7 forx in the second equation.3(4y - 7) - 2y = 112y - 21 - 2y = 110y - 21 = 110y = 22y = 2210 = 2.2Find x:x = 4(2.2) - 7x = 8.8 - 7 = 1.8Solution: (1.8, 2.2)b) y = 34x - 1y = - 13x - 1Graphing: Normally graphing isnot the best way to solve a system ofequations, but since both equations arein y-form and if you happened tonotice that they have the same y-intercept,you can tell that they cross at(0, -1) the y-intercept. We did notactually graph here, but we used theprinciples of the graphs to solve thesystem of equations. Substitution willwork nicely as well.Solution: (0, -1)c) x + 2y = 16x - y = 2Elimination: Subtract the secondequation from the first.0 + 3y = 143y = 14y = 143Find x by substituting y = 143into the second equation:x - 143 = 2x = 2 + 143 = 203Solution: (203 , 143 )d) x + 3y = 43x - y = 2Elimination with a multiplicationfirst. Multiply the bottom equation by 3and add it to the top equation.x + 3y = 4+ 9x - 3y = 610x = 10x = 1Find y by substituting x = 1 into thesecond equation:3(1) - y = 23 - y = 21 = ySolution: (1, 1)Now we can go back and look at the original problem.Solve this system of linear equations in two variables: 5x - 4y = 72y + 6x = 22You may use substitution or elimination but both methods need a little work to get started.502 MILEPOSTS 9. Substitution method:Before we can substitute we need toisolate one of the variables. Solve thesecond equation for y and it becomesy = 11 - 3x.Now substitute 11 - 3x for y in the firstequation and solve.5x - 4(11 - 3x) = 75x - 44 + 12x = 717x - 44 = 717x = 51x = 3Solve for y;y = 11 - 3(3) = 2Solution (3, 2)Elimination method:Before we can eliminate we need torearrange the second equation so thatthe variables line up.5x - 4y = 76x + 2y = 22Now multiply the second equation by2 and add to eliminate y.5x - 4y = 7+ 12x + 4y = 4417x = 51x = 3Solve for y in the first equation:5(3) - 4y = 7-4y = -8y = 2Solution (3, 2)Here are some more to try. Find the solution to these systems of linear equations. Use the method ofyour choice.1 y = 3x - 12x - 3y = 10 2. x = - 12y + 48x + 3y = 313. 2y = 4x + 106x + 2y = 104. 3x - 5y = -14x + 5y = 225. 4x + 5y = 112x + 6y = 166. x + 2y = 5x + y = 57. y = 2x - 3x - y = -48. y + 2 = x3x - 3y = x +149. 2x + y = 7x + 5y = 1210. y = 35x - 2y = x10 + 111. 2x + y = -2x + 53x + 2y = 2x + 3y12. 4x - 3y = -10x = 14y - 1Answers:1. (-1, -4) 2. (7/2, 1) 3. (0, 5) 4. (2, 4) 5. ( -1,3) 6. (5, 0)7. (7, 11) 8. (-8, -10) 9. ( 23/9, 17/9) 10. (6, 1.6) 11. (1, 1) 12. (-14 , 3)Skill Building Resources 503 10. Milepost Number 5FX-112Multiplying PolynomialsThe product of polynomials can be found by using the distributive property or using genericrectangles. If you are multiplying two binomials, you can also use the F.O.I.L. method.Let us look at an example for each of the three methods before returning to the original problem.In multiplying binomials, such as (3x - 2)(4x + 5), you might use a generic rectangle.4x+53x - 23x(4x)3x(5)-2(4x)-2(5)4x+53x - 212x2 -8x15x -10= 12x 2 + 7x - 10You might view multiplying binomials with generic rectangles as a form of doubledistribution. The 4x is distributed across the first row of the generic rectangle and thenthe 5 is distributed across the second row of the generic rectangle. Some people write itthis way:(3x - 2)(4x + 5) = (3x - 2)(4x) + (3x - 2)(5) = 12x2 - 8x + 15x - 10= 12x2 + 7x - 10.Another approach to multiplying binomials is to use the mnemonic F.O.I.L. F.O.I.L. isan acronym for First, Outside, Inside, Last:F. multiply the FIRST terms of each binomial (3x)(4x) = 12x2O. multiply the OUTSIDE terms (3x)(5) = 15xI. multiply the INSIDE terms (-2)(4x) = -8xL. multiply the LAST terms of each binomial (-2)(5) = -10Finally, we combine like terms: 12x2 + 15x - 8x - 10 = 12x2 + 7x - 10.504 MILEPOSTS 11. Now we can go back and try the original problem again using a variety of methods.Multiply and simplify.a) (x + 1)(2x2 - 3)We can use F.O.I.L. here.F O I L(x)(2x2) (x)(-3) (1)(2x2) (1)(-3)2x3- 3x + 2x2 - 3 = 2x3+ 2x2- 3x -3b) (x + 1)(x - 2x2 + 3)Using generic rectanglesx 2 - 2x + 3x3 -2x2 3xx 2 -2x 3x - x + x + 3 3 2x+1c) 2(x + 3)2Write our the factors and distribute.2(x +3)(x +3) = (2x + 6)(x + 3)(2x + 6)(x) + (2x + 6)(3)2x2 + 6x + 6x +18 = 2x2+12x +18d) (x + 1)(2x - 3)2Write out the factors. Multiply two ofthe factors together and then multiplythat answer by the third factor.(x + 1)(2x - 3)(2x - 3)(2x2 - x - 3)(2x - 3)4x3 - 6x2 - 2x2 + 3x - 6x + 94x3 - 8x2 - 3x + 9Here are some more to try. Multiply and simplify.1. (2x + 3)(x - 7) 2. (4x - 2)(3x + 5) 3. (x - 2)(x2 + 3x + 5)4. (x + 8)(x - 12) 5. 4(3x - 5)2 6. (2x + y)(2x - y)7. (2x + 3)2 8. (5x - 8)(2x + 7) 9. (x + 3)(x2 - 4x + 7)10. (x + 7)(x - 11) 11. -8x3(5x2 + 7) 12. (2x + y)(x + 1)2Answers:1. 2x2-11x -21 2. 12x2 +14x-10 3. x3+x2-x-10 4. x2-4x-965. 36x2-120x+100 6. 4x2- y2 7. 4x2+12x +9 8. 10x2 +19x-569. x3-x2-5x+21 10. x2-4x-77 11. -40x5- 56x312. 2x3+4x2+2x+x2y+2xy +ySkill Building Resources 505 12. Milepost Number 6FX-139Factoring Quadratic ExpressionsFactoring quadratics means changing the expression into a product of factors or to find thedimensions of the generic rectangle. You can use diamond problems with generic rectangles or justguess and check with F.O.I.L. or the distributive property. Here are some examples using diamondsand generic rectangles:Diamond Problems can be used to help factor easier quadratics like x2 + 6x + 8.82 46x2 4x2x 8x + 4x2 4x2x 8x+2(x + 4)(x + 2)We can modify the diamond method slightly to factor problems that are a little different inthat they no longer have a 1 as the coefficient of x2. For example, factor:2x 2 + 7x + 3multiply6? ?766 17x + 32x 2+12 2x 6x2x 6x1x 31x 3(2x + 1)(x + 3)Another problem: 5x2 - 13x + 6. Note that the upper value in the diamond is the productof 5 and 6.5x 5x30 2x - 22-10x 5x-10x6-3 -10 (5x - 3)( x - 2)-3x-13 6-3-3xNow we can go back and try the original problem. Factor each quadratic.4x - 1 2multiply-4 -4? ?02x - 12x 2+12 4x4x2x2x(2x + 1)(2x - 1)2 -20-2x-1-2x-1a)506 MILEPOSTS 13. 4x + 4x + 1 2multiply4? ?1412 22x + 122 4x 2x4x 2x2x 12x 12x+1(2x +1)(2x +1)or (2x +1)2b)22y + 5y + 2multiply45? ?2y + 1y 2+22 2y y2y y4y 24y 2(2y+1)(y + 2)454 1c)3m - 5m - 2 2multiply-6 -6 3m 3m? ?3m + 12 2 m+mm-2 -6m -2-2(3m+1)(m - 2)-6 1-5 -5-6md)Here are some more to try. Factor each expression1. 2x2 + 7x - 4 2. 7x2 + 13x - 23. 3x2 + 11x + 10 4. x2 + 5x - 245. 2x2 + 5x - 7 6. 3x2 - 13x + 47. 64x2 + 16x + 1 8. 5x2 + 12x - 99. 8x2 + 24x + 10 10. 6x3 + 31x2 + 5xAnswers:1. (x + 4)(2x - 1) 2. (7x - 1)(x + 2) 3. (3x + 5)(x + 2)4. (x + 8)(x - 3) 5. (2x + 7)(x - 1) 6. (3x - 1)(x - 4) 7. (8x + 1)28. (5x - 3)(x + 3) 9. 2(4x2 + 12x + 5) = 2(2x + 1)(2x + 5)10. x(6x2 + 31x + 5) = x(6x + 1)(x + 5)Skill Building Resources 507 14. Milepost Number 7PG-30Solving for One Variable in an Equationor Formula with Two or More VariablesWhen you solve for one variable in an equation with two or more variables it usually helps to start bysimplifying, for example, removing parentheses and fractions. Then isolate the desired variable in thesame way as you solve an equation with only one variable. Here are two examples.Solve for y: x!-!3y4 + 2(x + 1) = 7First multiply all terms by 4 to removethe fractions and then simplify.(4) x!-!3y4 + (4)2(x + 1) = (4)7x - 3y + 8x + 8 = 289x - 3y = 20Then solve for y.-3y = -9x + 20y = -9x!+!20-3 = 3x - 203Solve for y: x + 2 y!+!1! = 3x + 4First isolate the radical.x + 2 y!+!1! = 3x + 42 y!+!1! = 2x + 4y!+!1! = x + 2Then remove the radical by squaring bothsides. Remember (x + 2)2 = (x+2)(x+2).( y!+!1! )2 = (x + 2)2y + 1 = x2 + 4x + 4y = x2 + 4x + 3Now we can go back and look at the original problem.Rewrite the following equations so that you could enter them into the graphing calculator.In other words, solve for y.a) x - 3(y + 2) = 6x - 3y - 6 = 6x - 3y = 12-3y = -x + 12y = -x!+!12-3 or y = 13x - 4b) 6x!-!1y - 3 = 26x!-!1y = 5(y)6x!-!1y = (y)56x - 1 = 5yy = 6x!-!15 or y = 65x - 15c) y!-!4 = x + 1( y!-!4 )2 = (x + 1)2y - 4 = (x + 1)2y = (x + 1)2 + 4 or x2 + 2x + 5d) y!+!4 = x + 2( y!+!4 )2 = (x + 2)2y + 4 = x2 + 4x + 4y = x2 + 4x508 MILEPOSTS 15. Here are some more to try. Solve for y.151. 2x - 5y = 7 2. 2(x + y ) + 1 = x - 43. 4(x - y) + 12 = 2x - 4 4. x = y - 25. x = y2 + 1 6.5x!+!2y - 1 = 57. y!+!3 = x - 2 8. (y + 2)2 = x2 + 99.x!+!24 +4!-!y2 = 3 10. 2y!+!1 = x + 311 x =24!-!y 12. x =y!+!1y!-!1Answers:1. y = 25x - 75 2. y = -12x - 523. y = 12x + 4 4. y = 5x + 10 5. y = x!-!16. y =56 x +137. y = x2 - 4x + 1 8. y = x2!+!9 - 2 9. y = 12x - 110. y = 12x2 + 3x + 4 11. y = 4x!-!2x 12. y =x!+!1x! - !1Skill Building Resources 509 16. Milepost Number 8PG-53Find the Slope of the Line Through Two Given Pointsand the Distance Between the Two PointsTo compute either the slope or the distance determined by two points, a generic right triangle providesa good diagram. The slope is the ratio of the vertical leg over the horizontal leg. (Remember to checkwhether it is negative or positive.) The distance is the length of the hypotenuse which is found byusing the Pythagorean theorem. Here are two examples.Use a generic triangle to find the slope and the distance between the given points.a) (-2, 3) and (3, 5) b) (-7, 20) and (3, -5)(-2,3)(3,5)(3,-5)yx25(-7,20)yx2510Using an accurate graph the length of thelegs can be seen as 2 and 5. The lineslopes in the positive direction.slope = vertical!changehorizontal!change = 25. Thedistance is found by d2 = 22 + 52 = 29.So the distance d = 29 " 5.39.A point moving from left to right alongthis line moves up, a positive direction.Sketching a generic graph, the verticalchange or difference in y-values is 25. Thehorizontal change or difference in x-valuesis 10. The line slopes in the negativedirection.slope = vertical!changehorizontal!change =- 2510 = - 52. Thedistance is found by d2 = 252 + 102 = 725.So the distance d = 725 " 26.92.A point moving along this line from left toright moves down in a negative direction.510 MILEPOSTS 17. We can now go back and try the original problem. Use a generic triangle to find the slope of the linethrough the two given points and then find the distance between the two points.a) b)(0,0)(4,4)(4,7)yx(--2,4)yx4463slope = 44= 1d = 42+!42 = 32 " 5.66slope = 36= 12d = 62+!32 = 45 " 6.71c) Sometimes people like to use formulasthat represent the diagram. Using thepoints(12, 18) and (-16, -19):slope =y2! - !y1x2! - !x1= -16!-!12-19!-!18= -28-37 = 2837 .distance = (y2! - !y1)2+!(x2! - !x1)2= (-28)2+!(-37)2 .= 2153 " 46.40.y2 - y1 and x2 - x1 represent thelengths of the vertical and horizontallegs respectively.d) Using the formulas again, this timewith the points (0, 0) and (25, 25):slope = 25!-!025!-!0 = 1distance = (25)2+!(25)2= 1250 " 35.35.Here are some more to try. Find the slope of the line through the two given points and then find thedistance between the two points.1. (1, 2) and (4, 5) 2. (7, 3) and (5, 4) 3. (-6, 8) and (-4, 5)4. (5, 0) and (0, 1) 5. (10, 2) and (2, 24) 6. (-6, 5) and (8, -3)7. (-3, 5) and (2, 12) 8. (-6, -3) and (2, 10) 9. (-15, 39) and (29, -2)Answers:1. m = 1 d = 18 2. m = -12d = 5 3. m = -32d = 134. m = -15 d = 26 5. m = -114 d = 548 6. m = -47d = 2607. m = 75d = 74 8. m = 138 d = 233 9. m = -4144 d = 3617Skill Building Resources 511 18. Milepost Number 9PG-149Using Function Notation and Calculating the Domain and RangeWe should first review some vocabulary and notation.An equation is called a FUNCTION if there exists no more than one output for eachinput. If an equation has two or more outputs for a single input value, it is not afunction. The set of possible inputs of an equation is called the DOMAIN, while theset of all possible outputs of an equation is called the RANGE.Functions are often given names, most commonly f,x = 2g or h. The notation f(x) represents the output of afunction, named f, when x is the input. It is pronouncedf of x. The notation g(2), pronounced g of 2,represents the output of function g when x = 2.f(x) = 3x + 4Similarly, the function y = 3x + 4 and f(x) = 3x + 4represent the same function. Notice that this notation isinterchangeable, that is, y = f(x). In some textbooks,3x + 4 is called the RULE of the function. The graph off(x) = 3x + 4 is a line extending forever in both the x andy directions so the domain and range of f(x) are both allreal numbers.f(2) = 10For each function below, tell the domain and range. Then find f(2) and solve f(x) = 3.f(x) = | x - 1 | - 2y44 44xSince you can use any real numberfor x in this equation, the domainis all real numbers. The smallestpossible result for y is -2, so therange is y $ -2. By looking at thegraph or substituting x = 2 into theequation, f(2) =| 2 - 1 | - 2 = -1.To solve f(x) = 3, find the pointswhere the horizontal line y = 3intersects the graph or solve theequation.3 = | x -1 | - 2,x = -4 or 6.y4f(x) is an unknownequation. 4 44xAny real number can replace x, so thedomain is all reals. The y-values arebetween -2 to +2 so the range is-2% y %2. By inspection f(2) " 1.8.Since -2 % y % 2, f(x) = 3 has nosolution.512 MILEPOSTS 19. f(x) = x!+!3y44 44xYou can only use numbers -3 orlarger for x-values so the domain isx $ 3. The smallest possibley-value is zero. The range is y $ 0.Looking at the graph gives anapproximate answer when x = 2,y " 2.25 or substituting x = 2 into theequation, f(2) = 2!+!3 = 5 .To solve f(x) = 3, find the point wherey = 3 intersects the graph or solve3 = x!+!3x = 6Now we can go back and try the original problem.Given g(x) = 2(x + 3)2. State the domain and range. The graph is a parabola openingupward with locator point (-3, 0). The domain is all real numbers and the range is y $ 0.a) g(-5) = 2(-5 + 3)2 = 2(-2)2 = 8 b) g(a + 1) = 2(a + 1 + 3)2 = 2(a + 4)2=2(a2 + 8a + 16) = 2a2 + 16a + 32c) If g(x) = 32, then32 = 2(x + 3)216 = (x + 3)24 = x + 3x = 1 or -7d) If g(x) = 0, then0 = 2(x + 3)20 = (x + 3)20 = x + 3x = -3Here are some more to try.For each graph, tell the domain and range1. 2. 3.4. If f(x) = 3 - x2 , find f( 5); find f(3a) 5. If g(x) = 5 - 3x2, find g(-2); find g(a+2)6. If f(x) = x!+!32x!-!5 , find f(2); find f(2.5) 7. If f(x) = x2 + 5x + 6, solve f(x) = 08. If g(x) = 3(x - 5)2 , solve g(x ) = 27 9. If f(x) = (x + 2)2 , solve f(x) = 27Skill Building Resources 513 20. Answers:1. D: all x & -2, R all y & 0. 2. D: all x, R: all y $ -5 3. D: all x, range y > -24. -22, 3 - 9a2 5. -7, -3a2 - 12a - 7 6. -5, not possible7. -2, -3 8. 8, 2 9. -2 27Milepost Number 10LS-53Writing the Equation of a Line Given Two PointsThe equation of a line is y = mx + b where m represents the slope and b represents the y-intercept.One way to find the equation is to calculate the slope and then solve for the y-intercept. A secondmethod is to use the two points to write two equations involving m and b and then solve the system.Here is an example of each method.Find an equation of the line passing through (6, 5) and (9, 2).Method One33Using the points we makea generic slope triangleand find the slope to bem = - = -1. Theequation now looks likey = -1x + b.(9, 2)(6, 5)33We can use either one of the twooriginal points here; I will use (9, 2).Substitute the x and y values into theequation to give:2 = -1(9) + bwhich we can solve for b.2 = -9 + b11 = bTherefore the equation of the line isy = -1x + 11.Method TwoSubstitute the 2 points in for x and yin the equation y = mx + b:5 = 6m + b2 = 9m + bSubtracting the second equation fromthe first gives:3 = -3m so m = -1. We can now findb by substituting m = -1 into eitherequation. Using the first equation:5 = 6(-1) + bwhich we can solve for b.5 = -6 + b11 = bTherefore the equation of the line isy = -1x + 11.514 MILEPOSTS 21. Now we can go back and try the original problem. Write an equation for each line defined below.a) The line through points (-1, 4) and (2, 1).b) The line through points (6, 3) and (5, 5).33Using the first method for part(a). For the points we make ageneric slope triangle and findthe slope to bem = - = -1. The equationnow looks likey = -1x + b.(2, 1)(-1, 4)33We can use either one of the twooriginal points here; I will use (-1, 4).Substitute the x and y values into theequation to give:4 = -1(-1) + bwhich we can solve for b.4 = 1 + b3 = bTherefore the equation of the line isy = -1x + 3.Using the second method for part (b).Substitute the 2 points in for x and y in theequation y = mx + b:3 = 6m + b5 = 5m + bSubtracting the second equation fromthe first gives:-2= 1m so m = -2.We can now find b by substituting m =-2 into either equation. Using the firstequation:3 = 6(-2) + bwhich we can solve for b.3 = -12 + b15 = bTherefore the equation of the line isy = -2x + 15.Here are some more to try. Find an equation of the line through the given points.1. (2, 3) and (1, 2) 2. (-3, -5) and (-1, 0) 3. (4, 2) and (8, -1)4. (1, 3) and (5, 7) 5. (0, 4) and (-1, -5) 6. (-3, 2) and (2, -3)7. (4, 2) and (-1, -2) 8. (3, 1) and (-2, -4) 9. (4, 1) and (4, 10)Answers:1. y = x+1 2.!y = 52x +523. y = - 34x +5 4. y = x + 2 5. y = 9x + 46. y = -x -1 7. y = 45x - 658. y = x - 2 9. x = 4Skill Building Resources 515 22. Milepost Number 11LS-110Slopes of Parallel and Perpendicular LinesParallel lines have the same slopes. For perpendicular lines, the product of the slopes equal negativeone. Here are two examples of problems involving parallel and perpendicular lines.Find the equation of the line that isparallel to y = 12x - 5 and passes throughthe point (4, 10).Any line parallel to the line y = 12x - 5which has slope 12, must also haveslope 12. The equation must looklike y = 12x + b. Substituting thegiven point in place of x and y we have:10 = 12(4) + b12Solving we find that b = 8 so thatthe equation is y = x + 8.Find the equation of the line that isperpendicular to y = 12x - 5 and passesthrough the point (4, 10).Since the original line has slope 12, aperpendicular line must have slope -2. Theequation must look like y = -2 x + b.Substituting the given point in place of xand y we have:10 = -2(4) + bSolving we find that b = 18 so that theequation is y = -2x + 18.We will now go back and solve the original problemFind an equation for each of the lines described below.The line with slope 13through the13point (0, 5).The slope and the y-intercept of theline are both given so the equation is:y = x + 5.The line parallel to y = 2x - 5 through thepoint (1, 7).The slope must be 2 so the equation mustlook like y = 2x + b. Substituting the pointfor x and y we have:7 = 2(1) + b.Solving we find that b = 5 so that theequation is y = 2x + 5.516 MILEPOSTS 23. 12The line perpendicular to y = 2x - 5through the point (1, 7).The slope must be - so theequation must look like y = - 12x + b.Substituting the coordinates of the point forx and y we have:7 = - 12(1) + b.Solving we find that b = 712so thatthe equation is y = - 12x + 712.The line through the point (0,0) so that thetangent of the angle it makes with thex-axis is 2.The tangent of the angle is the same as theslope. The y-intercept is given so theequation must be:y = 2xHere are some more to try. Find an equation for each of the lines described below.1. The line with slope 13passingthrough (-2, 5)2. The line through the point (3, 0) sothat the tangent of the angle it makeswith the x-axis is -2.3. The line parallel to y = 23x + 5passing through (3, 2)4. The line perpendicular toy = 23x + 5 passing through (3, 2).5. The line parallel to 3x + 4y = 4passing through (-4, 2).6. The line perpendicular to 3x + 4y = 4passing through (-4, 2).7. The line parallel to the linedetermined by (-3, -2) and (2, 4)passing through (0, -1).8. The line perpendicular to 2x - 3y = 6passing through (0, 3).9. The line parallel to y = 7 passingthrough (-2, 5).10. The line perpendicular to y = 7 passingthrough (-2, 5).Answers:1. y = 13x + 5 23 2. y = -2x + 6 3. y = 23x 4. y = - 32x + 6 125. y = - 34 x + 1 6. y = 43x + 223 7. y = 65x - 1 8. y = - 32x + 39. y = 5 10. x = -2Skill Building Resources 517 24. Milepost Number 12LS-149Graphing Linear InequalitiesGraphing inequalities is very similar to graphing equations. First you graph the line. Withinequalities you also need to determine if the line is solid (included) or dashed (not included) andwhich side of the line to shade. Here are two examples together in a system of inequalities.On graph paper, graph and shade the solution for each of the systems of inequalities below.Describe each resulting region.y % 25xy > 5 - xFirst, we treat the inequality as an equation: y = 25x. We cangraph it using the slope, 25, and the y-intercept, (0, 0). Thisline divides the grid into two regions. Choose a point in eitherregion and check whether or not it makes the original inequalitytrue. If the point (0, 1) is the test point.1 % ?25(0) FALSE!Since points above the line make the inequality false, pointsbelow the line must make it true. Therefore we shade all theyx55 Test point(0, 1)-5-5points below the line to represent the solution.Next we do the same thing for the second inequality. First treatit as an equation and graph it on the same set of axes. Theslope is -1 and the y-intercept is (0, 5). This line is dashed(not solid) because the inequality is strictly greater than, notgreater than or equal to. The line has divided the grid into tworegions and we will chose a point on one side as a test point.You can use the same point as last time. (This is not necessaryand may not always be feasible. Here it is a convenient point.)1 >? 5 - 0 FALSE!Since our test point made the inequality false, the opposite sidewould make the inequality true. For the second inequality weyx55 Test point(0, 1)-5-5shade above the line and to the right.Putting these two inequalities together with their overlappingshading gives us the solution to the system of inequalities. Inthis case the solution is the darkest region, with a solid lineabove the region and a dashed line bordering below (left).yx55-5-5518 MILEPOSTS 25. Now we can go back to the original problem. For this system of inequalities:y % -2x + 3 y $ x x $ -1a) Draw the graph. b) Find the area of the shaded region.12Start by looking at the equation of the line that marks the edgeof each inequality. The first has slope -2 and y-intercept(0, 3). Checking (0, 1) gives a true statement so we shade belowthe solid line. The second has slope 1 and y-intercept(0, 0). Again checking (0, 1) gives a true statement so we shadeabove the solid line. The third is a vertical line at x = -1.Checking the point tells us to shade the right side. Theoverlapping shading is a triangle with vertices (-1, 5), (1, 1), and(-1, -1). The shaded area = (6)(2) .yxHere are a few more to try. Graph and shade the solution for the system of inequalities below.1. y % -x + 2y % 3x - 6 2. y > 23x + 4y < 712 x + 53. x < 3y $ -24. y % 4x + 16y > - 43x - 4Answers:1. 2. 3. 4.y = 3x - 6y = -x + 2yxy = 2/3 x + 4yy = 7/12 x + 5xx = 3y = - 2yxxy = -4/3 x - 4yy = 4x + 16Skill Building Resources 519 26. Milepost Number 13CC-42Multiplication and Division of Rational ExpressionsTo multiply or divide rational expressions you follow the same procedures as you did with numericalfractions. However, you need to first factor in order to simplify. Here are two examplesProblem A: Multiply x2 + 6x(x + 6)2 !x2 + 7x + 6x2 - 1and simplify your result.After factoring, our expression becomes: x(x + 6)(x + 6)(x + 6)!(x + 6)(x +1)(x + 1)(x - 1)After multiplying, reorder the factors: (x!+!6)(x!+!6) . (x!+!6)(x!+!6) . x(x!-!1) . (x!+!1)(x!+!1)Since (x + 6)(x + 6) = 1 and (x + 1)(x + 1) = 1, simplify: 1 1 xx! - !1 1 => xx! - !1 .Problem B: Divide x2 - 4x - 5x2 - 4x + 4 x2 - 2x - 15x2 + 4x - 12and simplify your result.First, change to a multiplication expression: x2 - 4x - 5x2 - 4x + 4 !x2 + 4x -12x2 - 2x - 15After factoring, we get: (x - 5)(x + 1)(x - 2)(x - 2)!(x + 6)(x - 2)(x - 5)(x + 3)Reorder the factors: (x - 5)(x - 5)!(x - 2)(x - 2)!(x + 1)(x - 2)!(x + 6)(x + 3)Since (x - 5)(x - 5) = 1 and (x - 2)(x - 2) = 1, simplify: (x + 1)(x + 6)(x - 2)(x + 3)Thus,x2 - 4x - 5x2 - 4x + 4 x2 - 2x - 15x2 + 4x - 12 = (x + 1)(x + 6)or x2 + 7x + 6(x - 2)(x + 3) x2 + x - 6Now we can go back and try the original problemThink about factoring first, then perform each operation. Simplify when it is helpful.a) x2 + 5x + 6x2 - 4x!4xx + 2=( x+2 )( x+3)x (x -4) !4x(x +2) =4(x +3)( x- 4)c) x2 - 2xx2 - 4x + 44x2x - 2 = x (x -2)(x -2)( x- 2)!(x -2)4x!x = 14x520 MILEPOSTS 27. Here are some more to try. Perform each operation.1. x 2 -16(x - 4)2 x 2 - 3x - 18x 2 - 2x - 242. x2 - x - 6x2 + 3x -10x2 + 2 x - 15x2 - 6x + 9x2 + 4x - 21x2 + 9x +143. x2 - x - 6x2 - x - 20 x 2 + 6x + 8x2 - x - 64. x2 - x - 30x2 +13x + 40x2 +11x + 24x2 - 9x + 185. 15!-!5xx2!- !x! - !6 5xx2!+!6x!+!86. 17x!+!119x2!+!5x!-!14 9x!-!1x2! - !3x!+!27. 2x2! - !5x! - !33x2!-!10x!+!3 9x2! - !14x2!+!4x!+!18. x2! - !1x2! - !6x! - !7 x3!+! x2! - !2xx! - !7Answers:1. (x+3)/(x-4) 2. (x-3)/(x-2) 3. (x+2)/(x-5) 4. (x+3)/(x-3)5. -(x + 4)/x 6. 17(x - 1)/(9x - 1) 7. (3x+1)/(2x+1) 8. 1/x(x+2)Skill Building Resources 521 28. Milepost Number 14CC-76Solving Rational EquationsTo solve rational equations (equations with fractions) usually it is best to multiply everything by thecommon denominator to remove the fractions. This process is called FRACTION BUSTERS. Thensolve the equation in the usual ways. Here are two examples.24x!+!1 = 161Multiply both sides by the commondenominator (x + 1)(x + 1)( 24x!+!1 )= (x + 1)( 161 )Then simplify.24 = 16(x + 1)24 = 16x + 168 = 16x816 = 16x16x = 1252x +16= 8Multiply each term by the commondenominator 6x.52x +16= 86x( 52x +16) = 6x(8)Then simplify.6x( 52x ) + 6x(16) = 48x15 + x = 48x15 = 47xx =1547Now we can go back any try the original problem. Solve each of the following rational equations.a)x3= 4x(3x)( x3) = (3x)( 4x)x2 = 12x = 12 = 2 3b)xx! - !1 = 4xx(x-1))( xx! - !1 ) = x(x-1))( 4x)x2 = 4(x - 1)x2 - 4x + 4 = 0(x - 2)(x - 2) = 0x = 2522 MILEPOSTS 29. c)1x+ 13x = 63x(1x+ 13x ) = 3x(6)3x(1x) + 3x( 13x ) = 18x3 + 1 = 18x4 = 18xx = 29d)1x+ 1x!+!1 = 3x(x+1)(1x+ 1x!+!1 ) = x(x+1)3x(x+1)(1x) + x(x+1)( 1x!+!1 ) =x(x+1)3x +1 + x = 3x2 + 3x0 = 3x2 + x - 1using the quadratic formulax " -0.43, -0.77Here are some more to try. Solve each of the following rational equations.1. 3x5 = x! - !24 2. 4x!-!1x = 3x3. 2x5 - 133 4. 4x!-!1= 137x!+!1 = x - 15. x3= x + 4 6. x-15=3x+17. x!+!63 = x 8. 2x!+!36 + 12 = x29. 3x+ 5x! - !7 = -2 10. 2x!+!34 - x! - !76 = 2x!-!312Answers:1. - 10/7 2. 1/3, 1 3. 115 4. 0, 4 5. -66. 4 7. 3 8. 6 9. 3 512 10. -13Skill Building Resources 523 30. Milepost Number 15CC-124Addition and Subtraction of Rational ExpressionsAddition and subtraction of rational expressions is done the same way as addition and subtraction ofnumerical fractions. You change to a common denominator (if necessary), combine the numerators,and then simplify. Here is an example:The Least Common Multiple (lowestcommon denominator) of(x + 3)(x + 2) and (x + 2) is(x + 3)(x + 2).42x(x!+!2)(x!+!3) + x!+!2The denominator of the first fractionalready is the Least Common Multiple.To get a common denominator in thesecond fraction, multiply the fractionby x!+!3x!+!3 , a form of one (1).= 4(x!+!2)(x!+!3) + 2xx!+!2 (x!+!3)(x!+!3)Multiply the numerator anddenominator of the second term: = 4(x!+!2)(x!+!3) + 2x(x!+!3)(x!+!2)(x!+!3)Distribute the numerator.(x!+!2)(x!+!3) + 2x2!+!6x(x!+!2)(x!+!3)= 4Add, factor, and simplify.= 2x2!+!6x!+!4(x!+!2)(x!+!3) = 2(x!+!1)(x!+!2)(x!+!2)(x!+!3) = 2(x!+!1)(x!+!3)Now we can go back and try the original problem. Add or subtract and simplify.a) 2! - !xx!+!4 + 3x!+!6x!+!4 = 2!-!x!+!3x!-!6x!+!4= 2x!-!4x!+!4= 2(x!-!2)x!+!4b) 3(x!+!2)(x!+!3) + x(x!+!2)(x!+!3)= 3!+!x(x!+!2)(x!+!3)= 1(3--!-+--!--x)(x!+!2)(x--!-+--!--3)= 1x!+!2524 MILEPOSTS 31. c) 3x! - !1 - 2x! - !2= (x!-!2)(x!-!2) 3x! - !1 - (x!-!1)(x!-!1) 2x! - !2= 3(x!-!2)(x!-!1)(x!-!2) - 2(x!-!1)(x!-!1)(x!-!2)= 3x!-!6!-!2x!+!2(x!-!1)(x!-!2)= x! - !4(x!-!1)(x!-!2)d) 8x- 4x!+!2 = (x!+!2)(x!+!2) 8x- xx4x!+!2= 8(x!+!2)x(x!+!2) - 4xx(x!+!2)= 8x!+!16!-!4xx(x!+!2)= 4x!+!16x(x!+!2)= 4(x!+!4)x(x!+!2)Here are some more to try. Add or subtract and simplify.1. x(x!+!2)(x!+!3) + 2(x!+!2)(x!+!3) 2. 8x!+!32x!+!3 - 2x!-!62x!+!3 .3. 6x(x!-!3) + 2x!+!3 4. 3x!+!1x2!- !16 - 3x!+!5x2!+!8x!+!165. 7x!-!1x2! - !2x! - !3 - 6xx2! - !x! - !26. 3x! - !1 + 41! - !x + 1x7. 3y9y2! - !4x2 - 13y!+!2x 8. 2x!+!4 - x! - !4x2! - !169. 5x!+!9x2! - !2x! - !3 + 6x2!-!7x!+!12 10. x!+!4x2!-!3x!-!28 + x! - !5x2!+!2x!-!35Answers:1.1x+3 2.3(2x+3)2x+3 = 3 3. 2(x2!+! 9 )x(x!-!3)(x!+!3) 4. 4(5x!!+!!6)(x!-!4)(x!+!4)25. x!+!2(x!-!3)(x!-!2) 6. -1x(x!-!1) 7. 2x(3y!+!2x)(3y!-!2x) 8. 1x!+!49. ! 5(x!+!2)(x!-!4)(x!+!1) 10. 2x(x!+!7)(x-7)Skill Building Resources 525 32. Milepost Number 16CC-136Integral and Rational ExponentsThere are some basic rules for integral and rational exponents. The patterns are summarized belowwith some examples:x0 = 1. Examples: 20 = 1, (-3)0 = 1, ( 14)0 = 1.xn . Examples: x-3 = 1x-n = 1x3 , y-4 = 1y4 , 4-2 = 142 = 116 .1x-n = xn. Examples: 1x-5 = x5,1x-2 = x2,13-2 = 32 = 9xab = (xa)1b = b xa Examples: 512 = 5,or 1634 = (4 16)3 = 23 = 8xab = (x1b )a = (b x )a 423 = 3 42 = 3 16 = 23 2We can now go back and try the original problem. Use integral or rational exponents to write each ofthe following as a power of x.a) 5 x = x1/5(using the fourth property above)b) 1x3 = x-3(using the second property above)c)3x2 = x2/3(using the fourth property above)d) 1x!= 1x1/2 = x-1/2(using properties four and two above)526 MILEPOSTS 33. Here are some more to try. Use integral or rational exponents to simplify each expression. Youshould not need a calculator for any of these.1. x-5 2. m0 3. 4-14. y-3 5. 5-2 6. 507. y-7 8. (x3y4)-2 9. x-1y-810. x-4y-2 (x-3y-6)0 11. 251/2 12. 25-1/213. 21/214. 127-1/315. x3/216. 93/217. ( x3y6)1/318. 16-3/419. (m2 !) -3/220. ( x3y6)1/221. ( 9x3y6)-2Answers:1. 1x5 2. 1 3. 144. 1y35. 152 6. 1 7. 1y7 8. 1x6y89. 1xy8 10. 1x4y2 11. 5 12. 151813. 2 14. 3 15. x3 = x x 16. 2717. xy2 18. 19. 1m3 20. xy3 x21. 181x6y12Skill Building Resources 527 34. Milepost Number 17CT-5Completing the Square and Locator Point for a ParabolaIf a parabola is in graphing form then the locator point or vertex is easily found and a sketch of thegraph can quickly be made. If the equation of the parabola is not in graphing form the equation needsto be rearranged. One way to rearrange the equation is by completing the square. Here are threeexamples:Note that first two examples are in your book problem PG-46 shown with algebra tiles. You mightlike that way better because you can "see it" or the way shown below which just gives a rule.First recall that y = x2 is the parent equation for parabolas and then the graphing equationfor that function is given byy = a(x - h)2 + kwhere (h, k) is the locator point, and relative to the parent graph the function has been:vertically stretched, if the absolute value of a is greater than 1vertically compressed, if the absolute value of a is less than 1reflected across the x-axis, if a is less than 0.Example 1: y = x2 + 8x +10We need to make x2 + 8x into a perfectsquare. Taking half of the x coefficientand squaring it will accomplish the task.The 16 that was put into theparenthesis must be compensated forby subtracting 16.Factor and simplifyy = x2 + 8x +10y = (x2 + 8x + ?) + 10? = 822= 16y = (x2 + 8x + 16) + 10 - 16y = (x + 4)2 - 6. The locator is (-4, -6).Example 2: y = x2 + 5x + 2We need to make x2 + 5x into a perfectsquare. Again, taking half of thex coefficient and squaring it will alwaysaccomplish the task.The 254 that was put into the parenthesis2m5ust be compensated for by subtracting4 .Factor and simplify.y = x2 + 5x + 2y = (x2 + 5x + ?) + 2? = 522= 254y = (x2 + 5x + 254 ) + 2 - 254y = (x + 52)2 - 4 14. The locator is (- 52, - 4 14).528 MILEPOSTS 35. Example 3: y = 2x2 - 6x + 294This problem is a little different because wehave 2x2. First wemust factor the 2 out of thex-terms. Then we make x2 - 3xinto a perfect square as before.The that was put into the parenthesis92must be 94compensated for by subtracting2( ) = .Factor and simplifyy = 2x2 - 6x + 2y = 2(x2 - 3x ) + 2y = 2(x2 - 3x + ? ) + 2? = -322= 94y = 2(x2 - 3x + 94) + 2 - 92y = 2(x - 32)2 - 52. The locator is (32, - 52). and there is a stretch factor of 2.We can now go back to the original problem. Write the equation in graphing form and sketch a graphof y = 2x2 - 4x + 5.First we must factor the 2 out ofthe x-terms. Then we make x2 - 2x intoa perfect square as before.The 1 that was put into theparenthesis must be compensated for bysubtracting 2(1) = 2Factor and simplifyYou might also do this problemby completing two perfect squares usingthe algebra tiles. You will have threeextra ones, giving the +3 on the end ofthe graphing form.y = 2x2 - 4x + 5y = 2(x2 - 2x ) + 5y = 2(x2 - 2x + ? ) + 5? = -222= 1y = 2(x2 - 2x + 1) + 5 - 2y = 2(x -1)2 + 3. The locator is (1, 3) andthere is a stretch factor of 2.The graph of the parabola passes through(1, 3), (2, 5), (3, 13), (0, 5), and (-1, 13).Here are some more to try. Write the equation in graphing form. Tell the locator and stretch factor.1. y = x2 - 6x + 9 2. y = x2 + 3 3. y = x2 - 4x4. y = x2 + 2x - 3 5. y = x2 + 5x +1 6. y = x2 - 13x7. y = 3x2 - 6x + 1 8. y = 5x2 + 20x - 16 9. y = -x2 - 6x + 10Answers:1. (3, 0) normal size 2. (0, 3) normal 3. (2, -4) normal 4. (-1, -4) normal5. (- 52, - 5 14.) normal 6. ( 16, - 136 .) normal 7. (1, -2) stretch by 38. (-2, -36) stretch by 5 9. (-3, 19) normal size but upside downSkill Building Resources 529 36. Milepost Number 18CT-76Absolute Value Equations and InequalitiesAbsolute value means the distance from a reference point. There is a pattern used to solve absolutevalue equations and two patterns used for the different inequalities. They are shown below and thensome examples are solved.Examplesx = k means: x = k or x = -kx < k means: -k < x < kx > k means: x > k or x < -kx = 5 means x = 5 or x = -5(5 or -5 are 5 units from zero)x < 5 means -5 < x < 5(the numbers between -5 and 5 areless than 5 units from zero)x > 5 means x > 5 or x < -5(the numbers greater than 5 or lessthan -5 are more than 5 units fromzero)If the expression inside the absolute value is more complicated, you still follow one of the three basicpatterns above. Also % and $ use the same patterns as the pure inequality.If 2x + 3 = 7 , then the quantity(2x + 3) must equal 7 or -7.2x + 3 = 7 or 2x + 3 = -72x = 4 or 2x = -10x = 2 or x = -5If 2 x+3 7, then the quantity(2x + 3) must be between -7 and 7.-7 % 2x + 3 % 7-10 % 2x % 4-5 % x % 2530 MILEPOSTS 37. Now we can go back and try the original problem. Solve each absolute value equation or inequality.2 2x + 3 = 10(isolate the absolute value)2 x+3 =5(using the first pattern )2x +3 = 5 or 2x + 3 = -52x = 2 or 2x = -8x = 1 or x = -4- x + 3 < 10(isolate the absolute value)x+3 > - 10(dividing by a negative changes thesign) Since any absolute value isnever negative, the solution is allnumbers.Here are some more to try. Solve each absolute value equation or inequality.1. | x - 2 | + 10 = 8 2. 15 - | x + 1 | = 3 3. -3 . | x + 6 | + 12 = 04. | 2x + 7 | = 0 5. |x + 4| $ 7 6. |x| - 5 % 87. 4r-2 > 8 8. -2|x - 3| + 6 < -4 9. 4-d 7Answers:1. No solution2. x = 11, -13 3. x = -2, -10 4. x = - 725. x $ 3 or x % -11 6. -13 % x % 13 7. r < - 32or r > 528. x > 8 or x < -29. -3 % d % 11Skill Building Resources 531 38. Milepost Number 19CT-111Writing and Solving Exponential EquationsExponential functions are equations of the form y = kmx where k represents the initial value, mrepresents the multiplier, x represents the time. Some problems just involve substituting in theinformation and doing the calculations. If you are trying to solve for the time (x), then you will usuallyneed to use logarithms. If you need to find the multiplier (m), then you will need roots. Here aresome examples.Lunch at our favorite fast food stand now cost $6.50. The price has steadily increased 4% peryear for many years.What will lunch cost in 10 years?The initial value is $6.50, the multiplieris 1.04, and the time is 10 years.Substituting into the formula:y = 6.50(1.04)10 = $9.62What did it cost 10 years ago?y = 6.50(1.04)-10 = $4.39How long before lunch costs $10?The initial value is $6.50, the multiplier is1.04, and the time is unknown but the finalvalue is $10. Substituting into the formula:10 = 6.50(1.04)xThis time we must solve an equation.(1.04)x = 106.50 = 1.538x = log(1.538)log(1.04) " 11 yearsTickets for the big concert first went on sale three weeks ago for $60. This week people arecharging $100.What was the weekly multiplier andweekly percent increase?The initial value is $60, the time is 3weeks, and the final value is $100.Substituting into the formula:100 = 60k3k3 = 10060 " 1.667k = 3 1.667 " 1.186The multiplier was about 1.186 so it was aweekly increase of about 18.6%.Write an equation that represents the cost ofthe tickets w weeks from the time that theywent on sale. Assume that they continue toincrease in the same way.The initial value is $60 and the multiplier is1.186 so the equation is:y = 60(1.186)w532 MILEPOSTS 39. We can now go back and solve the original problem parts (b), (c), and (d).When rabbits were first brought to Australia, they had no natural enemies. There were about80,000 rabbits in 1866. Two years later, in 1868, the population had grown to over 2,400,000!b) Write an exponential equation for the number of rabbits t years after 1866.For 1866, 80,000 would be the initial value, time would be 2 years, and the final amountwould be 2,400,000. Here is the equation to solve:2,400,000 = 80,000m230 = m2 so the multiplier m = 30 " 5.477.The desired equation is: R = 80,000(5.477)tc) How many rabbits do you predict would have been present in 1871?The initial value is still 80,000, the multiplier " 5.477 and now the time is 5 years.80,000(5.477)5 " 394 milliond) According to your model, in what year was the first pair of rabbits introduced intoAustralia?Now 2 is the initial value, 80,000 is the final value, the multiplier is still 5.477 but thetime is not known. Here is the equation to solve:80,000 = 2(5.477)x40,000 = (5.477)xx = log(40000)log(5.477) " 6.23 years, so some time during 1859.Here are some more to try.1. A video tape loses 60% of its value every year it is in the store. The video cost $80 new.Write a function that represents its value in t years. What is it worth after 4 years?2. Inflation is at a rate of 7% per year. Janelle's favorite bread now costs $1.79. What didit cost 10 years ago? How long before the cost of the bread doubles?3. Find the initial value if five years from now, a bond that appreciates 4% per year will beworth $146.4. Sixty years ago when Sam's grandfather was a kid he could buy his friend dinner for$1.50. If that same dinner now costs $25.25 and inflation was consistent, write anequation that will give you the costs at different times.5. A two-bedroom house in Omaha is now worth $110,000. If it appreciates at a rate of2.5% per year, how long will it take to be worth $200,000?6. A car valued at $14,000 depreciates 18% per year. After how many years will the valuehave depreciated to $1000?Answers:1. y = 80(.4)t , $2.05 2. $.91, 10.2 years 3. $1204. y = 1.50(1.048)x 5. 24.2 years 6. 13.3 yearsSkill Building Resources 533 40. Milepost Number 20CT-144Finding the Equation for the Inverse of a FunctionTo find the equation for the inverse of a function just interchange the x and y variables and thensolve for y. This also means that the coordinates of points that are on the graph of the function will bereversed on the inverse. Here are some examples:If y = 2(x + 3) then the inverse is:x = 2(y + 3).Solving for y to get the final answer:(y + 3) = y = - 3x2x2If y = - 23x + 6 then the inverse is:x = - 23y + 6 .Solving for y to get the final answer:- 23y = x - 6y = - 32(x - 6) = - 32x + 9If y = 12(x + 4) 2 + 1 the inverse is:x = 12(y + 4) 2 + 1.Solving for y to get the final answer:12(y + 4) 2 = x - 1(y + 4)2 = 2x - 2y + 4 = 2x!-!2y = 2x!-!2 - 4Note that because of the , this inverseis not a function.If y = x!-!2 + 5 then the inverse is:x = y!-!2 + 5.Solving for y to get the final answer:y!-!2 = x - 5y - 2 = (x - 5)2y = (x - 5)2 + 2Note that since the original function is onehalf of a parabola, the graph of the inversefunction is also only one half of a parabola.534 MILEPOSTS 41. We can now go back and try the original problem:Find the equation for the inverse of the following function: y = 2 3(x -1) +5 . Sketch thegraph of both the original and the inverse.Interchanging x and y we get x = 2 3( y-1) + 5. Solving for y to get the final answer:2 3( y-1) = x - 53(y-1) = x-5223(y-1) = (x-5)4y-1 = (x -5)212y =(x-5)212 + 1For the original function:Domain: x $ 1; Range: y $ 5Some points on the original graph are:(1, 5), ( 73, 9) , (4, 11)--half a parabola.For the inverse function:Domain: x $ 5; Range: y $ 1.Some points on the inverse graph are:(5, 1), ( 9, 73), (11, 4)--half a parabola.Here are some more to try. Find the equation for the inverse of each function.1. y = 3x - 22. y =x+143. y = 13x+24. y = x3+1 5. y =1 + x+5 6. y =3(x+2)2 -77. y =2 x-1 + 3 8. y = 12+x9. y =log3( x-2)Answers:1. y =x + 23 2. y=4 x-1 3. y=3 x-64. y=3 x-1 5. y=(x -1)2-5 6. y=x +73 - 27. y=x -322+1 8. y=1x-2 9. y=3x +2Skill Building Resources 535 42. Milepost Number 21ST-12Solving a System of Equations in Three VariablesTo solve a system of equations in three variable using elimination you use the same basic process asyou do with two variables only you have to do it twice. Choose any variable to eliminate and then youare left with two equations in two variables. Continue to solve in the usual way. To solve usingmatrix multiplication you need to change the system into matrices and then isolate the variable matrixby using the inverse matrix on the graphing calculator. Here is an example of each method.Solve for (x, y, z) 5x - 4y - 6z = -19-2x +2y +z = 53x - 6y -5z =-16Method One--EliminationChoose a variable to eliminate.Any is possible. We choose z. Use thefirst two equations. Multiply the secondequation by 6 and addit to the first.Then we must also eliminate z using twoother equations. Multiply the second by 5and add it to the third.Now we have two equations with twovariables. Using lines (**) and (***) wecan subtract the second line from the first toeliminate x and find y.Using our answer for y in (**) we canfind x.Finally go back to any of the originalequations to find z. Using the first one:The solution is (-1, 12, 2)6(-2x +2y +z =5) = -12x +12y +6z =305x - 4y - 6z = -19(**) -7x + 8y = 115(-2x +2y +z =5) = -10x +10y +5z =253x - 6y -5z =-16(***) -7x + 4y = 9(**) -7x + 8y = 11(***) -7x + 4y = 94y = 2y = 12-7x + 8( 12) = 11-7x = 7x = -15(-1) - 4( 12) - 6z = -19-7 - 6z = -19-6z = -12z = 2536 MILEPOSTS 43. Method Two--MatricesWrite the problem as a matrix equation.12Left multiplying both sides of the equationby the inverse of the coefficient matrixgives:(You may need to refer to you calculatordirections or your resource page from unit 5for help with enteringthis in your graphing calculator.The solution is (-1, , 2)5 -4 -6-2 2 13 -6 -5!xyz=-195-16xyz=5 -4 -6-2 2 13 -6 -5-1!-195-16xyz=-1.52We can now go back and solve the original question.Use elimination or matrix multiplication to solvethis system of equations:x + y - z = 123x + 2y + z = 62x + 5y - z = 10Method OneAdding equations one and two eliminates z.x + y - z = 123x + 2y + z = 64x + 3y =18 (**)Adding equations two and threealso eliminates z.3x + 2y + z = 62x + 5y - z = 105x + 7y = 16 (***)Now we have two equations in twovariables. Multiplying (**) by 5and (***) by -4 eliminates x.5(4x + 3y =18) = 20x + 15y = 90-4(5x + 7y = 16) = -20x - 28y = -64-13y = 26y = -2Using y = -2 into (**) gives x = 6.Using y = -2 and x = 6 in any ofthe original equations gives z = -8The solution is (6, -2, -8).Method TwoWrite the system in matrices.1 1 -13 2 12 5 -1!xyz=12610Isolate the variable matrix.xyz=1 1 -13 2 12 5 -1-1!12610Use the graphing calculator to multiply.xyz=6-2-8x = 6 y = -2 z = -8Skill Building Resources 537 44. Here are some more to try. Use elimination or matrix multiplication to solvethese system of equations. Most teachers expect their students to be able to use both methodssuccessfully.1. x + y + z = 343x + 2y +4z = 95x + 2y + 3z = 562. x - 2y + 3z = 82x + y + z = 6x + y + 2z = 123. 5x + y + 2z = 63x - 6y - 9z = -48x - 2y + z = 124. 4x - y + z = -52x + 2y +3z = 105x - 2y + 6z = 15. x + y = 2 - z-y + 1 = - z - 2x3x - 2y + 5z = 166. a - b + 2c = 2a + 2b - c = 12a + b + c = 47. -4x = z - 2y + 12y + z = 12 - x8x - 3y + 4z = 18. 3x + y - 2z = 6x + 2y + z = 76x + 2y - 4z = 129. 4x + 4y - 5z = -22x - 4y + 10z = 6x + 2y + 5z = 0Answers:151. (17, 34 1212, 5) 2. (-1, 3, 5) 3. (-1, -3, 7) 4. (-1, 3, 2)5. (-3, 0, 5) 6. no solution 7. (-3, 5. 10) 8. infinite solutions9. ( , - , )538 MILEPOSTS