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Name: ______________________________________________ LESSON 2-2 Creating and Solving Equations Reteach To write an equation for a real-world problem, look for words that will help you solve the problem and translate them into parts of the equation. For example, the sum of two times a number and 8 is 18 as shown below. 2n is the same as “a number times 2.” 2n 8 18 18 is the same as “is 18.” 8 is the same as “the sum of...and 8.” Example The Lions and Tigers played a football game. The Lions scored three more than three times the number of points the Tigers scored. The total number of points scored by both teams is 31. How many points were scored by each team? points the Tigers scored t Lions scored three more than three times 3t 3 total number of points scored by both teams is 31 t (3t 3) 31 So the equation is t (3t 3) 31. You can simplify and use Properties of Equality to solve. Write an equation and solve each problem. 1. Mako runs m miles each day. De’Anthony runs twice as many miles as Mako. Altogether, they run 18 miles each day. How many miles does Mako run each day? ________________________________________________________________________________________ 2. Bella’s cell phone costs $18 per month plus $0.15 for every minute she uses the phone. Last month her bill was $29.25. For how many minutes did she use her phone last month? ________________________________________________________________________________________ 3. Eric and Charlotte collected donations for the food bank. Charlotte collected 3 times the amount of Eric’s donations minus $20. Eric and Charlotte combined their donations and spent half of the money on canned food. The rest of the money, $50, was donated to the food bank directly. How much money did each student collect? Solving for a Variable LESSON 2-3 Reteach Solving for a variable in a formula can make the formula easier to use. You can solve a formula, or literal equation, for any one of the variables. To solve a literal equation or formula, underline the variable you are solving for, and then undo what has been done to that variable. Use inverse operations in the same way you do when solving an equation or inequality. The formula for finding the circumference of a circle when you know diameter is C d. If you know the circumference, you could find the diameter by using a formula for d. Examples Solve C d for d. C d C C d or d 9 Solve F C 32 for C. 5 Since d is multiplied by , use division to undo this. Divide both sides by . What has been done to C? First undo adding 32. 9 F 32 C 32 32 5 Subtract 32 from both sides. 5 5 9 9 F 32 9 5 C 5 Multiply both sides by , the reciprocal of 9 5 9 F 32 C Simplify. 9 5 . Solve each formula for the indicated variable. 1. A = 1 bh, for b 2 _______________________ 4. P a b c, for c _______________________ 2. A lw, for l 3. R ________________________ 5. I = 1 prt, for t 2 2s 6t 5 , for s 2 ________________________ 6. G ________________________ HJ , for H K ________________________ Solve each equation for the indicated variable. 7. m n p q 360, for n _______________________ 8. t rs s, for s ________________________ 9. a c , for a b d ________________________ LESSON 2-4 Creating and Solving Inequalities Reteach An inequality, such as 2x 8, has infinitely many numbers in its solution. 1 For example, 2, 0, , and 350 are all values of x that make this 2 inequality true. Solve an inequality by UNDOING what has been done to x, using the same steps as you would use for an equality, with the exception given below. Solving inequalities has one special rule different from solving equations. Multiplying or dividing an inequality by a NEGATIVE number REVERSES the inequality sign. Try some positive and negative numbers in 3x 15. Example 3x15 Note that 6 and 8 make this true, but it is not true for 6 or 8. 3 x 15 3 3 Dividing by 3 REVERSES the inequality sign. x 5 This is still true for 6 and 8, and not true for 6 or 8. Solve each inequality. Show your work. 1. 3e 10 4 2. c 8 11 2 _______________________________________ ________________________________________ _______________________________________ ________________________________________ _______________________________________ ________________________________________ Solve each inequality. 3. 15 3 4s _______________________________________ 5. 8c 4 4(c 3) _______________________________________ 7. 8 4a 12 2a 10 4. 3 j14 4 ________________________________________ 6. 5(x 1) 3x 10 8x ________________________________________ 8. 0.6t 3 15 Graphing Functions LESSON 3-4 Reteach To check whether an equation is a function, isolate the y variable on one side of the equation and simplify. 4x 2y 8 2y 4x 8 Subtract 4x from both sides. y 2x 4 Divide each term by 2. Write the equation as a function: f(x) 2x 4. If the domain {2, 3, 4}, substitute each value into the function and simplify to find the values of the range. f(x) 2(2) 4 f(x) 2(3) 4 f(x) 2(4) 4 f(x) 4 4 f(x) 6 4 f(x) 8 4 f(x) 8 f(x) 10 f(x) 12 Use a table to find the ordered pairs. Graph the ordered pairs. x y 2 8 3 10 4 12 This is a function because each value of x has only one value of y. Any vertical line would pass through only one point. Is this equation a function? Use the domain values to find the values for the range. Graph the points and tell whether it is a function or not. 1. f(x) 3x 2 for the domain {1, 2, 3, 4} x y function? _________________ Use the vertical line test to tell if each graph shows a function. 2. 3. function? ____________ 4. function? ____________ function? _____________ LESSON 5-2 Using Intercepts Reteach Doug has $12 to spend on popcorn and peanuts. The peanuts are $4 and popcorn is $2. If he spends all his money, the equation 4x 2y 12 shows the amount of peanuts, x, and popcorn, y, he can buy. Here is the graph of 4x 2y 12. To find the y-intercept, substitute x 0. Solve for y. He can buy 6 boxes of popcorn (y) if he buys 0 peanuts. To find the x-intercept, substitute y 0. Solve for x. He can buy 3 bags of peanuts (x) if he buys 0 popcorn. 4(0) 2y 12 y6 4x 2(0) 12 x3 Find each x- and y-intercept. 1. 2. _______________________ 3. ________________________ ________________________ Find each intercept. Use these two points to graph each line. 4. 3x 9y 9 5. 4x 6y 12 6. 2x y 4 LESSON 5-3 Interpreting Rate of Change and Slope Reteach Find the rate of change, or slope, for the graph of a straight line by finding change in y . change in x Step 1: First choose any two points on the line. Step 2: Begin at one of the points. Step 3: Count vertically until you are even with the second point. This is the rise. If you go down the rise will be negative. If you go up the rise will be positive. Step 4: Count over until you are at the second point. This is the run. If you go left the run will be negative. If you go right the run will be positive. Step 5: Divide to find the slope. rise 6 slope 3 run 2 The slope of a horizontal line is zero. A horizontal line has no steepness at all. The slope of a vertical line is undefined. A vertical line is infinitely steep. Find the slope of each line. 1. 2. _______________________ 4. 3. ________________________ 5. _______________________ ________________________ 6. ________________________ ________________________ LESSON 5-4 Direct Variation Reteach When a situation can be described by an equation of the form y kx, where k is a constant, we say that y varies directly with x. The equation y kx is an equation of direct variation. When k is greater than zero (k is a positive number) the equation y kx implies that as x increases, y also increases. k is called the variation constant and can be found if one pair of (x, y) values are known. Example In the graph, y 4 when x 1. Substituting these values in the equation y kx gives 4 k x 1 so k 4. The equation of direct variation for this graph is y 4x. This equation can be used to find values for y for known values of x. Use the equation to complete the table. y 4x 8 42 12 4 3 x 2 3 5 10 y 8 12 ? ? ? 4 5 20 ? 4 10 40 Determine k, the constant of variation for each (x, y) pair, and write the equation of direct variation. 1. (3, 6) 2. (6, 24) _______________________ 3. (4, 28) ________________________ ________________________ Determine k, the constant of variation that relates x and y for each table, and write the equation of direct variation. 4. x 1 3 5 y 6 18 30 5. _______________________ x 2 4 8 y 6 12 24 6. ________________________ x 1 4 9 y 5 20 45 ________________________ For each table, y varies directly with x. Determine the missing value. 7. x 2 4 6 y 8 16 ? _______________________ 8. x 3 7 11 y 12 28 ? ________________________ 9. x 5 7 9 y 10 14 ? ________________________ LESSON 6-1 Slope-Intercept Form Reteach An equation is in slope-intercept form if it is written as: m is the slope. b is the y-intercept. y mx b. You can use the slope and y-intercept to graph a line. Write 2x 6y 12 in slope-intercept form. Then graph the line. Step 3: Graph the line. Step 1: Solve for y. Then count 1 down (because the rise is negative) and 3 right (because the run is positive) and plot another point. 2x 6y 12 2x Subtract 2x from both sides. 2x Plot (0, 2). Draw a line connecting the points. 6y 2 x 12 6y 2 x 12 6 6 1 y x2 3 Plot (0, 2). Divide both sides by 6. Count 1 down. Simplify. Step 2: Find the slope and y-intercept. slope: m 1 1 3 3 Count 3 right. y-intercept: b 2 Find each slope and y-intercept. Then graph each equation. 1. y 1 x 3 2 2. 3x y 2 3. 2x y 3 slope:______________ slope: ____________ slope: __________________ y-intercept: _________ y-intercept: ________ y-intercept: _____________ LESSON 6-2 Point-Slope Form Reteach An equation is in slope-intercept form if it is written as: m is the slope. b is the y-intercept. y mx b. A line has a slope of 4 and a y-intercept of 3. Write the equation in slope-intercept form. y mx b Substitute the given values for m and b. y 4x 3 Find values for m and b from a table or graph. change in y -values between any two points on the line. change in x -values At the y-intercept, x 0 and y b. Slope m Example: The linear function in this graph is f(x) 2x 5. Example: The linear function in this table is f(x) 3x 6. x 0 1 2 3 f(x) 6 3 0 3 Write an equation for each linear function f(x) using the given information. 1 1. slope ,y-intercept 3 4 __________________________________ 2. slope 5, y-intercept 0 __________________________________ 3. slope 7, y-intercept 2 __________________________________ 4. __________________________________ 5. x 1 0 1 2 f(x) 4 2 0 2 __________________________________ LESSON 6-4 Transforming Linear Functions Reteach For a linear function f(x) mx b, changing the value of b moves the graph up or down. Description Equation Parent function Translate up 2 Translate down 4 f(x) x g(x) x 2 h(x) x 4 y-intercept or b 0 2 4 Changing the absolute value of the slope m makes the line more or less steep. If m is positive, the line goes up from left to right. If m is negative, the line goes down from left to right. Predict the change in the graph from f(x) to g(x). Then graph both lines to check your prediction. 1. f(x) x; g(x) x 5 _______________________________________ 2. f(x) 3x 1; g(x) 3x 1 ________________________________________ LESSON 7-1 Parallel and Perpendicular Lines Reteach The chart shows how parallel and perpendicular lines differ. Lines Slopes y-intercept parallel same different perpendicular negative reciprocals of each other can be the same or different The examples show equations for parallel and perpendicular lines in slope-intercept and point-slope form. Examples – parallel slope-intercept point-slope y 3x 6 y 5 3(x 2) y 3x 4 y 6 3(x 1) Examples – perpendicular slope-intercept point-slope y 4x 2 y 3 2(x 4) y 1 x3 4 1 y 2 ( x 3) 2 Find each slope and y-intercept. 1. Are the lines whose equations are y 2x 4 and y 1 x 3 parallel or perpendicular? 2 Explain. ________________________________________________________________________________________ ________________________________________________________________________________________ 2. Are the lines whose equations are y 4 7(x 3) and y 2 7(x 5) parallel or perpendicular? Explain. ________________________________________________________________________________________ ________________________________________________________________________________________ LESSON 7-2 Using Functions to Solve One-Variable Equations Reteach Having several methods to solve one-variable equations can help. Depending on the problem, often one method is easier than the others. Algebraic Solution is often fastest and can handle decimals and fractions most easily. f(x) 19 8x and g(x) 17 10x Solve using algebra. Set f(x) g(x) and solve for x. 19 8x 17 10x 36 18x 2x Graphic Solution is often easier with graphing calculators and can give a visual understanding of a problem. This method also works well with decimals. Solve using graphs. 1 f ( x ) x 2 and g( x ) 3x 3 2 For f(x) slope 1 and y-intercept 2 2 For g(x) slope 3 and y-intercept 3 Plot f(x) and g(x) and find the intersection. Tables work well with graphing calculators and are usually easiest for integers. Solve with a table. f(x) 3 2x and g(x) 12 3x x f (x) 3 2x g(x) 12 3x 0 3 12 1 1 9 2 1 6 3 3 3 4 5 0 1. In the algebraic example above, if x 2, what does f(x) equal? _____ What does g(x) equal? ______ 2. In the graphic example above, at what point do the lines for the equations intersect? ______ 3. What does 3, 3 in the table indicate? ____________________________________________________ LESSON 7-3 Linear Inequalities in Two Variables Reteach To graph a linear inequality: Step 1: Solve the inequality for y. Step 2: Graph the boundary line. If or use a solid line. If or use a dashed line. Step 3: Determine which side to shade. Graph the solutions of 2x y 4. Step 1: Solve for y. 2x y 4 2 x 2 x y 2 x 4 Step 2: Graph the boundary line. Use a solid line for . Step 3: Determine which side to shade. Substitute (0, 0) into 2x y 4. 2x y 4 ? 2(0) 0 4 ? 0 4. The statement is true. Shade the side that contains the point (0, 0). Summary for Graphing Linear Inequalities in Two Variables • The boundary line is solid for and and dashed for and . When the inequality is written with y alone on the left, then: • shade below and to the left of the line for and , • shade above and to the right of the line for and . Graph the solutions of each linear inequality. In the first one, the boundary line is already drawn. 1. y 2x 9 2. y x 3 3. x y 2 0 LESSON 8-1 Scatter Plots and Trend Lines Reteach Correlation is one way to describe the relationship between two sets of data. Positive Correlation Data: As one set increases, the other set increases. Graph: The graph goes up from left to right. Negative Correlation Data: As one set increases, the other set decreases. Graph: The graph goes down from left to right. No Correlation Data: There is no relationship between the sets. Graph: The graph has no pattern. Example Correlation Correlation Coefficient (estimated ) 1st graph above strong positive 1 2nd graph above strong negative 1 3rd graph above no correlation 0 4th graph beside weak positive 0.5 5th graph beside weak negative 0.5 Estimate the correlation coefficient for each scatter plot as 1, 0.5, 0, 0.5, or 1. 1. 2. _______________________ 3. ________________________ ________________________ LESSON 9-1 Solving Linear Systems by Graphing Reteach The solution to a system of linear equations can be found by graphing. Write both equations so that they are in slope-intercept form and draw their lines on a coordinate graph. The point of intersection is the solution. If the lines have the same slope but different y-intercepts they won’t intersect and there is no solution. If the graphs are the same line then there are an infinite number of solutions. Example Solve the system. y 2x 2 y 2x 6 4 x 2y 4 6 x 3y 18 Rewrite each equation in slope-intercept form. Graph the lines and look for the point of intersection. The lines intersect at (2, 2). The solution to the system is (2, 2). Solve each linear system of equations by graphing. x y 1 1. 4 x 2y 16 3 x y 10 2. 2 x 4 y 0 ____________ 2 x y 4 3. x y 2 ____________ 6 x 3 y 12 4. 2 x 2y 10 ____________ ____________ LESSON 9-2 Solving Linear Systems by Substitution Reteach You can use substitution to solve a system of equations if one of the equations is already solved for a variable. y x 2 Solve 3 x y 10 Step 1: Choose the equation to use as the substitute. Use the first equation y x 2 because it is already solved for a variable. Step 2: Solve by substitution. x2 3x y 10 3x (x 2) 10 4x 2 10 2 2 4x 8 4x 8 4 4 x2 Step 3: Now substitute x 2 back into one of the original equations to find the value of y. y x 2 y22 y 4 The solution is (2, 4). Check: Substitute (2, 4) into both equations. y x2 3xy10 Substitute x 2 for y. Combine like terms. 4 22 ? 4 4 ? 3(2)4 10 ? 64 10 ? 10 10 ? You may need to solve one of the equations for a variable before solving with substitution. Solve each system by substitution. y x 2 1. y 2x 5 _______________________________________ x y 3 3. 2 x y 12 _______________________________________ x y 10 2. x 2y 3 ________________________________________ y x 8 4. 5 x 2y 9 ________________________________________ LESSON 9-3 Solving Linear Systems by Adding or Subtracting Reteach To use the elimination method to solve a system of linear equations: 1. Add or subtract the equations to eliminate one variable. 2. Solve the resulting equation for the other variable. 3. Substitute the value for the known variable into one of the original equations. 4. Solve for the other variable. 5. Check the values in both equations. 3 x 2y 7 The y-terms have Use the elimination opposite coefficients, 5 x 2y 1 method when the so add. coefficients of one of 3 x 2y 7 the variables are the Add the equations. 5 x 2y 1 same or opposite. 8x 8 Solve for x. x1 Substitute x 1 into 3x 2y 7 and solve for y: 3x 2y 7 3(1) 2y 7 2y 4 y2 The solution to the system is the ordered pair (1, 2). Check using both equations: 3x 2y 7 5x 2y 1 3(1) 2(2) 7 5(1) 2(2) 1 ? 7 7 ? 1 1 Solve each system by adding or subtracting. 2 x y 5 1. 3 x y 1 __________________________________ x y 12 3. 2x y 6 __________________________________ 3 x 2y 10 2. 3 x 2y 14 ________________________________ 2x y 1 4. 2 x 3 y 5 ________________________________ LESSON 9-4 Solving Linear Systems by Multiplying First Reteach To solve a system by elimination, you may first need to multiply one of the equations to make the coefficients match. 2 x 5 y 9 x 3 y 10 Multiply bottom equation by 2. 2x 5y 9 2( x 3 y ) 2(10) 2x 5y 9 2 x 6 y 20 0 11y 11 Solve for y: 11y 11 11 11 y 1 Substitute 1 for y in x 3y 10. x 3(1) 10 x 3 10 3 3 x7 The solution to the system is the ordered pair (7, 1). You may need to multiply both of the equations to make the coefficients match. 5 x 3 y 2 4 x 2y 10 Multiply the top by 2 and the bottom by 3. -2(5 x + 3y = 2) 3(4 x + 2y = 10) 10 x ( 6 y ) 4 12 x 6 y 30 2 x 0 26 x 13 The solution to this system is the ordered pair (13, 21). After you multiply, add or subtract the two equations. Solve for the variable that is left. Substitute to find the value of the other variable. Check in both equations. Solve each system by multiplying first. Check your answer. 2 x 3 y 5 1. x 2y 1 __________________________________ 2 x 5 y 22 3. 10 x 3 y 22 3 x y 2 2. 8 x 2y 4 ________________________________ 4 x 2y 14 4. 7 x 3 y 8 LESSON 10-2 Graphing Systems of Linear Inequalities Reteach You can graph a system of linear inequalities by combining the graphs of the inequalities. Graph of y 2x 3 Graph of y x 6 All solutions are in this double shaded area. Graph of the system y 2 x 3 y x 6 Two ordered pairs that are solutions: (3, 4) and (5, 2) Solve each system of linear inequalities by graphing. Check your answer by testing an ordered pair from each region of your graph. y x 3 1. y x 6 _______________________ y x 2. y 2x 1 ________________________ y 2x 2 3. y 2x 3 ________________________ LESSON 10-3 Modeling with Linear Systems Reteach Mrs. Hathaway bought a total of 12 items made up of some sticky notes and some pens. The sticky notes cost $4 each and the pens cost $2 each. She spent a total of $40 on all items. How many pens and how many sticky notes did she buy? Organize the information. Number of Items Cost Sticky Notes n 4n Pens p 2p Total 12 40 Write two equations. Use the information in each row of the chart. Number of Items n p 12 n p 12 Cost 4n 2p 40 4n 2p 40 Write each equation in slope-intercept form. n p 12 n p 12 4n 2p 40 4n 2p 40 1 n p 10 2 Set the equations equal to each other and solve. 1 p 12 - p 10 2 1 12 p 10 2 1 2 p 2 4p She bought 4 pens. n p 12 n 4 12 n8 She bought 8 sticky notes. Solve. 1. Tia has 25 china figures in her collection. The horse figures cost $2 each, and the cat figures cost $1 each. She paid $39 for all the figures in the collection. How many horses and how many cats does she have? Equations: ___________________________________ Solution: ____________________________________ 2. Mr. Wallace has 32 models of antique cars. The Hupmobile models cost $5 each, and the Duesenberg models cost $18 each. He paid a total of $264 for all the models. How many Hupmobile models and how many Duesenberg models does he have? Equations: ___________________________________ Solution: ____________________________________ LESSON 12-1 Understanding Geometric Sequences Reteach It is important to understand the difference between arithmetic and geometric sequences. Arithmetic sequences are based on adding a common difference, d. Geometric sequences are based on multiplying a common ratio, r. • If the first term of an arithmetic sequence, a1, is 2 and the common difference is 3, the arithmetic sequence is: 2, 5, 8, 11, … • If the first term of an arithmetic sequence, a1, is 72 and the common difference is 3, the arithmetic sequence is: 72, 69, 66, 63, … • If the first term of a geometric sequence, a1, is 2 and the common ratio is 3, the geometric sequence is: 2, 6, 18, 54, … • If the first term of a geometric sequence, a1, is 72 and the 1 8 common ratio is , the geometric sequence is: 72, 24, 8, , … 3 3 Complete each table. 1. An arithmetic sequence has a1 4 and d 3: an a1 a2 a3 a4 a5 Value 2. A geometric sequence has a1 4 and r 3: an a1 a2 a3 a4 a5 Value 3. An arithmetic sequence has a1 96 and d 4: an a1 a2 a3 a4 a5 Value 4. A geometric sequence has a1 96 and r an Value a1 a2 a3 a4 a5 1 : 4 LESSON 13-2 Modeling Exponential Growth and Decay Reteach In the exponential growth and decay formulas, y final amount, a original amount, r rate of growth or decay, and t time. Exponential growth: y a (1 r)t Exponential decay: y a (1 r)t The population of a city is increasing at a rate of 4% each year. In 2000, there were 236,000 people in the city. Write an exponential growth function to model this situation. Then find the population in 2009. The population of a city is decreasing at a rate of 6% each year. In 2000, there were 35,000 people in the city. Write an exponential decay function to model this situation. Then find the population in 2012. Step 1: Identify the variables. Step 1: Identify the variables. a 236,000——r 0.04 Step 2: Substitute for a and r. a 35,000——r 0.06 Step 2: Substitute for a and r. y a (1 r)t y a (1 r)t y 236,000 (1 0.04)t y 35,000 (1 0.06)t The exponential growth function is y 236,000 (1.04)t. Growth;the growth factor is greater than 1. Step 3: Substitute for t. y 236,000 (1.04) The exponential decay function is y 35,000 (0.94)t. Decay; the growth factor is less than 1 and greater than 0. Step 3: Substitute for t. 9 335,902 The 2009 population was about 335,902 people. y 35,000 (0.94)12 16,657 The 2009 population was about 16,657 people. Write an exponential growth function to model each situation. Then find the value of the function after the given amount of time. 1. Annual sales at a company are $372,000 and increasing at a rate of 5% per year; 8 years y _________(1 ______) ___ 2. The population of a town is 4200 and increasing at a rate of 3% per year; 7 years y _________(1 ______) ___ Write an exponential decay function to model the situation. Then find the value of the function after the given amount of time. 3. Monthly car sales for a certain type of car are $350,000 and are decreasing at a rate of 3% per month; 6 months _______________________________________ y _________(1 ______) ___ LESSON 14-1 Understanding Polynomial Expressions Reteach Polynomials have special names based on the number of terms. POLYNOMIALS No. of Terms Name 1 2 Monomial Binomial 3 4 or more Trinomial Polynomial The degree of a monomial is the sum of the exponents in the monomial. The degree of a polynomial is the degree of the term with the greatest degree. Examples Find the degree of 8x 2 y 3 . Find the degree of 4ab 9a 3 . 8x 2 y 3 The exponents are 2 and 3. 4ab 9a 3 2 3 The degree of the binomial is 3. The degree of the monomial is 2 3 5. Identify each polynomial. Write the degree of each expression. 1. 7m3n5 _______________________ 2. 4x2y3 y4 7 3. x5 x5y ________________________ ________________________ You can simplify polynomials by combining like terms. The following are like terms: 4y and 7y 8x2 and 2x2 7m5 and m5 same variables raised to same power The following are not like terms: Examples Add 3x2 4x 5x2 6x. 3x2 5x2 4x 6x 8x 10x 2 3x2 and 3x same variable, different exponent 47 and 7y one with variable, one constant 8m and m5 same variable but different power Identify and rearrange like terms so they are together. Combine like terms. Simplify each expression. 4. 2y2 3y 7y y2 _______________________ 5. 8m4 3m 4m4 ________________________ 6. 12x5 10x4 8x4 ________________________ LESSON 14-2 Adding Polynomial Expressions Reteach You can add polynomials by combining like terms. These are examples of like terms: 4y and 7y 8x2 and 2x2 m5 and 7m5 These are like terms because they have the same variables and same exponent. These are not like terms: 3x2 and 3x 4y and 7 same variable but different exponent one with a variable, one is a constant 8m and 8n different variables Add (5y2 7y 2) (4y2 y 8). (5y2 7y 2 ) (4y2 y 8 ) Identify like terms. (5y2 4y2) ( 7y y ) ( 2 8 ) Rearrange terms so that like terms are together. 9y2 8y 10 Combine like terms. Add (5y2 7y 2) (4y2 y 8). (5y2 7y 2 ) (4y2 y 8 ) Identify like terms. (5y2 4y2) ( 7y y ) ( 2 8 ) Rearrange terms so that like terms are together. 9y2 8y 10 Combine like terms. Add. 1. (6x2 3x) (2x2 6x) ____________________________________________________ 2. (m2 10m 5) (8m 2) ____________________________________________________ 3. (6x3 5x) (4x3 x2 2x 9) ____________________________________________________ 4. (2y5 6y3 1) (y5 8y4 2y3 1) ____________________________________________________ Subtracting Polynomial Expressions LESSON 14-3 Reteach To subtract polynomials, you must remember to add the opposites. Find the opposite of (5m3 m 4). (5m3 m 4) (5m3 m 4) Write the opposite of the polynomial. 5m3 m 4 Write the opposite of each term in the polynomial. Subtract (4x3 x2 7) (2x3). (4x3 x2 7) (2x3) Rewrite subtraction as addition of the opposite. (4x3 x2 7) (2x3) Identify like terms. (4x3 2x3) x2 7 Rearrange terms so that like terms are together. 2x3 x2 7 Combine like terms. Subtract (6y4 3y2 7) (2y4 y2 5). (6y4 3y2 7) (2y4 y2 5) Rewrite subtraction as addition of the opposite. (6y4 3y2 7 ) (2y4 y2 5) Identify like terms. (6y4 2y4 ) (3y2 y2) (7 5) Rearrange terms so that like terms are together. 4y4 4y2 12 Combine like terms. Subtract. 1. (9x 3 5x) (3x) _______________________________________ 2. (6t 4 3) (2t 4 2) _______________________________________ 3. (2x3 4x 2) (4x3 6) _______________________________________ 4. (t 3 2t) (t 2 2t 6) _______________________________________ 5. (4c5 8c2 2c 2) (c3 2c 5) LESSON 15-1 Multiplying Polynomial Expressions by Monomials Reteach To multiply monomial expressions, multiply the constants, and then multiply variables with the same base. Example Multiply (3a2b) (4ab3). (3a2b) (4ab3) (3 4) (a2 a) (b b3) Rearrange so that the constants and the variables with the same bases are together. 12a3b4 Multiply. To multiply a polynomial expression by a monomial, distribute the monomial to each term in the polynomial. Example Multiply 2x(x2 3x 7). 2x(x2 3x 7) (2x)x2 (2x)3x (2x)7 Distribute. 2x 6x 14x Multiply. 3 2 Multiply. 1. (5x 2 y 3 ) (2xy) 2. (2xyz) (4x 2 yz) ___________________________ 3. (3x) (x 2 y 3 ) ___________________________ ___________________________ Fill in the blanks below. Then complete the multiplication. 4. 4(x 5) x 5. 3x(x 8) 5 ___________________________ x 6. 2x(x 2 6x 3) 8 ___________________________ x2 6x ___________________________ Multiply. 7. 5(x 9) ___________________________ 10. 3(5 x 2 2) 8. 4x(x 2 8) ___________________________ 11. (5a 3 b) (2ab) 9. 3x 2 (2x 2 5x 4) ___________________________ 12. 5y(y 2 7y 2) 3 LESSON 15-2 Multiplying Polynomial Expressions Reteach Use the Distributive Property to multiply binomial and polynomial expressions. Examples Multiply (x 3) (x 7). (x 3) (x 7) x(x 7) 3(x 7) Distribute. (x)x (x)7 (3)x (3)7 Distribute again. x 7x 3x 21 Multiply. x2 4x 21 Combine like terms. 2 Multiply (x 5) (x2 3x 4). (x 5) (x2 3x 4) x(x2 3x 4) 5 (x2 3x 4) Distribute. (x)x2 (x)3x (x)4 (5)x2 (5)3x (5)4 x3 3x2 4x 5x2 15x 20 x3 8x2 19x 20 Distribute again. Multiply. Combine like terms. Fill in the blanks below. Then finish multiplying. 1. (x 4) (x 5) __ (x 5) __ (x 5) _______________________ 2. (x 2) (x 8) 3. (x 3) (x 6) __ (x 8) __ (x 8) _______________________ __ (x 6) __ (x 6) ________________________ Multiply. 4. (x 2) (x 3) _______________________ 5. (x 7) (x 7) 6. (x 2) (x 1) _______________________ ________________________ Fill in the blanks below. Then finish multiplying. 7. (x 3) (2x2 4x 8) __ (2x2 4x 8) __ (2x2 4x 8) _______________________________________ 8. (x 2) (6x2 4x 5) __ (6x2 4x 5) __ (6x2 4x 5) ________________________________________ LESSON 16-1 Characteristics of Quadratic Functions Reteach To analyze a function of the form y ax2, where a is not 0, you can take notes about the equation. Look. Think. Look. variable squared quadratic function U shaped graph y 3x2 y 3x 2 Think. number times variables squared Point (0, 0) is the highest or lowest point on the U. Remember: The line x 0 divides the U into left and right parts that are identical. The U is symmetric with x 0 as the line of symmetry. Now look at the number 3, the coefficient of x2. Look. 3 is greater than 0. Look. 3 is greater than 1. vertical stretch Think. The U opens upward. (0, 0) is the lowest point on the U. Think. The U is narrower than the U that represents the graph of y x2. In this example, the coefficient of x2 is positive. Use similar thinking when the coefficient is negative. The U will flip over the x-axis of the one shown here. Answer each question about y 3x2. 1. Does the graph open up or down? __________________________ 2. Is (0, 0) the highest (maximum) or lowest (minimum) point on the graph? __________________ Answer each question about y 0.1x2. 3. Is the graph wider or more narrow than the graph of y x2? __________________________ 4. What is an equation of the axis of symmetry of the graph? __________________________ Answer each question about y 0.1x2. 5. Does the graph open up or down? __________________________ LESSON 16-2 Transforming Quadratic Functions Reteach A parabola has the equation f(x) a(x h)2 k. Identify: a. a, a stretch if a 1 or compression if 0 a 1 b. h, the horizontal translation c. k, the vertical translation The vertex is (h, k) and the parabola opens up if a 0 and opens down if a 0. In parabola f(x) 4 (x 3)2 5, the stretch is 4, the horizontal translation is 3 to the right, and the vertical translation is up 5. The vertex is (3, 5), and the parabola opens up. Complete 1–4 for parabola f(x) 2(x 7)2 9. 1. Stretch or shrink? _______________ 2. Open up or down? ______________ 3. Horizontal translation? ____________ 4. Vertical translation? ______________ Complete 5–8 for parabola f ( x ) 1 ( x 4)2 8. 2 5. Stretch or shrink? _______________ 6. Open up or down? ______________ 7. Horizontal translation? ____________ 8. Vertical translation? ______________ For a parabola that opens up, the vertex represents the minimum point. For a parabola that opens down, the vertex represents the maximum point. The following graph is a translation of y x2. 9. The vertex is (_____, ______). 10. Is the vertex a maximum or a minimum? _______________________________________ 11. The quadratic equation for the graph is _______________________________________ . LESSON 16-3 Interpreting Vertex Form and Standard Form Reteach The equation of a parabola can be written in either vertex or standard form. Vertex Form Standard Form y a( x h) k y ax 2 bx c 2 Find the vertex of the quadratic equation y 2( x 1)2 4. The vertex is the lowest point of a parabola when the parabola opens up. When the equation is written in vertex form, the coordinates of the vertex are (h, k). y a( x h)2 k h 1, k 4 y 2( x 1)2 4 The vertex is the ordered pair (1, 4) and the line of symmetry is x 1. To change the equation from vertex form to standard form, do the following: y 2( x 1)2 4 y 2( x 1)( x 1) 4 Show the factors. y 2( x 2 2 x 1) 4 Expand. y 2x 4 x 2 4 Multiply. y 2x 2 4 x 2 Simplify. 2 ( b ) ( b ) , f When written in standard form, the coordinates of the vertex are . 2a 2a y ax 2 bx c (b ) ( 4) 1, and y 2(1)2 4(1) 2 4 a 2, b 4 , so x 2 2a 2(2) y 2x 4 x 2 The vertex is the ordered pair (1, 4) and the line of symmetry is x 1. Find the vertex and axis of symmetry of each quadratic equation. 1. y ( x 5)2 2 _______________________ 4. y 2x 2 12x 24 _______________________ 2. y x 2 6x 8 ________________________ 5. y 8( x 9)2 5 ________________________ 3. y 2( x 4)2 1 ________________________ 6. y 4x 2 16x 1 ________________________ LESSON 17-3 Applying the Zero Product Property to Solve Equations Reteach Quadratic equations in factored form can be solved by using the Zero Product Property. If the product of two quantities equals zero, at least one of the quantities must equal zero. If (x) (y) 0, then If (x 3) (x 2) 0, then x 0 or x 3 0 or y0 x20 You can use the Zero Product Property to solve any quadratic equation written in factored form, such as (a b)(a b) 0. Examples Find the zeros of (x 5)(x 1) 0. x 5 0 or x 1 0 Set each factor equal to 0. x 5 or x 1 Solve each equation for x. Solve (x 7)(x 2) 0. x 7 0 or x 2 0 Set each factor equal to 0. x 7 or x 2 Solve each equation for x. Use the Zero Product Property to solve each equation by filling in the blanks below. Then find the solutions. Check your answer. 1. (x 6) (x 3) 0 2. (x 8) (x 5) 0 x ________ or x ________ x ________ or x _________ _______________________________________ ________________________________________ Use the Zero Product Property to solve each equation. 3. (y 7)(y 3) 0 _______________________ 6. (t 9)(t 3) 0 _______________________ 9. (z 6)(z 4) 0 _______________________ 4. 0 (x 6)(x 3) ________________________ 7. (n 5)(n 3) 0 ________________________ 10. 0 (x 4)(x 2) ________________________ 5. (x 4)(x 3) 0 ________________________ 8. (a 10)(a 3) 0 ________________________ 11. 0 (g 3)(g 3) ________________________ LESSON 18-1 Solving Equations by Factoring x2 bx c Reteach To find the factors for a trinomial in the form x2 bx c, answer these 2 questions. 1. What numbers have a product equal to c? 2. What numbers have a sum equal to b? Find numbers for which the answer to both is yes. Factor x2 5x 6. What numbers have a product equal to c, 6? 1 and 6 1 and 6 2 and 3 2 and 3 What numbers have a sum equal to b, 5? 1 and 6 1 and 6 2 and 3 2 and 3 The factors of x2 5x 6 are (x 2) and (x 3). Solve the trinomial by setting it equal to 0. Factor and use the Zero Product Property to solve. Example Solve x2 5x 6 0. x2 5x 6 0 (x 2)(x 3) 0 Factor x2 5x 6. x 2 0 or x 3 0 Set each factor equal to 0. x 2 or x 3 Solve each equation for x. Complete the factoring. 1. x2 x 2 What numbers have a product equal to c, _____? What numbers have a sum equal to b, _____? Factors: Factor. 2. x2 4x 4 _______________________ 3. x2 4x 3 ________________________ 4. x2 3x 10 ________________________ Solve. 5. x2 12x 35 0 _______________________ 6. x2 9x 18 0 ________________________ 7. x2 x 20 0 ________________________ LESSON 18-2 Solving Equations by Factoring ax2 bx c Reteach When a factorable quadratic expression is written in standard form, ax2 bx c 0, you can use the Zero Product Property to solve the equation. To solve a quadratic equation, move all terms to the left side of the equation to get 0 on the right side. Example Solve 3x2 4x 8 6x by factoring. 3x2 4x 8 6x 3x2 4x 6x 8 0 Subtract 8 and add 6x to both sides. 3x2 10x 8 0 Simplify. (3x 2)(x 4) 0 Factor the quadratic expression. 3x 2 0 or x 4 0 Set each factor equal to 0. x 2 or x 4 3 Solve each equation. Sometimes you can factor out a common factor. Example Solve 3x2 12x 12 0 by factoring. 3x2 12x 12 0 3 (x2 4x 4) 0 3 (x 2) (x 2) 0 Factor out a common factor. Factor the quadratic expression. Set each factor equal to 0. 30 Solve each equation. or x20 x2 Use the Zero Product Property to find the solutions. 1. (2x 3)(x 9) 0 2. (5x 1)(x 2) 0 _______________________ ________________________ 3. 2(3x 1)(3x 1) 0 ________________________ Solve the equations by factoring. 4. 2x2 5x 3 2x _______________________ 7. 18x2 24x 8 0 _______________________ 5. 6x2 3x 2 4x ________________________ 8. 10x2 25x 15 0 ________________________ 6. 7x2 8x 10x 11 ________________________ 9. 6x2 96 ________________________ LESSON 19-1 Solving Equations by Taking Square Roots Reteach These equations have something in common. They have the same roots. This comes from adding 2x2 5 13 2x2 18 5 to each side of 2x 2 5 13. But 2x2 18 is easier to read and solve. 2x 2 18 x2 9 This comes from dividing each side of 2x2 18 by 2. Now x2 9 is very easy to solve. x2 9 x 9 x 3 This comes from taking the square roots of 9. Here is another example. Given 3x 7 13 2 Simpler 3x2 6 Simpler still x2 2 Done x 2 Identify the reason for each step in the solution. 1. 4x2 1 15 ® 4x2 16 ® x2 4 ® x 2 ________________________________________________________________________________________ 2. 2x2 3 9 ® 2x2 6 ® x2 3 ® x 3 ________________________________________________________________________________________ Solve using square roots. 3. x2 9 _______________________ 6. x2 400 0 _______________________ 9. (x 6)2 144 _______________________ 12. (x 3)2 121 _______________________ 4. x2 16 ________________________ 7. x2 49 0 ________________________ 10. (x 5)2 81 ________________________ 13. (x 1)2 36 ________________________ 5. x2 1 ________________________ 8. x2 64 0 ________________________ 11. (x 4)2 100 ________________________ 14. (x 2)2 4 ________________________ LESSON 19-2 Solving Equations by Completing the Square Reteach To solve a quadratic equation, complete the square. Here is an example. Solve x2 10x 24. Leave room for adding a number to each side of the equation. ¯ ¯ x2 10x _______ 24 _______ What number? 2 Answer The square of one half of 10, the coefficient of x 1 2 2 10 5 25 Now fill in the blanks with this number. ¯ ¯ 25 25 x2 10x _______ 24 _______ x2 10x 25 1 Now you have a simpler equation to work with. (x 5)2 1 x2 10x 25 is a perfect square trinomial. It equals (x 5)2. ( x 5) 1 2 x 5 1 Finish. ¯ x 5 1 x 4 Remember . There are two square roots. ¯ x 5 1 x 6 Two equations to solve. The solutions are 4 and 6. Solve by completing the square. 1. x2 6x 7 _______________________ 4. x2 4x 32 _______________________ 2. x2 8x 12 ________________________ 5. x2 14x 24 ________________________ 3. x2 2x 63 ________________________ 6. x2 6x 9 ________________________ 7. The product of two consecutive positive integers is 56. What are they? ________________________________________________________________________________________ Using the Quadratic Formula to Solve Equations LESSON 19-3 Reteach Write the quadratic equation in standard form ax 2 bx c 0. Use the quadratic formula. x b b2 4ac 2a Solve 2x2 5x 12 0 using the quadratic formula. Step 3: Simplify. 2x2 5x 12 0 x ( 5) ( 5)2 4(2)( 12) 2(2) Step 1: Identify a, b, and c. x 5 25 ( 96) 4 x 5 121 4 x 5 11 4 a2 b 5 c 12 Step 2: Substitute into the quadratic formula. x ( 5) ( 5)2 4(2)( 12) 2(2) Step 4: Write two equations and solve. 5 11 5 11 x or x 4 4 3 x4 or x 2 Solve using the quadratic formula by filling in the blanks below. 1. x2 2x 35 0 2. 3x2 7x 2 0 a ____; b ____; c ____ x 2 4 2 Simplify: 3. x2 x 20 0 2 4 2 ________________________________________ 4. 2x2 9x 5 0 a ____; b ____; c ____ x x Simplify: _______________________________________ a ____; b ____; c ____ 2 4 2 Simplify: _______________________________________ a ____; b ____; c ____ x 4 2 2 Simplify: ________________________________________ LESSON 19-4 Choosing a Method for Solving Quadratic Equations Reteach The method you choose to solve a quadratic equation depends on the form of the equation. If there are two terms and both have the same variable try factoring: 3x 2 6x 0; 3x( x 2) 0, x 0 of x 2 If there are two terms and one term is a constant try taking square roots: x 2 25 0; x 2 25, x 2 25, x 5 of x 5 If there is a binomial squared equal to a constant try taking square roots: ( x 5)2 9; ( x 5)2 9, x 5 3, x 2 or x 8 If there are three terms try factoring into two binomials: x 2 7x 6 0; ( x 1)( x 6) 0, x 1 of x 6 If the equation does not seem to factor try writing the equation in standard form ax 2 bx c 0 and use the quadratic formula 2 x 2 5 x 6 0; b b2 4ac : 2a ( 5) ( 5)2 4(2)( 6) 5 25 48 5 73 2(2) 4 4 Solve each quadratic equation by any means. Identify the method and explain why you chose it. 1. 4 x 2 12x 0 2. 5 x 2 30 x 0 3. x 2 49 0 _______________________ ________________________ ________________________ _______________________ ________________________ ________________________ _______________________ ________________________ ________________________ 4. 4 x 2 64 5. (2x 5)2 49 6. ( x 5)2 81 _______________________ ________________________ ________________________ _______________________ ________________________ ________________________ _______________________ ________________________ ________________________ 7. x 2 14x 49 0 8. 2x 2 14 x 20 9. 3x 2 4 x 20 _______________________ ________________________ ________________________ _______________________ ________________________ ________________________ _______________________ ________________________ ________________________