Mathematics Placement The ACT COMPASS math test is a self-adaptive test, which potentially tests students within four different levels of math including pre-algebra, algebra, college algebra, and trigonometry. As you answer questions correctly, you will move into more difficult levels of math. Similarly, if you answer questions incorrectly, the computerized test will begin to ask questions from a lower level of math. Multiple-choice items in each of the five mathematics placement areas test the following: basic skills performing a sequence of basic operations application applying sequences of basic operations to novel settings or in complex ways analysis demonstrating conceptual understanding of principles and relationships in mathematical operations Students are permitted to use approved calculators when completing the COMPASS mathematics placement or diagnostic tests. An online calculator is available for those students who wish to access it via Microsoft Windows. Because this is an adaptive test, you may change your answer while you are still on a problem, but once you go on to another problem, you may not go back to a question.
Math Diagnostics Test Depending on the results of you Math Placement Test, you may be required to take the Math Diagnostic Test. This math test evaluates students' skill levels in 15 subareas in Pre-Algebra and Algebra: Pre-Algebra Integers Decimals Exponents, square roots, and scientific notation Fractions Percentages Averages (means, medians, and modes) Algebra Substituting values Setting up equations Factoring polynomials Exponents and radicals Basic operations/polynomials Linear equations/one variable Linear equations/two variables Rational expressions
Mathematics Placement Sample Questions (PreAlgebra) Following are 16 sample Algebra Placement Test Questions taken from the ACT COMPASS website. First you will see the question, then the following slide will have the answer. If you need some additional refreshers, the remainder of the slides cover the content from the Algebra section.
Algebra Placement Test Sample Questions
Algebra Placement Test Sample Questions This is an example of Substituting Values into Algebraic Expressions. The correct answer is A (-4). You would need to be familiar with Evaluating Expressions to correctly answer this question (see Evaluating Expressions slides for additional information on this topic). To solve: Step 1: Substitute value into expression Step 2: Solve ( 3) 3 2 1 1 9 1 2 8 2 4
Algebra Placement Test Sample Questions 2. Doctors use the term maximum heart rate (MHR) when referring to the quantity found by starting with 220 beats per minute and subtracting 1 beat per minute for each year of a person s age. Doctors recommend exercising 3 or 4 times each week for at least20 minutes with your heart rate increased from its resting heart rate (RHR) to its training heart rate (THR), where THR = RHR +.65(MHR RHR) Which of the following is closest to the THR of a 43-year-old person whose RHR is 54 beats per minute? A. 197 B. 169 C. 162 D. 134 E. 80
Algebra Placement Test Sample Questions 2. Doctors use the term maximum heart rate (MHR) when referring to the quantity found by starting with 220 beats per minute and subtracting 1 beat per minute for each year of a person s age. Doctors recommend exercising 3 or 4 times each week for at least 20 minutes with your heart rate increased from its resting heart rate (RHR) to its training heart rate (THR), where THR = RHR +.65(MHR RHR) Which of the following is closest to the THR of a 43-year-old person whose RHR is 54 beats per minute? A. 197 B. 169 C. 162 D. 134 E. 80 This is an example of Substituting Values into Algebraic Expressions. The correct answer is D (134). You would need to be familiar with Evaluating Expressions to correctly answer this question (see Evaluating Expressions slides for additional information on this topic). To solve: THR = 54 +.65[(220-1(43)) 54] = 54+.65(123) = 54+79.95 = 133.95
Algebra Placement Test Sample Questions 3. When getting into shape by exercising, the subject s maximum recommended number of heartbeats per minute (h) can be determined by subtracting the subject s age (a) from 220 and then taking 75% of that value. This relation is expressed by which of the following formulas? A. h =.75(220 a) B. h =.75(220) a C. h = 220.75a D..75h = 220 a E. 220 =.75(h a)
Algebra Placement Test Sample Questions 3. When getting into shape by exercising, the subject s maximum recommended number of heartbeats per minute (h) can be determined by subtracting the subject s age (a) from 220 and then taking 75% of that value. This relation is expressed by which of the following formulas? A. h =.75(220 a) B. h =.75(220) a C. h = 220.75a D..75h = 220 a E. 220 =.75(h a) This is an example of Setting Up Equations for Given Situations. The correct answer is A (h =.75(220 a)). You would need to be familiar with Writing Equations to correctly answer this question (see Writing Equations slides for additional information on this topic). To solve: Remember to change the percent to a decimal: 75% =.75. Other key words are subtract from (-) and of (x).
Algebra Placement Test Sample Questions 4. An airplane flew for 8 hours at an airspeed of x miles per hour (mph), and for 7 more hours at 325 mph. If the average airspeed for the entire flight was 350 mph, which of the following equations could be used to find x? A. x + 325 = 2(350) B. x + 7(325) = 15(350) C. 8x 7(325) = 350 D. 8x + 7(325) = 2(350) E 8x + 7(325) = 15(350)
Algebra Placement Test Sample Questions 4. An airplane flew for 8 hours at an airspeed of x miles per hour (mph), and for 7 more hours at 325 mph. If the average airspeed for the entire flight (which was 15 hours total) was 350 mph, which of the following equations could be used to find x? A. x + 325 = 2(350) B. x + 7(325) = 15(350) C. 8x 7(325) = 350 D. 8x + 7(325) = 2(350) E. 8x + 7(325) = 15(350) This is an example of Setting Up Equations for Given Situations. The correct answer is E (8x + 7(325) = 15(350) ). You would need to be familiar with Writing Equations to correctly answer this question (see Writing Equations slides for additional information on this topic). To solve: See key words in the question.
Algebra Placement Test Sample Questions 5. Which of the following is equivalent to 3a + 4b ( 6a 3b)? A. 16ab B. 3a + b C. 3a + 7b D. 9a + b E. 9a + 7b
Algebra Placement Test Sample Questions 5. Which of the following is equivalent to 3a + 4b ( 6a 3b)? A. 16ab B. 3a + b C. 3a + 7b D. 9a + b E. 9a + 7b This is an example of Basic Operations with Polynomials. The correct answer is E (9a + 7b). You would need to be familiar with Combining Like Terms to correctly answer this question (see Basic Operations with Polynomials slides for additional information on this topic). To solve: Distribute the negative through the parenthesis and then combine like terms: 3a + 4b ( 6a 3b) = 3a + 4b + 6a + 3b = (3a + 6a) + (4b + 3b) = 9a + 7b
Algebra Placement Test Sample Questions
Algebra Placement Test Sample Questions This is an example of Basic Operations with Polynomials. The correct answer is A (3a 2 b ab 2 + 3a 2 b 2 ). You would need to be familiar with Adding Polynomials and Combining Like Terms to correctly answer this question (see Basic Operations with Polynomials slides for additional information on this topic). To solve: Sum means add the two polynomials, then combine like terms: (3a 2 b + 2a 2 b 2 )+ (-ab 2 +a 2 b 2 ) = 3a 2 b - ab 2 + (2a 2 b 2 + a 2 b 2 ) = 3a 2 b ab 2 + 3a 2 b 2
Algebra Placement Test Sample Questions 7. Which of the following is a factor of the polynomial x 2 x 20? A. x 5 B. x 4 C. x + 2 D. x + 5 E. x + 10
Algebra Placement Test Sample Questions 7. Which of the following is a factor of the polynomial x 2 x 20? A. x 5 B. x 4 C. x + 2 D. x + 5 E. x + 10 This is an example of Factoring Polynomials. The correct answer is A (x-5). You would need to be familiar with Factoring Polynomials to correctly answer this question (see Factoring Polynomials slides for additional information on this topic). To solve: Factor the quadratic equation: x 2 x 20 = (x 5)(x + 4). Check using FOIL.
Algebra Placement Test Sample Questions 8. Which of the following is a factor of x 2 5x 6? A. (x + 2) B. (x 6) C. (x 3) D. (x 2) E. (x 1)
Algebra Placement Test Sample Questions 8. Which of the following is a factor of x 2 5x 6? A. (x + 2) B. (x 6) C. (x 3) D. (x 2) E. (x 1) This is an example of Factoring Polynomials. The correct answer is B (x - 6). You would need to be familiar with Factoring Polynomials to correctly answer this question (see Factoring Polynomials slides for additional information on this topic). To solve: Factor the quadratic equation: x 2 5x 6 = (x 6)(x + 1). Check using FOIL.
Algebra Placement Test Sample Questions
Algebra Placement Test Sample Questions This is an example of Linear Equations in One Variable. The correct answer is E (-½). You would need to be familiar with Solving Linear Equations to correctly answer this question (see Solving Linear Equations slides for additional information on this topic). To solve: Solve for x: 2( x x x 5 5) 11 2 11 2 5 11 11 2 10 2 1 2
Algebra Placement Test Sample Questions
Algebra Placement Test Sample Questions This is an example of Linear Equations in One Variable. The correct answer is C (-1). You would need to be familiar with Solving Linear Equations to correctly answer this question (see Solving Linear Equations slides for additional information on this topic). To solve: Solve for x: 4 5 8 10 x 5 10 5 10 3 10 3 10 x 15 10 15 10 x x 1 1 2 10 10 3 2 1
Algebra Placement Test Sample Questions
Algebra Placement Test Sample Questions This is an example of Exponents. The correct answer is B. You would need to be familiar with Rules for Exponents to correctly answer this question (see Rules for Exponents slides for additional information on this topic). To solve: Apply rules for exponents: 3 5 16r tz 3 2 4rt z (16 4)( r 3 r)( t t 3 )( z 5 z 2 ) 4r 2 t 2 z 3 4r t 2 2 z 3
Algebra Placement Test Sample Questions
Algebra Placement Test Sample Questions x 3 x y x 3 x y 3 x y 3 x y 9x y This is an example of Exponents. The correct answer is C. You would need to be familiar with Rules for Exponents to correctly answer this question (see Rules for Exponents slides for additional information on this topic). To solve: First, you want to get rid of the radical signs in the denominator. To do that you would multiply the numerator and the denominator by the expression. 3x xy
Algebra Placement Test Sample Questions
Algebra Placement Test Sample Questions This is an example of Rational Expressions. The correct answer is B (x + 8). You would need to be familiar with Rational Expressions to correctly answer this question (see Rational Expressions slides for additional information on this topic). To solve: x 12x 32 x 4 ( x 4)( x 8) x 4 x 8 2 Step 1: Factor the numerator. Step 2: Cancel common terms.
Algebra Placement Test Sample Questions
Algebra Placement Test Sample Questions This is an example of Rational Expressions. The correct answer is E (-x 3). You would need to be familiar with Rational Expressions to correctly answer this question (see Rational Expressions slides for additional information on this topic). To solve: 9 x 2 x 3 ( x ( x ( x x 3)( x x 3 3) 2 x 9) 3 3) 3 Step 1: Factor the numerator. Step 2: Cancel common terms. Step 3: Distribute the neg. back through
Algebra Placement Test Sample Questions
Algebra Placement Test Sample Questions This is an example of Linear Equations in Two Variables. The correct answer is D ( 2/3). You would need to be familiar with Linear Equations to correctly answer this question (see Linear Equations slides for additional information on this topic). Use the slope intercept form: y=mx+b where m=slope and b=y-intercept 2x+3y+6 = 0 (standard form) Solve for y: 3y = -2x 6 y = (-2/3)x 2 Therefore, the slope is 2/3
Algebra Placement Test Sample Questions
Algebra Placement Test Sample Questions This is an example of Linear Equations in Two Variables. The correct answer is E ((8, 1)). You would need to be familiar with Linear Equations to correctly answer this question (see Linear Equations slides for additional information on this topic). Graph the point and the line. A perpendicular (cross at 90 degree angle) bisector (cuts the line segment into two equal length segments). Also, for future reference, parallel lines have the same slope, and perpendicular lines have negative reciprocal slope.
Algebra Review The following slides review the concepts found on the COMPASS Algebra Placement Test.
Substituting Values 1. Evaluating Variable Expressions
Evaluating Variable Expressions Section 2.1 Variable - a letter that is used to stand for a quantity that is unknown or may change Variable Expression - an expression that contains variables Terms - addends of a variable expression Ex: the expression 4x 2 + 3x - 2y + 5 has four terms that include three variable terms (4x 2, 3x, 2y) and one constant term (5) Coefficient is the number part of the variable term. In the term 4x 2 the 4 is the coefficient.
Note x = 1x -x = -1x xy = x times y
Evaluating the Variable Expression Evaluating the Variable Expression - replace the variable with numbers and then simplify EX: xy - 3 where x = 5, y = 4 replace 5(4) - 3 = 20-3 = 17 Don t forget Order of Operations - PEMDAS Parenthesis, Exponents, Mult/Div, Add/Subtract
Practice: Evaluate Ex1: ab 2 a when a = 2 and b = -2 Ex2: a 2 b 2 when a = 3 and b = -4 a - b Ex3: a 2 + b 2 when a = 5 and b = -3 a + b
Practice: Answers Ex1: ab 2 a when a = 2 and b = -2 2( 2) 2 2 2(4) 2 8 2 6 Ex2: a 2 b 2 when a = 3 and b = -4 a - b 2 3 3 ( ( 4) 4) 2 9 3 16 4 7 7 1 Ex3: a 2 + b 2 when a = 5 and b = -3 a + b 2 5 5 ( ( 3) 3) 2 25 5 ( 9 3) 34 2 17
You Try: Evaluate Ex1: 3a + 2b when a = 2 and b = 3 Ex2: 2a (c + a) 2 when a = 2 and c = -4 Ex3: a 2 5a 6 when a = -2 For Ex 4-7 use: a = -2, b = 4, c = -1, d = 3 Ex4: (b + c) 2 + (a + d) 2 Ex5: b 2a bc 2 -d Ex6: b 2 - a ad + 3c Ex7: 1/3(d 2 ) 3/8(b 2 )
You Try: Answers Ex1: 3a + 2b when a = 2 and b = 3 3(2) 2(3) 12 Ex2: 2a (c + a) 2 when a = 2 and c = -4 2(2) ( 4 2) Ex3: a 2 5a 6 when a = -2 ( 2) 2 5( 2 2) 6 6 2(2) 6 2(2) 4 4 4 0 For Ex 4-7 use: a = -2, b = 4, c = -1, d = 3 Ex4: (b + c) 2 + (a + d) 2 2 2 (4 1) ( 2 3) 4 ( 2) 2 10 6 8 3 2 1 2 9 1 10 Ex5: b 2a bc 2 -d 4 4( 2( 1) 2 2) 3 4 4(1) 4 3 8 4 3 8 1 8 Ex6: b 2 - a ad + 3c 4 2 2(3) ( 2) 3( 1) 16 6 ( ( 2) 3) 18 3 6 Ex7: 1/3(d 2 ) 3/8(b 2 ) 1 2 2 (3 3 ) 3 (4 8 ) 1 3 (9) 3 8 (16) 3 6 3
Setting Up Equations 1. Phrases that mean Addition 2. Phrases that mean Subtraction 3. Phrases that mean Multiplication 4. Phrases that mean Division 5. Phrases that mean Exponents 6. Translating Words into Expressions
Phrases that mean Addition Section 2.6 added to 6 added to y y + 6 more than 8 more than x x + 8 the sum of the sum of x and z x + z increased by t increased by 9 t + 9 the total of the total of 5 and y 5 + y plus b plus 17 b + 17 Other words that mean addition: older, greater, additional, longer
Phrases that mean Subtraction Section 2.6 minus x minus 2 x - 2 less than 7 less than t t - 7 less 7 less t 7 - t subtracted from 5 subtracted from d d - 5 decreased by m decreased by 3 m - 3 the difference between the difference between y and 4 y - 4 Additional words that mean subtraction: younger, fewer, shorter Note: The book uses less than and difference frequently. Make sure you know the difference between them!
Phrases that mean Multiplication Section 2.6 times 10 times t 10t of one half of x (1/2)x the product of the product of y and z yz multiplied by y multiplied by 11 11y twice twice n 2n
Phrases that mean Division Section 2.6 divided by x divided by 12 x/12 the quotient of the quotient of y and z y/z the ratio of the ratio of t to 9 t/9 Additional words that mean division: divided among, split between
Phrases that mean Exponents (Powers) Section 2.6 the square of the square of x x 2 the cube of the cube of a a 3
Translating Phrases into Expressions Section 2.6 Note: Remember that addition and multiplication are commutative, so order doesn t matter. However, subtraction and division are not commutative, so order DOES matter.
S e c t i o n 2. 6 Word Phrase Variable (can be any letter) 4 less than a number a a 4 A number decreased by 9 b b 9 50 minus a number z 50 z A number more than 5 x 5 + x A number increased by 7 y y + 7 the sum of 6 and a number c 6 + c 3 plus a number k 3 + k The difference of a number and 8 n n 8 7 times a number m 7m The product of 12 and a number t 12t 3/4 of a number k ¾k A number times 8 d d x 8 A number divided by 2 y y/2 6 divided by a number s 6/s Expression
Translate into a variable expression The total of five times b and c 5b + c Identify words that indicate mathematical operations. Use the operations to write the variable expression.
Translate into a variable expression The quotient of eight less than n and fourteen n 8 14 Identify words that indicate mathematical operations. Use the operations to write the variable expression.
Translate into a variable expression Thirteen more than the sum of seven and the square of x Identify words that indicate mathematical operations. (7 + x 2 ) + 13 Use the operations to write the variable expression.
Your turn! Translate into a variable expression eighteen less than the cube of x y decreased by the sum of z and nine the difference between the square of q and the sum of r and t
Your turn! (answers) Translate into a variable expression eighteen less than the cube of x x 3-18 y decreased by the sum of z and nine y (z + 9) the difference between the square of q and the sum of r and t q 2 (r + t)
More Examples Translate a number multiplied by the total of six and the cube of the number into a variable expression. Break into parts The unknown number: n The cube of a number: n 3 The total of six and the cube of the number: 6 + n 3 Put it together n(6 + n 3 )
More Examples Translate the quotient of twice a number and the difference between the number and twenty into a variable expression. Break into parts The unknown number: n Twice the number: 2n The difference between the number and twenty: n - 20 Put it together 2n n - 20
Your turn! Translate a number added to the product of five and the square of the number into a variable expression. Translate the product of three and the sum of seven and twice a number into a variable expression.
Your turn! (answers) Translate a number added to the product of five and the square of the number into a variable expression. 5n 2 + n Translate the product of three and the sum of seven and twice a number into a variable expression. 3(7 + 2n)
More Examples After translating a verbal expression into a variable expression, simplify the variable expression by using the Properties of Real Numbers. Translate and simplify the total of four times an unknown number and twice the difference between the number and eight. Break into parts The unknown number: n Four times the number: 4n Twice the difference between the number and eight: 2(n 8) Put it together: 4n + 2(n - 8) Simplify: 4n + 2n 16 = 6n - 16
Your turn! Translate and simplify a number minus the difference between twice the number and seventeen. n (2n 17) n 2n + 17 -n + 17
Factoring Polynomials 1. Greatest Common Factor and Factoring by Grouping 2. Factoring Trinomials: x 2 + bx + c 3. More on Factoring Trinomials: ax 2 + bx + c 4. Factoring Special Products: Difference of Two Squares and Perfect Square Trinomials
Greatest Common Factor Factoring is the process of breaking a product into smaller parts called factors. Recall that the product is the answer to a multiplication problem and the factors are the terms being multiplied. Examples: For 2 5 = 10, the factors are 2 and 5. For (x - 2)(x + 4) = 0, the factors are x - 2 and x + 4. Common Factor: If all terms in the equation have a common factor, you can factor it out and have a simpler equation to solve. Try to find the greatest common factor (GCF) to factor out.
Greatest Common Factor Find the GCF for each of the following: Ex 1: 16, 40, 56 16 = 2x2x2x2 = 2 4 40 = 2x2x2x5 = 2 3 5 56 = 2x2x2x7 = 2 3 7 so the GCF = 2 3 = 8 Ex 2: 45x 2 y 2 z 2 and 75xy 2 z 3 45x 2 y 2 z 2 = 3 2 5 x 2 y 2 z 2 75xy 2 z 3 = 3 5 2 x y 2 z 3 so the GCF = 3 5 x y 2 z 2 = 15xy 2 z 2
Greatest Common Factor Example: Factor the expression 8x - 12. The common factors are 1, 2, and 4 because they all divide evenly into both 8x and 12. The GCF is 4 and can be factored out. Divide 8x by 4, which is 2x. Divide 12 by 4, which is 3. The factored equation is 4(2x - 3). TIP: Factoring out a common factor is the opposite of applying the distributive property.
Greatest Common Factor Factor each of the polynomials by finding the greatest common monomial factor: Ex 1: 14x + 21 Ex 2: Ex 3: Ex 4: -3x 2 + 6x -6ax + 9ay 16x 4 y 14x 2 y Ex 5: 34x 4 y 6 51x 3 y 5 + 17x 5 y 4
Greatest Common Factor Factor each of the polynomials by finding the greatest common monomial factor: Ex 1: 14x + 21 = 7(2x + 3) Ex 2: -3x 2 + 6x = -3x(x 2) Ex 3: -6ax + 9ay = -3a(2x - 3y) Ex 4: 16x 4 y 14x 2 y = 2x 2 y(8x 2 7) Ex 5: 34x 4 y 6 51x 3 y 5 + 17x 5 y 4 = 17x 3 y 4 (2xy 2 3y + x 2 ) or 17x 3 y 4 (x 2 3y + 2xy 2 )
Factor by Grouping Consider the expression y(x + 2) + (x + 2) as the sum of two terms, y(x + 2) and 1(x +2). Each of these terms have a common binomial factor of (x + 2). Factoring out this common binomial factor by using the distributive property gives y(x + 2) + (x + 2) = (y + 1)(x + 2)
Factor by Grouping Section 7.1 Factor each expression by factoring out the common binomial factor: Ex 1: 7y 2 (y + 3) + 2(y + 3) Ex 2: 9a(x + 1) (x + 1) Ex 3: 2x 2 (x + 5) + (x + 5) Ex 4: 10y(2y + 3) 7(2y + 3) Ex 5: a(x 2) b(x 2)
Factor by Grouping Factor each expression by factoring out the common binomial factor: Ex 1: 7y 2 (y + 3) + 2(y + 3) = (7y 2 + 3)(y + 3) Ex 2: 9a(x + 1) (x + 1) = (9a 1)(x + 1) Ex 3: 2x 2 (x + 5) + (x + 5) = (2x 2 + 1)(x + 5) Ex 4: 10y(2y + 3) 7(2y + 3) = (10y 7)(2y + 3) Ex 5: a(x 2) b(x 2) = (a b)(x 2)
Factor by Grouping Factor each polynomial by grouping. If the polynomial cannot be factored, write not factorable: Ex 1: bx + b + cx + c Ex 2: x 3 + 3x 2 + 6x + 18 Ex 3: 24y 3yz + 2xz 16x Ex 4: 10xy + x y 1 Ex 5: x 2 5 + x 2 y + 5y
Factor by Grouping Factor each polynomial by grouping. If the polynomial cannot be factored, write not factorable: Ex 1: bx + b + cx + c = b(x + 1) + c(x + 1) = (b + c)(x + 1) Ex 2: x 3 + 3x 2 + 6x + 18 = x 2 (x + 3) + 6(x + 3) = (x 2 + 6)(x + 3) Ex 3: 24y 3yz + 2xz 16x = 3yz + 24y + 2xz 16x = = -3y(z 8) + 2x(z 8) = (2x 3y)(z 8) Ex 4: 10xy + x y 1 = 10x(y + 1) (y + 1) = (10x 1)(y + 1) Ex 5: x 2 5 + x 2 y + 5y = (x 2 5) + y(x 2 + 5) not factorable
Factor Trinomials: x 2 + bx + c Quadratic equations are written in this form y = ax 2 + bx + c where ax 2 is the quadratic term (to the 2 nd power) bx is the linear term (to the 1 st power) c is the constant term Notice that the variable terms are listed in decreasing order.
Factor Trinomials: x 2 + bx + c Example: Multiply (x - 1)(x + 4) FOIL (First, Outside, Inside, Last) 1x 2 + 4x - x - 4 x 2 + 3x - 4 Combine like terms. Result is a trinomial Notice that the result of multiplying binomials is a trinomial. Factoring a trinomial requires finding the two binomials that were originally multiplied together to give the trinomial. The process of factoring involves working backwards from the terms in the trinomial.
Factor Trinomials: x 2 + bx + c To factor a trinomial with leading coefficient 1, find the two factors of the constant term whose sum is the coefficient of the middle term. x 2 + (a + b)x + ab Sum of constants a and b Product of constants a and b
Factor Trinomials: x 2 + bx + c Ex: Factor: x 2 + 7x + 12 To factor x 2 + 7x + 12, find the two factors of 12 whose sum is 7. x 2 + (3 + 4)x + 3(4) Sum of constants a and b Product of constants a and b Therefore, x 2 + 7x + 12 = (x + 3)(x + 4)
Factor Trinomials: x 2 + bx + c Ex: Factor: x 2-8x + 16 To factor x 2-8x + 16, find the two factors of 16 whose sum is -8. x 2 + (-4 + -4)x + -4(-4) Sum of constants a and b Product of constants a and b Therefore, x 2-8x + 16 = (x - 4)(x - 4)
Factor Trinomials: x 2 + bx + c Ex: Factor: x 2 + 4x 21 To factor x 2 + 4x 21, find the two factors of -21 whose sum is +4. x 2 + (7 + -3)x + 7(-3) Sum of constants a and b Product of constants a and b Therefore, x 2 + 4x 21 = (x +7)(x 3)
Factor Trinomials: x 2 + bx + c We have seen from the previous 3 examples that a rule can be formed: To factor x 2 + bx + c, if possible, find an integer pair of factors of c whose sum is b. 1. If c is positive, then both factors must have the same sign. a. Both will be positive if b is positive b. Both will be negative if b is negative 2. If c is negative, then one factor must be positive and the other negative.
Factor Trinomials: x 2 + bx + c Factor the trinomials, if the trinomial cannot be factored, write not factorable: Ex 1: x 2 x 12 Ex 2: y 2 3y + 2 Ex 3: y 2 + 12y + 35 Ex 4: a 2 + a + 2
Factor Trinomials: x 2 + bx + c Section 7.2 Factor the trinomials, if the trinomial cannot be factored, write not factorable: Ex 1: x 2 x 12 = (x + 3)(x - 4) Ex 2: y 2 3y + 2 = (y 1)(y 2) Ex 3: y 2 + 12y + 35 = (y + 5)(y + 7) Ex 4: a 2 + a + 2 is not factorable
Factoring Trinomials: ax 2 + bx + c A polynomial is completely factored if none of its factors can be factored. To factor ax 2 + bx + c, where a 1, look for a common monomial factor. If there is a common monomial factor, factor out this monomial factor and factor the remaining trinomial, if possible. We will discuss other options in section 7.3.
Factoring Trinomials: ax 2 + bx + c Completely factor the polynomials: Ex 1: 5x 2 5x 60 Ex 2: 7y 3 70y 2 + 168y Ex 3: 3x 2 18x + 30 Ex 4: a 4 + 30a 3 + 81a 2 Ex 5: 20a 2 + 40ab + 20b 2
Factoring Trinomials: ax 2 + bx + c Completely factor the polynomials: Ex 1: 5x 2 5x 60 = 5(x 2 x 12) = 5(x 4)(x + 3) Ex 2: 7y 3 70y 2 + 168y = 7y(y 2 10y + 24) = 7y(y 4)(y 6) Ex 3: 3x 2 18x + 30 = 3(x 2 6x + 10) Ex 4: a 4 + 30a 3 + 81a 2 = a 2 (a 2 + 30a + 81) = a 2 (a + 3)(a + 27) Ex 5: 20a 2 + 40ab + 20b 2 = 20(a 2 +2ab + b 2 ) = 20(a + b)(a + b)
Factoring Trinomials: ax 2 + bx + c Factoring using the ac-method: 1. Multiply a by c 2. Find two integers whose product is ac and whose sum b 3. Rewrite the middle term bx using the two numbers found in Step 2 as coefficients 4. Factor by grouping the first two terms and the last two terms 5. Factor out the common binomial factor to find two binomial factors of the trinomial ax 2 + bx + c
Factoring Trinomials: ax 2 + bx + c Factoring 4x 2-5x - 6 using the ac-method: 1. Multiply a by c: 4(-6) = -24 2. Find two integers whose product is ac and whose sum b: 3(-8) and 3 + -8 3. Rewrite the middle term bx using the two numbers found in Step 2 as coefficients: 4x 2-8x + 3x - 6 4. Factor by grouping the first two terms and the last two terms: 4x(x - 2) + 3(x 2) 5. Factor out the common binomial factor to find two binomial factors of the trinomial ax 2 + bx + c: (4x + 3)(x 2)
Factoring Trinomials: ax 2 + bx + c Factoring using the Trial by Error Method: 1. The key is using the FOIL method (First, Outside, Inside, Last)
Factor: Factoring Trinomials: ax 2 + bx + c Ex 1: 2x 2 3x 2 Ex 2: 12y 2 15y + 3 Ex 3: -5y 2 + 40y - 60
Factoring Trinomials: ax 2 + bx + c Factor: Ex 1: 2x 2 3x 2 = 2x 2 4x + 1x 2 = 2x(x 2) + (x 2) = (2x + 1)(x 2) Ex 2: 12y 2 15y + 3 = 3(4y 2 5y + 1) = 3[4y 2 4y 1y + 1] = 3[4y(y 1) (y 1)] = 3(4y 1)(y 1) Ex 3: -5y 2 + 40y 60 = -5(y 2-8y + 12) = -5(y - 2)(y - 6)
Difference of Squares When factoring the difference of two squares: x 2 a 2 = (x + a)(x a) Ex 1: x 2 25 = (x + 5)(x 5) Ex 2: 36 y 2 = (6 + y)(6 y) Ex 3: a 6 100 = (a 3 + 10)(a 3 10)
Perfect Square Trinomials When factoring perfect square trinomials: x 2 + 2ax + a 2 = (x + a) 2 Ex 1: x 2 + 20x + 100 = (x + 10)(x + 10) = (x + 10) 2 Ex 2: 49 + 14y + y 2 = (7 + y) 2 Ex 3: 4a 6 + 20a 3 + 25 = (2a 3 + 5) 2
Perfect Square Trinomials Section 7.4 When factoring perfect square trinomials: x 2 2ax + a 2 = (x - a) 2 Ex 1: x 2 20x + 100 = (x 10)(x 10) = (x 10) 2 Ex 2: 49 14y + y 2 = (7 y) 2 Ex 3: 4a 6 20a 3 + 25 = (2a 3 5) 2
Factoring Trinomials: ax 2 + bx + c Procedures to follow when factoring Polynomials: 1. Factor out any common monomial factor. 2. Check the number of terms: a. Two terms: 1) Difference of two squares? factorable 2) Sum of two squares? not factorable b. Three terms: 1) Perfect square trinomial? 2) Use trial-and-error method? 3) Use the ac-method? c. Four terms: 1) Group terms with a common factor 3. Check the possibility of factoring any of the factors
Factoring Trinomials: ax 2 + bx + c We can add to a binominal the constant term that makes it a perfect-square trinomial ; this process is called Completing the Square. In this case, (½ of our coefficient of our linear term) 2 = constant term. To complete the squares we want: y 2 + 20y + = ( ) 2 to be in the form x 2 + 2ax + a 2 = (x + a) 2 so ½ (b) = ½ (20) = 10. Therefore, we would have y 2 +20y + 100 = (y + 10) 2
Factoring Trinomials: ax 2 + bx + c Complete the squares for the following: Ex 1: x 2 6x + = ( ) 2 Ex 2: x 2 4x + = ( ) 2 Ex 3: x 2 + + 16 = ( ) 2 Ex 4: x 2 - + 81 = ( ) 2
Factoring Trinomials: ax 2 + bx + c Complete the squares for the following: Ex 1: x 2 6x + 9 = (x - 3) 2 Ex 2: x 2 4x + 4 = (x - 2) 2 Ex 3: x 2 + 8x + 16 = (x + 4) 2 Ex 4: x 2-18x + 81 = (x - 9) 2
Exponents and Radicals 1. Properties of Exponents 2. Simplifying Radicals 3. Properties of Radicals
Exponents Exponent: In the expression x n (read: x to the nth power), where x is the base and n is the exponent. x n the base (x) is multiplied by itself exponent (n) number of times Ex: 5 2 = 5 x 5 = 25 2 3 = 2 x 2 x 2 = 8
Be careful (-3) 2 does not equal -3 2 it means (-3)(-3) = 9-3 2 means (3 x 3) = -9
Exponents When dealing with particular operations, there are rules for simplifying or evaluating an exponential expression. Exponent of 1: a 1 = a Exponent of 0: a 0 = 1, when a 0 Negative Exponents: a -n = 1 a n = 1/a n, where a 0 1 to a Power: 1 n = 1 To add and subtract exponential expressions, like bases with like exponents are required. Example: LIKE bases with LIKE exponents can be simplified. a 2 + a 2 = 2a 2 5b 2-3b 2 = 2b 2
Properties of Exponents Rule: a m a n = a m+n To multiply two or more exponential expressions that have the same base, add the exponents. Example: 5 3 x 5 2 = 5 (3+2) = 5 5 Rule: a m a n = a m /a n = a m-n where a 0 Rule: (a m ) n = a m(n) Rule: (ab) m = a m b m To divide two or more exponential expressions that have the same base, subtract the exponents. Example: 5 3 5 2 = 5 (3-2) = 5 1 = 5 To raise a power to a power, multiply the exponents. Example: (a 6 ) 5 = a 6x5 = a 30 To raise a product to a power, raise each factor to that power. Example: (ab) 5 = a 5 b 5 Rule: m m a a m b b where b 0 To raise a fraction to a power, raise both the numerator and the denominator to that power. Example:
Simplifying Radicals An expression with radicals is in simplest from if the following are true. 1. No radicands (expressions under radical signs) have perfect square factors other than 1. Example: 50 25 2 5 2 2. No radicands contain fractions. Example: 3 4 3 4 3 2 3. No radicals appear in the denominator of a fraction. Example: 1 1 2 Note: To simplify this expression, multiply the numerator and 2 2 2 denominator by 2. This is 1 2 2 algebraically justified because it is equivalent to multiplying the original 2 2 2 fraction by 1.
Properties of Radicals Rule: ab a b The square root of a product equals the product of the square roots of the factors. Example: 50 25 2 5 2 Rule: a b a b The square root of a quotient equals the quotient of the square roots of the numerator and denominator. Example: 3 4 3 4 3 2
Examples: 1. 8 3 8 0 = 8 3 1 or 8 3+0 = 8 3 = 8 8 8 = 512 2. x 3 x 2 = x 3+2 = x 5 3. 3y y 4 = 3y 1+4 = 3y 5 4. x -3 = 1/ x 3 5. -10x 5 /2x 2 = -5x 5-2 = -5x 3 6. 7. 5 0 36a b c 5 9a b x x 2 3 y 4a 5 ( 5) xy b 0 ( 3) c 2 4a 10 b 3 c 2 4a c 10 2 b 3 y y y y
Basic Operations of Polynomials 1. Addition with Polynomials 2. Subtraction with Polynomials 3. Multiplication with Polynomials
Polynomials Vocabulary A monomial is an algebraic expression that contains only one term (i.e. -2x 3 and 4a 5 ). A binomial is an algebraic expression that contains exactly two unlike terms (i.e. 3x + 5 and a 2 3). A trinomial is an algebraic expression that contains exactly three unlike terms (a 2 + 6a 7 and x 3 8x 2 + 12x). A polynomial is an algebraic expression that contains one or more unlike terms. The degree of the polynomial is the largest of the degrees of its terms. The coefficient of the term of the largest degree is called the leading coefficient. To simplify polynomials means to add or subtract any like terms and when possible write it in order of descending exponents.
Adding Polynomials Simplifying Polynomials The sum of two or more polynomials is found by combining like terms. Remember that like terms are constants or terms that contain the same variable raised to the same powers. Ex: Find the sum: (10x 2 + 17x - 91) + (12x 2 + 29x - 95) To easily see the like terms, rewrite the polynomials vertically so that the like terms line up. Then add the coefficients of the variables. 10x 2 + 17x - 91 +12x 2 + 29x - 95 22x 2 + 46x - 186 (This is written in simplest form because it contains no like terms.)
Adding Polynomials Simplifying Polynomials Ex: Find the sum: (7x 3 + 5x 2 + x 6) + (-3x 2 +11) + (-3x 3 x 2 5x +2) 7x 3 + 5x 2 + x 6-3x 2 +11-3x 3 x 2 5x +2 4x 3 + x 2 4x +7
Subtracting Polynomials Simplifying Polynomials Remember that a negative sign written in front of a parentheses, means to change the sign of every term within the parenthesis. Ex: -(12x 2 + 29x - 95) = -12x 2-29x + 95 The difference of two polynomials is found by changing the sign of each term of the second polynomial and then combining like terms. Ex: Find the difference: (10x 2 + 17x - 91) - (12x 2 + 29x - 95) To easily see the like terms, rewrite the polynomials vertically so that the like terms line up. Then add the coefficients of the variables. 10x 2 + 17x - 91-12x 2-29x + 95-2x 2-12x + 4 (This is written in simplest form because it contains no like terms.)
Adding & Subtracting Polynomials Simplifying Polynomials Ex 1: Simplify and write in descending order: 4x + 2(x 3) (3x + 7) 4x + 2x 6 3x 7 Distribute 3x 13 Combine like terms Ex 2: Simplify and write in descending order: -[6x 2 3(4 + 2x) + 9] (x 2 + 5) Distribute inner most parenthesis -[6x 2 12 6x + 9] x 2 5-6x 2 + 12 + 6x - 9 x 2 5 Combine like terms -7x 2 + 6x 2 Distribute negative through bracket
Multiplying Polynomials Multiply Polynomials Alg.: D To multiply a polynomial by a monomial, distribute each term to the others by multiplication. Example: Multiply 2z(z 2-5 + 8). 2z(z 2-5 + 8) Follow the colors. 2z 3-10z + 16z 2z 3 + 6z Combine like terms
Multiplying Polynomials Multiply Polynomials Multiplying Binomials: Example: (x - 1)(x + 4) FOIL (first, outside, inside, last) 1x 2 + 4x - x - 4 x 2 + 3x - 4 Combine like terms Notice that the result of multiplying binomials is a trinomial.
Multiplying Polynomials Multiply Polynomials Alg.: D Multiply and simplify: 8x 2 + 3x 2-2x + 7 56x 2 +21x 14 Multiply each term of the first trinomial by +7-16x 3 6x 2 + 4x Multiply each term of the first trinomial by -2x -16x 3 + 50x 2 + 25x 14 Combine like terms
Multiplying Polynomials Multiply Polynomials Alg.: D Simplify: (2t + 3)(2t + 3) (t 2)(t 2) 4t 2 + 6t + 6t + 9 [t 2 2t 2t + 4] 4t 2 + 12t + 9 [t 2 4t + 4] 4t 2 + 12t + 9 t 2 + 4t 4 3t 2 + 16t + 5 Use FOIL Combine like terms Distribute negative Combine like terms
Linear Equations in One Variable 1. Solving Linear Equations x+b=c and ax=c 2. Solving Linear Equations ax+b=c 3. More Linear Equations ax+b=dx+c
Linear Equations Equation expresses the equality of two mathematical expressions Where s the solution? Solution is a number when substituted for the variable results in a true equation
Linear Equations If a, b, and c are constants and a 0 then a linear equation in x is an equation that can be written in the form: ax + b = c. (Note: A linear equation in x is also called a first-degree equation in x because the variable x can be written with the exponent 1. That is x = x 1.
Solving Linear Equations To solve an equation means to find a solution of the equation. The simplest equation to solve is an equation of the form variable = constant, because the constant is the solution. (ex: x = 5) The objective of solving linear (or first degree) equations is to get the variable by itself (with a coefficient of 1) on one side of the equation and any constants on the other side.
Addition Property of Equations The same number or variable term can be added to each side of an equation without changing the solution of the equation. Think of an equation as a balance scale. If the weights added to each side of the equation are not the same, the sides no longer balance. Whatever you do to the one side of the equation, you have to do to the other side to keep it balanced!
Solve equations of the form x + a = b x + 5 = 9-5 -5 x = 4 The goal is to rewrite the equation in the form variable = constant. Add the opposite of the constant term 5 to each side. Simplify
Solve equations of the form x + a = b Try: x + 6 = 10 4 = 10 + x x 3 = 12-8 + x = 5 x 4 = 9 3 = x - 6
Solve equations of the form x + a = b Section 3.1 Answers: x + 6 = 10 x = 4 4 = 10 + x -6 = x x 3 = 12-8 + x = 5 x = 15 x = 13 x 4 = 9 3 = x 6 x = 13 9 = x
Multiplication Property of Equations Each side of an equation can be multiplied by the same nonzero number without changing the solution of the equation. The multiplication property is used to remove a coefficient from a variable term in an equation by multiplying each side of the equation by the reciprocal of the coefficient. Still think of an equation as a balance scale. What ever you do to one side of the equation, you must do to the other side to remain equal.
Solve equations of the form ax = b 2x = 6 ½ (2x) = ½ (6) x = 3 Multiply each side of the equation by the reciprocal of the coefficient. Simplify to variable = constant. Try: 4x = 8-3x = -9 3x = -27 y/8 = 2 1/3(x) = 5 x/3 = 5
Answers: Solve equations of the form ax = b Section 3.1 4x = 8-3x = -9 x = 2 x = 3 3x = -27 y/8 = 2 x = -9 y = 16 1/3(x) = 5 x/3 = 5 x = 15 x = 15
Solving Application Problems Section 3.1 1. The perimeter of a square is 4 times the length of a side (P = 4s). Find the length of a side if the perimeter of a square is 64 ½ meters. 4s = 64 ½ 4 4 s = 16.125
Solving Application Problems 2. A suit cost the store $675 and the owner said that he wants to make a profit of $180. What price should he mark on the suit? Profit = Revenue Cost 180 = R 675 +675 +675 855 = R Therefore, the price he would mark on the suit would be $855.
Solving Application Problems 3. How long will a truck driver take to travel 350 miles if he averages 50 mph? Distance = Rate x Time (d = rt) 350 = 50t 50 50 7 = t Therefore, it would take 7 hours
Solving Linear Equations: ax + b = c Section 3.2 The goal is to rewrite the equation in the form variable = constant, because the constant is the solution. This requires applying both the Addition and Multiplication Properties of Equations.
Solve equations of the form ax + b = c Section 3.2 3x 7 = -5 +7 +7 3x = 2 3 3 x = 2/3 Undo (add opposite) any addition or subtraction first. Then undo (multiply reciprocal) any multiplication or division. Simplify
Try these 4x + 3 = 11 7 x = 9 8y + 4 = 12 5x 6 = 9-35 = -6x + 1 3x 7 = -5 2 = 5x + 12-3m 21 = 0 5 = 9 2x (2/5)x 3 = -7 6a + 3 + 2a = 11 2x 6x + 1 = 9
Try these (answers) 4x + 3 = 11 x = 2 5x 6 = 9 x = 3 2 = 5x + 12 x = -2 (2/5)x 3 = -7 x = -10 7 x = 9 x = -2-35 = -6x + 1 x = 6-3m 21 = 0 m = -7 6a + 3 + 2a = 11 a = 1 8y + 4 = 12 y = 1 3x 7 = -5 x = 2/3 5 = 9 2x x = 2 2x 6x + 1 = 9 x = -2
Applications 1. One hundred eighty-four feet of fencing is needed to enclose a rectangular shaped garden plot. If the plot is 23 feet wide, what is the length? P = 2l + 2w 184 = 2l + 2(23) 184 = 2l + 46-46 -46 138 = 2l 2 2 69 = l Therefore, the length of the fence is 69 feet
Solve Linear Equations: ax + b = cx + d Section 3.3 4x 5 = 6x + 11-6x -6x -2x 5 = 11 +5 +5-2x = 16-2 -2 x = -8 Begin by rewriting the equation so there is only one variable term in the equation. Simplify to variable = constant.
Try these 8x + 5 = 4x +13 12x 4 = 9x 7 2b + 3 = 5b + 12 7x + 4 = 6x + 7 15x 2 = 4x 13 4x 7 = 5x + 1 5x 4 = 2x + 5 2x 3 = -11 2x 4y 8 = y - 8
Try these (answers) 8x + 5 = 4x +13 x = 2 7x + 4 = 6x + 7 x = 3 5x 4 = 2x + 5 x = 3 12x 4 = 9x 7 x = -1 15x 2 = 4x 13 x = -1 2x 3 = -11 2x x = -2 2b + 3 = 5b + 12 b = -3 4x 7 = 5x + 1 x = -8 4y 8 = y 8 y = 0
Solve equations containing parentheses 4 + 5(2x 3) = 3(4x 1) 4 + 10x 15 = 12x 3 10x 11 = 12x 3-12x -12x -2x - 11= -3 +11 +11-2x = 8-2 -2 x = -4 Begin by using the Distributive Property. Get variable terms on one side. Get constant terms on the other side. Simplify to variable = constant.
Try these 6y + 2(2y + 3) = 16 12x 2(4x 6) = 28 9x 4(2x 3) = 11 9 5x = 12 (6x + 7) 5[2 (2x 4)] = 2(5 3x) x + 5(3x 20) = 10(x 4) -4[x + 3(x 5)] = 3(8x + 20) 3[2 4(2x 1)] = 4x 10
Try these (answers) 6y + 2(2y + 3) = 16 y = 1 9x 4(2x 3) = 11 x = -1 5[2 (2x 4)] = 2(5 3x) x = 5-4[x + 3(x 5)] = 3(8x + 20) x = 0 12x 2(4x 6) = 28 x = 4 9 5x = 12 (6x + 7) x = -4 x + 5(3x 20) = 10(x 4) x = 10 3[2 4(2x 1)] = 4x 10 x = 1
Applications 1. A sail is in the shape of a triangle. Find the height of the sail if its base is 20 ft and its area is 300 ft 2. A = ½bh 300 = ½ (20) h 300 = 10h 10 10 30 = h Therefore, the height of the sail is 30 feet.
Applications 2. The height of a trapezoid is 10 ft and its area is 250 ft 2. If one of the bases is 2 ft more than 3 times the other base, what are the lengths of the bases? A = ½h(b + c) Let b = the 1 250 = ½(10)(b + 3b+2) st base Then the 2 nd base would be 3b + 2 250 = 5(4b + 2) 250 = 20b + 10 240 = 20b 12 = b = 1 st base so the 2 nd base = 3(12) + 2 = 38
Applications 3. If six times a number is increased by 15.35, the result is 12.5 less than the number. What is the number? Let x = the number 6x + 15.35 = x 12.5 -x -x 5x + 15.35 = -12.5-15.35-15.35 5x = -27.85 5 5 x = -5.57
Linear Equations in Two Variables 1. Intercepts 2. Graph by Plotting Points 3. Slope Intercept Form: y = mx + b 4. Point-Slope Form: y y 1 = m(x x 1 )
x-coordinate y-coordinate Ordered Pair (3, 4) Horizontal Distance (left & right) Vertical Distance (up & down)
Intercepts The point of intersection with the x-axis is called the x- intercept, the value of x on the coordinate plane where y = 0 (i.e. the point would look like (x, 0)). The point of intersection with the y-axis is called the y- intercept, the value of y on the coordinate plane where x = 0 (i.e. the point would look like (0, y)).
Linear Equations FORM Slope = rise/run Vertical Lines Horizontal Lines Slope-intercept EQUATION x = constant y = constant y = mx + b Point-slope y y 1 = m(x x 1 ) Standard Ax + By = C Parallel Lines Have same slope (m 1 =m 2 ) Perpendicular Lines m y x 2 2 y x 1 1 Have negative reciprocal slope (m 1 = -1/m 2 )
Linear Equations: Graph by Plotting Points Graph y = 2x + 1 Make a t-chart: By choosing any value for x and substituting that value into the linear equation, we can find the corresponding value of y. Usually we use x = 0 because it is easy to calculate. x y = 2x + 1 y 0 2(0) + 1 1 1 2(1) + 1 3-1 2(-1) + 1-1 Graph the ordered pair solutions (0, 1), (1, 3), and (-1, -1). Draw a line through the ordered-pair solutions.
Linear Equations: Slope-Intercept Form Slope-Intercept Form An equation of a line of the form y = mx + b, where m is the slope of the line and b is the y-intercept. The slope of a line is the measure of its slant. Slope = m = rise/run. The point of intersection with the y-axis is called the y- intercept, the value of y on the coordinate plane where x = 0 (i.e. the point would look like (0, y)).
Linear Equations Slope-Intercept Form Example: It is easy to graph a linear equation when it is in slope-intercept form. The following equation will be converted from standard form, -6x + y = -2, into slope-intercept form, y = mx + b. To do so the equation will need to be manipulated so that it is solved for y. Isolate y on the left-hand side by itself. -6x + y = -2 Add 6x to each side y = 6x - 2 Now in Slope-intercept form Identify the slope, which is represented by m in the equation and the y- intercept, which is represented by b in the equation. y = 6x - 2 When using the slope-intercept form to graph, make sure the equation is solved for y BEFORE identifying the slope and y-intercept.
Slope = m =rise/run m = ½ So you rise 1 and run 2. Graph a line using the slope-intercept form of a line: 1. Locate the y-intercept and plot the point 2. From this point, use the slope to find the second point and plot. 3. Draw a line that connects the two points.
Graph Solutions Example 1: Graph equation: y = ( -2 / 3 )x + 2 You can see from the equation and the graph that the slope of the line is ( -2 / 3 ), and the y-intercept is 2. The graph crosses the y-axis at (0, 2) Rise = -2 (so moving down 2 units) Run = 3 (so moving to the right 3 units)
Linear Equations: Point-Slope Form Point-Slope form An equation of a line that has slope m and contains the point whose coordinates are (x 1, y 1 ) can be found by the point-slope formula: y y 1 = m(x x 1 ) Example: Find the equation of a line that passes through the point whose coordinates are (-2, -1) and has slope 3/2. y y y y ( 1 1 3 2 1) 3 ( x 2 3 ( x 2 2) 3 x 2 3 x 2 ( 2)) Substitute values Distribute Subtract 1 from both sides
Example: Find the equation of the line that passes through (1, 2) and (-3, 4). Step 1: Find the slope. Let x 1 = 1 and y 1 = 2. Let x 2 = -3 and y 2 = 4. m y x 2 2 y x 1 1 4 3 2 1 2 4 1 2 Step 2: Substitute the slope (m) and a point, (1, 2) in the slope-intercept form of a line. Solve for the y-intercept (b). Substitute: x = 1, y = 2, and m = - ½ into y = mx + b and solve for b. If 2 = - ½ (1) + b then b = 5/2 Step 3: Substitute the slope (m) and the y- intercept (b) into the slope-intercept form of a line, y = mx + b. y = - ½ x + 5/2 OR use Point-Slope Formula y y y 2 2 1 2 x 1 2 1 2 ( x x 5 2 1) 1 2
Horizontal Lines y = 1 (or any number) Lines that are horizontal have a slope of zero. They have "run", but no "rise". The rise/run formula for slope always yields zero since rise = 0. y = mx + b y = 0x + 1 y = 1 This equation also describes what is happening to the y- coordinates on the line. In this case, they are always 1. Vertical Lines x = -1 (or any number) Lines that are vertical have no slope (it does not exist). They have "rise", but no "run". The rise/run formula for slope always has a zero denominator and is undefined. These lines are described by what is happening to their x- coordinates. In this example, the x-coordinates are always equal to -1. x = -1 y = 1
Slope of a Line The slope of a line is the measure of its slant. The symbol for slope is m. Slope = m = rise/run = Positive slope: slants upward Negative slope: slants downward Horizontal = 0 slope Vertical = undefined By knowing this information, you should have an idea of what your graph will look like before you plot any points. See http://regentsprep.org/regents/math/glines/llines.htm
Try These: Graph the following lines: y = ¾ x + 2 y = 3x 4 y = -⅓ x + 2 y = -2x + 3 y = -⅓ x + 4 2x + y = 4 x 3y = 6 y = -4 x = -8
Linear Equations: Parallel and Perpendicular Lines Parallel lines - Lines in the same plane that do not cross, the distance between the lines is constant. Two different, nonvertical lines with slopes m 1 and m 2 are parallel if and only if they have the same slope (m 1 = m 2 ). Perpendicular lines - Lines that intersect at one point forming 90 angles. Two different, nonvertical lines with slopes m 1 and m 2 are perpendicular if and only if have their slopes are negative reciprocals of each other (m 1 = -1/m 2 ). The product of their slopes is (-1).
Parallel Lines Parallel lines - Lines in the same plane that do not cross, the distance between the lines is constant. Two different, nonvertical lines with slopes m 1 and m 2 are parallel if and only if they have the same slope (m 1 = m 2 ). Determine if y = 2x + 5 and -2x + y = -2 are parallel. Equation Slope- Intercept Slope y = 2x + 5 y = 2x + 5 2-2x + y = -2 y = 2x 2 2 Since the slopes are the same (equal), the two lines are parallel.
Parallel Lines Example: Write the equation of the line that passes through the point Q(3,7) and is parallel to the line y = 6x + 6. Since the goal is to write an equation that is parallel to the line y = 6x + 6, the equations will have the same slope (m). The given equation is written in the slopeintercept form, therefore m = 6. To write an equation with m = 6 and passing through the point Q, the point-slope form is used. y - y 1 = m(x - x 1 ) For point Q(3,7), x 1 = 3 and y 1 = 7. y + 7 = 6(x + 3) y - 7 = 6(x - 3) Distribute the 6. (multiply) y - 7 = 6x - 18 Add 7 to each side y = 6x 11 Therefore, line y = 6x 11 passes through point Q(3,7) and is parallel to line y = 6x + 6.
Perpendicular Lines Perpendicular lines - Lines that intersect at one point forming 90 angles. Two different, nonvertical lines with slopes m 1 and m 2 are perpendicular if and only if have their slopes are negative reciprocals of each other (m 1 = -1/m 2 ). The product of their slopes is (-1). Determine if y = 2x + 5 and ½ x + y = -2 are perpendicular. Equation Slope- Intercept Slope y = 2x + 5 y = 2x + 5 2 ½ x + y = -2 y = - ½ x 2 - ½ Since the slopes are negative reciprocals of each other, the two lines are perpendicular.
Perpendicular Lines Example: Write the equation of the line that passes through the point V(3,11) and is perpendicular to the line y = -3/4x + 24. Since the goal is to write an equation that is perpendicular to the line y = -3/4x + 24, the equations will have slopes (m) that are negative reciprocals. The given equation is written in the slope-intercept form, therefore m = -3/4. The negative reciprocal is 4/3. To write an equation with m = 4/3 and passing through the point V, the pointslope form is used. y - y 1 = m(x - x 1 ) For point V(3,11), x 1 = 3 and y 1 = 11. y y y 11 11 4 3 x 4 ( x 3 4 x 3 15 4 3) Distribute the 4/3. (multiply) Add 11 to each side. Therefore, line y = (4/3)x + 15 passes through point V(3,11) and is perpendicular to line, y = (-3/4)x + 24.
Rational Expressions 1. Rules for Rational Numbers 2. Simplifying Rational Expressions
Rational Expressions A rational number is a fraction, so a rational expression is a fraction in which both the numerator and denominator are polynomials. Examples: 5x 7 2, m 2 m Remember, the denominator can never be 0. Division by 0 is undefined. 9 3
Rational Expressions Rules for Rational Numbers 0,, : : : 0, : : : 0 : : : 0 : : : 0, : 1: ; d c b where c b d a d c b a Division d b where d b c a d c b a Multiplication b where b c a b c b a Subtraction b where b c a b c b a Addition b a where a b b a and a b is b a reciprocal
Rational Expressions Rules for Rational Numbers If a and b are real numbers and b 0, then a b a b a b If a b, (a b) and (b a) are opposites a b b a a b 1( a b) 1
Rational Expressions Reduce each rational expression to lowest terms and state any restrictions: Ex 1: 4x 12 4x 4( x 4x 3) x 3, x 0 x Ex 2: 7x 2 14 x 7( x ( x 2) 2) 7; x 2 Ex 3: x 2 x 2 7x 10 25 ( x ( x 2)( x 5)( x 5) 5) x x 2 ; x 5 5, or,5
Algebra Practice Problems The following slides are additional practice problems for the COMPASS Algebra Placement Test.