1 1.01 Write equivalent forms of algebraic expressions to solve problems. a) Apply the laws of exponents. There are a few rules that simplify our dealings with exponents. Given the same base, there are ways that we can simplify various expressions. For instance: Multiplying Powers (same base) Simplify (x 3 )(x 4 ) Copyright Elizabeth Stapel All Rights Reserved Think in terms of what the exponents mean: (x 3 )(x 4 ) = (xxx)(xxxx) = xxxxxxx = x 7...which also equals x (3+4). This demonstrates a basic exponent rule: Whenever you multiply two terms with the same base, you can add the exponents: ( x m ) ( x n ) = x ( m + n ) Power to a Power Simplify (x 2 ) 4 Again, think in terms of what the exponents mean: (x 2 ) 4 = (x 2 )(x 2 )(x 2 )(x 2 ) = (xx)(xx)(xx)(xx) = xxxxxxxx = x 8...which also equals x ( 2 4 ). This demonstrates another rule: Whenever you have an exponent expression that is raised to a power, you can multiply the exponent and power: ( x m ) n = x m n 1 Dividing Powers (same base) For all numbers x (not zero) and all integers m and n, when you are dividing, and the bases are the same, you SUBTRACT the exponents." Example The bases are the same, so the exponents are subtracted. The numbers in front of the bases are divided. A negative exponent just means that the base is on the wrong side of the fraction line, so you need to flip the base to the other side. For instance, "x 2 " just means "x 2, but underneath, as in 1/(x 2 )". Write x 4 using only positive exponents b) Operate with polynomials. Adding polynomials is just a matter of combining like terms, with some order of operations considerations thrown in. As long as you're careful with the minus signs, and don't confuse addition and multiplication, you should do fine. Simplify (3x 3 + 3x 2 4x + 5) + (x 3 2x 2 + x 4) Horizontally: (3x 3 + 3x 2 4x + 5) + (x 3 2x 2 + x 4) = 3x 3 + 3x 2 4x x 3 2x 2 + x 4 = 3x 3 + x 3 + 3x 2 2x 2 4x + x = 4x 3 + 1x 2 3x + 1 Vertically: Rights Stapel All Reserved
2 . 2 Subtracting polynomials is quite similar to adding polynomials, but you have that pesky minus sign to deal with. Here are some examples, done both horizontally and vertically: Simplify (x 3 + 3x 2 + 5x 4) (3x 3 8x 2 5x + 6) The first thing I have to do is take that negative through the parentheses. Some students find it helpful to put a "1" in front of the parentheses, to help them keep track of the minus sign: Horizontally: (x 3 + 3x 2 + 5x 4) (3x 3 8x 2 5x + 6) = (x 3 + 3x 2 + 5x 4) 1(3x 3 8x 2 5x + 6) = x 3 + 3x 2 + 5x 4 3x 3 + 8x 2 + 5x 6 = x 3 3x 3 + 3x 2 + 8x 2 + 5x + 5x 4 6 = 2x x x 10 That is, FOIL tells you to multiply the first terms in each of the parentheses, then multiply the two terms that are on the "outside" (furthest from each other), then the two terms that are on the "inside" (closest to each other), and then the last terms in each of the parentheses. In other words, using the previous example: Use FOIL to simplify (x + 3)(x + 2) "first": (x)(x) = x 2 "outer": (x)(2) = 2x "inner": (3)(x) = 3x "last": (3)(2) = 6 So: Copyright Elizabeth Stapel All Rights Reserved Vertically: Stapel All Rights Reserved In the horizontal case, you may have noticed that running the negative through the parentheses changed the sign on each term inside the parentheses. The shortcut here is to not bother writing in the subtaction sign or the parentheses; instead, I'll change all the signs in the second row and add down: (x + 3)(x + 2) = x 2 + 2x + 3x + 6 = x 2 + 5x + 6 Simplify (x 4)(x 3) ( Elementary School Multiplication) So the answer is: x 2 7x + 12 Either way, I get the answer: 2x x x 10 "Box" Method This is a "table" version of the distributive method. This style applies to all polynomial multiplications. polynomial multiplication There is also a special method, useful ONLY for a twoterm polynomial times another two-term polynomial. The method is called "FOIL". The letters F-O-I-L come from the words "first", "outer", "inner", "last", and are a memory device for helping you remember how to multiply horizontally, without having to write out the distribution like I did, and without dropping any terms. Here is what FOIL stands for: To multiply by the grid method, place one binomial at the top of a 2x2 grid (for binomials) and the second binomial on the side of the grid. Place the terms such that each term with its sign lines up with a row or column of the grid. Multiply the rows and columns of the grid to complete the interior of the grid. Finish by adding together the entries inside the grid.
3 3 If c is positive, then the factors you're looking for are either both positive or else both negative. If b is positive, then the factors are positive If b is negative, then the factors are negative. In either case, you're looking for factors that add to b. x 2 +2x + 3x + 6 Answer: x 2 + 5x + 6 Square Roots In general, how do you figure out what can "come out" of a square root? Factor the innards, and any factor that occurs in pairs can come out. For example: Simplify: c) Factor polynomials. If c is negative, then the factors you're looking for are of alternating signs; that is, one is negative and one is positive. If b is positive, then the larger factor is positive. If b is negative, then the larger factor is negative. In either case, you're looking for factors that are b units apart. To factor a "hard" quadratic, we have to handle all three coefficients, not just the two we handled above. In this case, we first need to multiply "a" and "c", and then find factors of the product "ac" that add up to "b". For instance: Factor 2x 2 + x 6. Looking at this quadratic, we have a = 2, b = 1, and c = 6, so ac = (2)( 6) = 12. Then we need to find factors of 12 that add up to +1. The pairs of factors for 12 are 1 and 12, 2 and 6, and 3 and 4. Since 12 is negative, we need one factor to be positive and one to be negative (positve times negative is negative). Then we want to use the pair "3 and 4", and we want the 3 to be negative, because = +1 If your text or teacher has you factoring "by grouping", you'll find that it is very easy to make mistakes with the signs. You'll still have to find the numbers that add to the coefficient in the middle, but your steps would look like this: 2x 2 + x 6 Factor x 2 5x + 6. The constant term is 6, but the middle coefficient this time is negative. Since we multiplied to a positive six, then the factors must have the same sign. (Remember that two negatives multiply to a positive.) Since we're adding to a negative ( 5), then both factors must be negative. So rather than using 2 and 3, as in the first example, this time we will use 2 and 3: = 2x 2 + 4x 3x 6 = 2x(x + 2) 3(x + 2) = (x + 2)(2x 3) x 2 5x + 6 = (x 2)(x 3) Note that you can use clues from the signs to determine which factors to use, as we did in this last example above:
4 1.02 Use formulas and algebraic expressions, including iterative and recursive forms, to model and solve problems. At first glance, these problems appear to be much worse than your usual equation solving, but they really aren't that bad. You pretty much do what you've done all along for solving linear equations, except that, due to all the variables, you won't necessarily be able to simplify your answers as much as you're used to doing. Here's how "solving literal equations" works: Solve A = bh for b Solve d = rt for r Copyright Elizabeth Stapel All Rights Reserved Solve P = 2l + 2w for w 4 This next one involves a little "trick" to solve it. See if you can see what it is: Solve Q = 3a + 5ac for a 1.03 Model and solve problems using direct variation. When two variable quantities have a constant (unchanged) ratio, their relationship is called a direct variation. It is said that one variable "varies directly" as the other. The constant ratio is called the constant of variation. The formula for direct variation is y= kx, where k is the constant of variation. "y varies directly as x" Solve Q = (c + d) / 2 for d Solve V = 3k / t for t Solving for k: Example A salesman's commission varies directly as his sales. Write the formula relating this situation and denote the variables. If the commission is $100 for $1000 in sales, find the commission for $1750 in sales. C= Commission and S = Sales Then C=KS or C/S = K 100/1000 = C/ C = C =$ 175
5 2.01 Find the lengths and midpoints of segments to solve problems The Midpoint Formula works exactly the same way. If you need to find the point that is exactly halfway between two given points, just average the x-values and the y-values. Here are some typical examples: Find the midpoint between ( 1, 2) and (3, 6). Apply the Midpoint Formula: So the answer is P = (1, 2) Use the parallelism or perpendicularity of lines and segments to solve problems. Parallel lines have the same slope. The symbol to indicate parallel lines is two vertical bars. It looks something like the number 11. where l 1 and l 2 are lines m 1 and m 2 are slopes Technically, the Midpoint Formula is the following: But as long as you remember that you're averaging the two points' x- and y-values, you'll do fine Distance Formula: y = 3x + 5 y = 3x - 7 y = 3x y = 3x These lines are ALL parallel. They all have the same slope (m). (Remember y = mx + b.) Given the two points (x 1, y 1 ) and (x 2, y 2 ), the distance between these points is given by the formula: Don't let the subscripts scare you. They only indicate that there is a "first" point and a "second" point; that is, that you have two points. Whichever one you call "first" or "second" is up to you. Example : Find the distance between the points ( 2, 3) and ( 4, 4). Just plug them in to the Distance Formula: Perpendicular lines have negative reciprocal slopes. The symbol to indicate perpendicular is an upside-down capital T. where l 1 and l 2 are lines m 1 and m 2 are slopes These lines are perpendicular. Their slopes (m) are negative reciprocals. (Remember y = mx + b.) Then the distance is sqrt(53), or about 7.28, rounded to two decimal places.
6 3.01 Use matrices to display and interpret data Operate (addition, subtraction, scalar multiplication) with matrices to solve problems. Matrix addition is fairly simple, and is done entry-wise. For instance: Add the following matrices: 6 For the following matrix A, find 2A and 1A. To do the scalar multiplication, I just multiply a 2 on every entry in the matrix: Add the pairs of entries, and simplify for the Subtracti on works entry-wise, too. Copyright Elizabeth Stapel All Rights Reserved Given the following matrices, find A Band A C, or explain why you can not. A and B are the same size, each being 2 3 matrices, so I can subtract, working entry-wise: 3.03 Create linear models for sets of data to solve problems. a) Interpret constants and coefficients in the context of the data. b) Check the model for goodness-of-fit and use the model, where appropriate, to draw conclusions or make predictions. Graphing Calculator Solution: 1. Enter the data in the calculator lists. Place the data in L 1 and L 2. STAT, #1Edit, type values into the lists There are two types of multiplication for matrices: scalar multiplication and matrix multiplication. Scalar multiplication is easy. You just take a number (called a "scalar") and multiply it on every entry in the matrix. For example: 2. Prepare a scatter plot of the data. Set up for the scatterplot. 2 nd StatPlot - choose the first icon - choices shown at right. Choose ZOOM #9 ZoomStat.
7 Graph shown below Use linear functions or inequalities to model and solve problems; justify results. a) Solve using tables, graphs, and algebraic properties. b) Interpret constants and coefficients in the context of the problem 3. Have the calculator determine the line of best fit. STAT CALC #4 LinReg(ax+b) 4. Have Calculator graph it for you Y= Vars #5 EQ enter Graph y= x To get a predicted value within the window, hit 2 nd Trace #1,type the desired value. The screen below shows x = 22. Slope is a ratio and can be expressed as: change in y over or or or change in x. Zero Slope:Lines that are horizontal have zero slope. No Slope or Slope Undefined :Vertical lines have no slope, or undefined slope. Slope Intercept Form y = mx + b m = slope b = y-intercept Point Slope Form the line = any point on Use this form when you know the slope and the y-intercept (where the line crosses the y-axis). Use this form when you know a point on the line and the slope (or can determine the slope). Standard Form Ax + By = C Use this form when the coefficients are all integers. Slope = A/B Y-intercept = C/B
8 4.02 Graph, factor, and evaluate quadratic functions to solve problems. 8 The general form of a quadratic is "y = ax 2 + bx + c". For graphing, the leading coefficient "a" indicates how "fat" or "skinny" the parabola will be. For a > 1 (such as a = 3 or a = 4), the parabola will be "skinny", because it grows more quickly (three times as fast or four times as fast, respectively, in the case of our sample values of a). For a < 1 (such as a = 1 / 3 or a = 1 / 4 ), the parabola will be "fat", because it grows more slowly (one-third as fast or one-fourth as fast, respectively, in the examples). Also, if a is negative, then the parabola is upside-down. Graph the parabola y = x 2 + 6x - 1 To find the axis of symmetry, use the formula x = -b/2a In this example, a = 1 and b = 6. Substituting gives: x = -(6)/2(1) = -6/2 = -3 Axis of symmetry: x = -3 Since the x-coordinate of the turning point ( Point of Vertex) is -3, use this value as the middle value for x in the chart. Include 3 values above and below -3 in the chart. Substitute each value of x into the quadratic equation (the parabola), to find the corresponding values for y and complete the table. When the table is complete, plot the points and draw the graph. x y There's no magic to solving quadratic equations. Quadratic equations can be solved by factoring and also by graphing. Solve for x: Here are the steps you should follow: Move all terms to the same side of the equal sign, so the equation is set equal to 0. This places the equation in standard form. Factor the (x + 3) and (x + 2) are called algebraic factors. These are factors of the expression. expression x 2 - x - 6. Set each factor equal to 0. (This process
9 is called the "zero product property". If the product of two factors equals 0, then either one or both of the factors must be 0.) 4. Solve each resulting equation. x = 3 and x = -2 are called roots. These are roots of the equation x 2 - x - 6 = 0. OR use the Quadratic Formula The Quadratic Formula uses the "a", "b", and "c" from "ax 2 + bx + c", where "a", "b", and "c" are just numbers. The Formula is derived from the process of completing the square, and is formally stated as: For ax 2 + bx + c = 0, the value of x is given by: 4.03 Use systems of linear equations or inequalities in two variables to model and solve problems. Solve using tables, graphs, and algebraic properties; justify results. Using the substitution method: Solve this system of equations (and check): 3y - 2x = 11 y + 2x = Replace the "y" value in the first equation by what "y" now equals. Grab the "y" value and plug it into the other equation. 3. Solve this new equation for "x". 4. Place this new "x" value into either of the ORIGINAL equations in order to solve for "y". Pick the easier one to work with! 5. Check: substitute x = 2 and y = 5 into BOTH ORIGINAL equations. If these answers are correct, BOTH equations will be TRUE! 3(9-2x) - 2x = 11 (27-6x) - 2x = x - 2x = x = 11-8x = -16 x = 2 y + 2x = 9 or y = 9-2x y = 9-2(2) y = 9-4 y = 5 3y - 2x = 11 3(5) - 2(2) = = = 11 (check!) y + 2x = (2) = = 9 9 = 9 (check!) 1. Solve one of the equations for either "x =" or "y =". This example solves the second equation for "y =". 3y - 2x = 11 y = 9-2x Solve this system of equations using the addition or subtraction method.
10 1. Solve this system of equations and check: a. First, be sure that the variables are "lined up" under one another. In this problem, they are already "lined up". b. Decide which variable ("x" or "y") will be easier to eliminate. In order to eliminate a variable, the numbers in front of them (the coefficients) must be the same or negatives of one another. Looks like "x" is the easier variable to eliminate in this problem since the x's already have the same coefficients. x - 2y = 14 x + 3y = 9 x - 2y = 14 x + 3y = 9 x - 2y = 14 x + 3y = 9 10 e. Plug "y = -1" into either of the ORIGINAL equations to get the value for "x". f. Check: substitute x = 12 and y = -1 into BOTH ORIGINAL equations. If these answers are correct, BOTH equations will be TRUE! x - 2y = 14 x - 2(-1) = 14 x + 2 = 14 x = 12 x - 2y = (-1) = = = 14 (check!) x + 3y = (-1) = = 9 9 = 9 (check!) c. Now, in this problem we need to subtract to eliminate the "x" variable. Subtract ALL of the sets of lined up terms. (Remember: when you subtract signed numbers, you change the signs and follow the rules for adding signed numbers.) x - 2y = 14 -x - 3y = - 9-5y = 5 d. Solve this simple equation. -5y = 5 y = -1
11 Solve graphically: 4x - 6y = 12 2x + 2y = 6 To solve a system of equations graphically, graph both equations and see where they intersect. The intersection point is the solution. First, solve each equation for "y =". 4x - 6y = 12 2x + 2y = Graph and evaluate exponential functions to solve problems Exponential functions form Notice: The variable x is an exponent. As such, the graphs of these functions are not straight lines. In a straight line, the "rate of change" is the same across the graph. In these graphs, the "rate of change" increases or decreases across the graphs. slope = y-intercept = -2 slope = -1 y-intercept = 3 Graph the lines. The slope intercept method of graphing was used in this example. The point of intersection of the two lines, (3,0), is the answer Example: when a > 0 and the b is between 0 and 1, the graph will be decreasing (decaying). For this example, each time x is increased by 1, y decreases to one half of its previous value. Such a situation is called Exponential Decay.
12 Example: when a > 0 and the b is greater than 1, the graph will be increasing (growing). For this example, each time x is increased by 1, y increases by a factor of At the Algebra level, there are two functions that can be easily used to illustrate the concepts of growth or decay in applied situations. When a quantity grows by a fixed percent at regular intervals, the pattern can be represented by the functions, Growth: Decay: a = initial amount before measuring growth/decay r = growth/decay rate (often a percent) x = number of time intervals that have passed Such a situation is called Exponential Growth. Example: A bank account balance, b, for an account starting with s dollars, earning an annual interest rate, r, and left untouched for n years can be calculated as (an exponential growth formula). Find a bank account balance to the nearest dollar, if the account starts with $100, has an annual rate of 4%, and the money left in the account for 12 years.