An equation is a statement that states that two expressions are equal. For example:

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1 Section 0.1: Linear Equations Solving linear equation in one variable: An equation is a statement that states that two expressions are equal. For example: (1) 513 (2) 16 (3) 4252 (4) To solve the equation is to find all values of the variable that make the equation true and these values are called the solution or roots. 1) x = 8 is the solution 2) x = 4 and x = 4 are both solutions the solution set = 4, 4 3) There is no solution or referred to as the Null set and for 4) It is true for all real numbers, so the solution set =,, and this is referred to as an Identity, since it s true for all values of x. A linear equation in one variable, call it x, can be written in the form ax + b = 0; where a and b are real numbers and a 0. Solve the following examples: (x + 6) 2 = (x 8) multiply both sides of the equation by 6 (least common denominator LCD) 2(x + 6) 12 = 3(x 8) 2x = 3x 24 x = 24 x = 24 Verify your answer by substituting 24 for x and show that the left side = right side. Solving an equation that is a contradiction: 3(x 2) + 7x = 5(2x + 1) 3x 6 + 7x = 10x x 6 = 10x = 5; since 6 does not equal 5, then the original equation is a contradiction. Basic steps for solving a word problem 1. Read the problem and determine what is being asked to find/solve for. 2. Draw a picture, sometimes helps. 3. Assign a variable for what you are solving for, and use multiple variables if more solving for more than one item. 4. Set up equation(s) 5. Solve the equation(s) 6. Check your answer(s) Examples: 1. Alaska covers an area that is 230 square miles more than 500 times the size of Rhode Island. If the combined area of the 2 states is 616,460 mi 2, how many square miles does each cover? Let R = the area Rhode Island covers, then the area Alaska covers = 500R + 230, therefore: R + (500R + 230) = 616, R = 616,230 R = 1,230 square miles and Alaska = 1,230(500) = 615,230 mi The sum of 3 consecutive odd integers is 75, find the integers. Let n be the smallest integer, then n + 2 is the 2 nd and n + 4 is the 3 rd, we also know: n + (n + 2) + (n + 4) = 75 3n + 6 = 75 N = n = 23, and the other integers are 25 and 27. They are each odd integers and = 75. Page 1

2 3. Suppose you need to make 10 liters of a 15% acidic solution but you only have 10% and 30% acidic solutions. How many liters of each solution will you need to use. Let x = the quantity of 10% solution and let y = the quantity of 30% solution x + y = 10 liters y = 10 x. We also know that x(10%) + y(30%) = 10(15%) x(.10) + (10 x)(.30) = 10(.15) 0.10x x = = 0.2x x = 7.5 liters and y = 2.5 liters Linear Inequalities in One Variable Examples: 1. ( 3/4)x + 2/3 7/4 multiply both sides by 12 9x x 13 divide by 9 and flip the inequality sign x 13/9 x [ 13/9, ) Note: One way to check your answer is to pick a value within the interval, in this case pick x = 0 and you end up with 2/3 7/4 which is a true statement. 2. Compound inequality, solve for x: 1 > 6 multiply each part by 3 and flip the inequality signs 3 < 2x subtract 5 from all parts 8 < 2x 13 divide each part by 2 4 < x 13/2 x ( 4, 13/2] Determining the Domain of an Expression: Examples: 1. ; the denominator cannot be 0 2x x -3/2 in set notation {x x R, x 3/2} or in interval notation x (, 3/2) U ( 3/2, ). 2. denominator cannot be 0 and cannot take the square root of a negative number x 3 > 0 x > 3 in set notation {x x > 3} or in interval notation x (3, ). Section 0.2: Quadratic Equations A quadratic equation is a 2 nd degree equation, standard form is ax 2 + bx + c = 0 where a, b and c are real numbers and a 0; a is called the leading coefficient, b is the coefficient of the first-degree term and c is the constant term. Solving a quadratic equation find the value(s) of x such that ax 2 + bx + c = 0; graphically, these would represent the x-intercepts of the function f(x) = ax 2 + bx + c Graphical representation of solving a quadratic equation: Page 2

3 Two Real Roots One Real Root No Real Roots but Two Complex Roots Complex Number Review: Imaginary numbers are of the form where k is a positive real number. The imaginary number 1; such that 1, it is the solution to the quadratic equation + 1 = 0. Rewrite an imaginary number: 40 = 410 =2 10 Complex numbers can be written in the form a + bi, where a and b are real numbers and 1. Adding or subtracting complex numbers: (3 + 5i) + ( 2 + 4i) combine the real and imaginary parts 1 + 9i. Multiplying complex numbers: (3 + 5i)( 2 + 4i) = i 10i + 20i i + 20( 1) = i. Product of complex conjugate: for the complex number a + bi, the complex conjugate is a bi. (a + bi)(a bi) = a 2 b 2 (i 2 ) = a 2 + b 2. Simplifying higher powers of i: i 22 = (i 2 ) 11 1; i 28 = (i 4 ) = 1; i 57 = (i 4 ) 14 i = i; i 75 = (i 2 ) 37 i = i Solve for x: x 2 + 6x + 13 = 0 x = 3 ± 2i. x = = = Division of complex numbers: = = = = = 6/25 17i/25 Methodologies for solving a Quadratic Equation Factoring o Trial and Error o Grouping/ AC method o Completing the Square Square Root Method Quadratic Formula; x = Graphing calculator, finding the zeros Solving a quadratic equation by factoring: Write the equation in standard form, one side of the equation must be 0. Factor the nonzero side of the equation. Set each factor to 0, if either factor = 0, then the product = 0. (Zero-product property) Solve for the variable and you might end up with 2 solutions Verify your answer(s) in the original equation (optional step, but recommended) Page 3

4 Page 4 Precalculus: MAC 1140 Examples (ac method): a. 2x 2 + 6x 8 = 0 factor out the GCF, in this case the GCF = 2 2(x 2 + 3x 4) = 0; divide both sides by 2 x 2 + 3x 4 = 0 now multiply 1 st and last coefficients (1)( 4) = 4, find two numbers whose product is 4 and whose sum is 3 (the coefficient for x). Numbers would 4 and 1, re-write the equation as follows: x 2 + 4x x 4 = 0 x(x + 4) (x + 4) = 0 (x + 4)(x 1) = 0 (x + 4) = 0 or (x 1) = 0 x = 4 or x = 1 Check both answers: (1) 4 = 0 and ( 4) 2 + 3( 4) 4 = 0 b. 2(3x 2 2) = 5x, re-write in standard form 6x 2 + 5x 4 = 0 multiply 1 st and last coefficients = 24, find two numbers whose product is 24 and whose sum is 5. Numbers would 8 and 3. 6x 2 + 8x 3x 4 = 0 2x(3x + 4) (3x + 4) = 0 (3x + 4)(2x 1) = 0 (3x + 4) = 0 or (2x 1) = 0 x = 4/3 or x = ½ Check both answers c. 20n n + 5 = 0 The GCF = 5, factor that out. 5(4n 2 + 4n + 1) = 0 (divide both sides of the equation by 5) 4n 2 + 4n + 1 = 0 multiply 1 st and last coefficients = 4, find 2 numbers whose product is 4 and whose sum = 4 2 and 2 4n 2 + 2n + 2n + 1 = 0 2n(2n + 1) + (2n + 1) = 0 (2n + 1)(2n + 1) Perfect square, recall (a + b)(a + b) = a 2 + 2ab + b 2 Solving Quadratic Equations by the Square Root Method If x 2 = a, then x = and x = - are the solutions or often written as x = ± Example: solve for x given x 2 27 = 0 x 2 = 27 x = ± 27 x = ±93, if possible, factor out squares x = ±3 3, Example: solve for x, given (x + ½) 2 = 8 x + ½ = ± 8 x + ½ = ±2 2 x = 2 2 ½ and 2 2 ½ (check and both answers are valid) Solve Quadratic Equations by Completing the Square To complete the square implies creating a perfect square trinomial. Recall: a 2 + 2ab + b 2 = (a + b)(a + b) a 2 2ab + b 2 = (a b)(a b) Completing the square method: Given ax 2 + bx + c = 0 and a 0, divide the equation by a so that the leading coefficient = 1. Isolate the constant term on one side of the equation. Determine the constant that completes the square, (b/2) 2 and add this to both sides of the equation. Factor the perfect square trinomial as a square of a binomial. Use the square root method to solve the equation. Verify your answer back in the original equation.

5 Completing the square method can be used to solve any quadratic equation Example: solve for x, given 6x x + 30 = 0 Make the leading coefficient 1; divide both sides by 6 x 2 + 8x + 5 = 0 Isolate the constant term by subtracting 5 from both sides x 2 + 8x = 5 Complete the square add (8/2) 2 = 16 to both sides x 2 + 8x + 16 = (x + 4) 2 = 11 x + 4 = ± 11 x = ± 11 4 or x = 11 4 or x = 11 4 Verify answer. Solving Quadratic Equations by the Quadratic Formula (derived via completing the square). Given ax 2 + bx + c = 0 and a 0, the quadratic formula is: x = Example: 3x + 5x 2 = 2, 1 st rewrite in standard form 5x 2 3x 2 = 0, then use the quadratic formula. x = = = = x = 10/10 or x = 4/10 x = 1 or x = 2/5 Quadratic Equations with Complex Solutions. Solve for x given 2x 2 + x + 1 = 0 x = = = = The Discriminant of a Quadratic Equation, ax 2 + bx + c = 0 and a 0, is given by If. If b 2 4ac > 0, then the equation has 2 distinct real solutions (or roots). If 2 4 = 0, then the equation has 1 distinct real solution (or root). If 2 4 < 0, then the equation has no real solutions but has 2 distinct complex solutions (or root). Section 0.3: Other Types of Equations A polynomial function equation of degree n can be expressed as: anx n + an-1x n a2x 2 + a1x + ao = 0 where each coefficient ak is a real number and an 0 and n is a non-negative integer. Factoring the Sum or Difference of two cubes (S-O-A-P; Same-Opposite-Always Positive): (a + b) (a 2 ab + b 2 ) = a(a 2 ab + b 2 ) + b(a 2 ab + b 2 ) a 3 a 2 b + ab 2 + a 2 b ab 2 + b 3 a 3 + b 3 (a b) (a 2 + ab + b 2 ) = a(a 2 + ab + b 2 ) b(a 2 + ab + b 2 ) a 3 + a 2 b + ab 2 a 2 b ab 2 b 3 a 3 b 3 Example: Factor a 3 27 = 0 a = 0 So we can use the SOAP rule: (a b) (a 2 + ab + b 2 ) (a 3) (a 2 + a ) = (a 3) (a 2 + 3a + 9) = 0 Solution: a = 3; and no other real number solutions since 3 2 4(1)(9) < 0 (discriminant < 0) Example: x 3 7x + 21 = 3x 2 x 3 3x 2 7x + 21 = 0; solve by grouping and then factoring. x 2 (x 3) 7(x 3) = 0 (x 2 7)(x 3) = 0 x = 3 and x = ± 7. A rational expression is the quotient of two polynomials such that the denominator is not equal to zero. Page 5

6 Rational equations include rational expressions on both sides of the equal sign. Solving rational equations: 1. Exclude any values that result in the denominator equaling zero. 2. Multiply both sides of the equation by the LCD to eliminate denominators. 3. Solve the resulting equation. 4. Check all solutions in the original equation. Page 6 Precalculus: MAC 1140 Example: solve for x given: x + = 1 + multiply both sides of the equation by the LCD: (x 3) x(x 3) + 12 = x 3 + 4x x 2 3x 5x = 0 x 2 8x + 15 = 0 (x 3)(x 5) = 0 the solutions are x = 3 and x = 5. Check answer and we see that x = 3 results in division by zero, which means x is an extraneous root and therefor x = 5 is the only valid solution. Radical equations and equations with rational exponents. A radical equation is any equation that contains terms with a variable in the radicand, for example, solve for x where: 37 Solving a radical equation: 1. Determine the domain. 2. Isolate the radical term to one side of the equation, if more than one radical, isolate either one. 3. Raise both sides of the equation to a power equivalent to the root of the radical. 4. Solve the resulting equation. If the equation still contains radicals, then repeat steps 2 and Check your answer(s) in the original equation algebraically or graphically to make sure there are no extraneous solutions. Power property of equality: if, then for n 2. Note: raising both sides of the equation to an even power can introduce an extraneous root. Example: solve the radical equation for x, given: + x = 4 1. Domain: x x 2 [ 2, ) 2. Isolate the radical on one side of the equation, if more than one radical, isolate either one. 2 = 4 x 3. Raise both sides of the equation to the power equivalent to the root of the radical. In this case 2 ( 2) 2 = (4 x) 2 4. Solve the resulting equation. If the equation still contains any radicals, repeat step #2. x + 2 = 16 8x + x 2 x 2 9x + 14 = 0 (x 7)(x 2) x = 7 and x = 2 5. Check answers to remove any extraneous solutions: x = = = 10 4; therefore x = 7 is not a solution. x = = = 4 = 4; therefore x = 2 is a solution. Example: solve the radical equation for x, given: = 1 1. Domain: x and x 4 0 x 4 (most restrictive) [4, ) 2. Isolate the radical on the left hand-side Square both sides x + 3 = (x 4) 4. Isolate the radical 6 = = 4 5. Square both sides 9 = x 4 x = Check answer = = 4 Power property of equality: For real-valued expression u and v, with positive integer s m and n, with m/n in lowest terms: If m is odd: if m is even:

7 and then and then u u Example: solve for x given x 3 = ± 8 x = 11 and x = 5. Check answer by substituting the results back into the original equation. For x: 11 3 = 8 = 4 (checks); for x = = 8 = 4 Note: be careful regarding the use of the ± sign. For example whereas, x 2 = 18 x = ± 18 and therefore x = ±3 2 Solving equations in quadratic form: solve by substitution. Example: let u = and now rewrite the equation as 3100 and you can factor that into (u + 5)(u 2) = 0 u = 5 and u = 2. And since u = x = 125 and x = 8. Both answers check. Word problem examples: 1. A legal size sheet of paper has a length equal to 3 inches less than twice its width. If the area of the paper is 119 in 2, find the length and width. Let L = length and W = width and we know that LW = 119, we also know that L = 2W 3, therefore: (2W 3)W = 119 2W 2 3W 119 = 0; factor or use the quadratic formula (2W 17)(W + 7) = 0 W = 17/2 or W = 7; since the width cannot be negative, the solution is W = 17/2 inches. 2. A printer sells for $300 and a company will sell 15 at that price per week. In addition, for each additional reduction of the price by $8, two additional printers are sold. What price will result in a weekly revenue of $6500? Let P equal the price and S equal the number of printers sold PS = And let x = the number of $8 price decrements P = 300 8x and S = x (300 8x)(15 + 2x) = = x + 600x 16x 2 16x 2 480x = 0 x 2 30x = 0 (x 5)(x 25) = 0 x = 5 or x = 25. If x = 5 P = 300 8(5) = $260 If x = 25 P = 300 8(25) = $100. In terms of profit, which would be a better choice P = $260. Solving Absolute Values Equations Examples: 1. Solve for x: 4 3x + 5 = 7 4 3x = 2 4 3x = 2 or 4 3x = 2 3x = 2 or 3x = 6 x = 2/3 or x = 2 Check answers by substituting back into the original equation. Section 0.4: Inequalities Expressions of Inequalities 1. Inequality Notation: a x < b 2. Solution Set: {x a x < b} 3. Interval Notation: [a, b) Page 7

8 4. Graph/Number Line: [ 1, 3) Unions and Intersections A U B is the union of two sets A and B and it includes all elements found in both A and B A U B = {x x is in A or B or both) A B is the intersection of the two sets A and B and it includes only those elements found in both sets A and B. A B = {x x is in both A and B} Solving a Linear Inequality Example: 14 5x 4 1 st Isolate the variable 5x 10 When dividing or multiplying by a negative number, flip the inequality sign. x 2 Interval notation: [2, ) Number line: Solving Double Linear Inequalities Example: 4 < 2x Subtract 6 from each part 4 6 < 2x Simplify 10 < 2x 8 Divide each part by 2 and flip the inequality signs 5 > x 4 This is equivalent to 4 x < 5 [ 4, 5) Interval notation Solving Quadratic Inequalities Example: Find the solution set for x 2 + 3x 10 0 First find the zeros (x-intercepts), set x 2 + 3x 10 = 0 and solve for x (x + 5)(x 2) = 0 x + 5 = 0 or x 2 = 0 x = 5 or x = 2, now test the following 3 intervals (, 5], [ 5, 2] and [2, ) by picking test points and substituting them into the original expression and see if it satisfies the inequality. 1. Pick 6 ( 6) 2 + 3( 6) 10 = 8 which is 0, so this interval is in the solution set 2. Pick 0 (0) 2 + 3(0) 10 = 10 which is < 0, so this interval is not in the solution set 3. Pick 3 (3) 2 + 3(3) 10 = 8 which is 0, so this interval is in the solution set Another approach, is use the factored form and you simply need to decide if it s factor is + or 1. Pick 6 ( )( ) which is +, so this interval is in the solution set 2. Pick 0 (+)( ) which is, so this interval is not in the solution set 3. Pick 3 (+)((+) which is +, so this interval is in the solution set (, 5] U [2, ) is the solution set. Solving Polynomial Inequalities Example: x 3 + 2x 2 15x < 0 Factor the polynomial, if possible Page 8

9 x(x 3)(x + 5) < 0 Set each part = 0 x = 0 or x = 3 or x = 5 The test intervals are (, 5), ( 5, 0), (0, 3), (3, ) 1. Pick 6 ( )( )( ) ( ) which is in the solution set 2. Pick 1 ( )( )(+) (+) which is not in the solution set 3. Pick 1 (+)( )(+) ( ) which is in the solution set 4. Pick 4 (+)(+)(+) (+) which is not in the solution set (, 5) U (0, 3) is the solution set. Solving Rational Inequalities Example: Solve the following inequality: 0 Factor the numerator and denominator, if possible. State any domain restrictions (i.e., division by 0). Divide the number line into intervals based on the zeros of the numerator and denominator Factored form: 0 x 5 and x 5 and the zeros are 5, 3 and 5 which means the intervals are (, 5), ( 5, 3), (3, 5), (5, ) 1. Pick 6 ( ) which is in the solution set 2. Pick 0 3. Pick 4 4. Pick 6 (+) which is not in the solution set ( ) which is in the solution set (+) which is not in the solution set (, 5) U (3, 5) is the solution set, and re-verify the restricted values are not in the solution set, 5 and 5, and they are not. Page 9

10 Solving Absolute Values Inequalities Examples: 1. Solve for x: 4 3x + 5 > 7 4 3x > 2 4 3x > 2 or 4 3x < 2 Note: this is an or condition because of the > 3x > 2 or 3x < 6 x < 2/3 or x > 2 (remember: flip the inequality sign whenever you divide or multiply by a negative number) x (-, 2/3) U (2, ) 2. Solve for x: 7 2x + 3 < 8 7 2x < 5 7 2x < 5 and 7 2x > 5 Note: this is an and condition because of the < 2x < 2 and 2x > 12 x > 1 and x < 6 (1, 6) Section 0.5: Graphing Equations Relations, Mapping Notation, and Ordered Pairs Relations: a correspondence between two sets. For example, there is a correspondence between the names of each student in pre-calculus and their GPA. Mapping notation: if N represents the set of names of each student and G represents the GPA s those students possess, then N G Ordered pair: an ordered pair defines the relation between two sets of data and is expressed with a parenthesis, where the first coordinate representing the independent variable and the second coordinate representing the dependent variable. For example: (Mike, 3.75). Domain: represents the set of all first coordinates. Range: represents the set of all second coordinates. The Graph of a Relation: relations can be expressed graphically on a rectangular coordinate system. Horizontal line (x-axis) and the vertical line (y-axis). The graph is split up into 4 quadrants, the origin is where the two lines intersect and is represented by (0, 0). Example: graph x = y 2 x y ordered pairs (x, y) 0 0 (0, 0) 1 1, 1 (1, 1) and (1, 1) 2 2, 2 (2, 2) and (2, 2) 3 3, 3 (3, 3) and (3, 3) 4 2, 2 (4, 2) and (4, 2) -1 not real n/a -2 not real n/a The Equation of a Circle: standard form of a circle, with center (h, k) and radius r is the set of points (x, y) such that Note: if the center is at the origin, (0, 0), then + Page 10

11 Distance Formula: the distance between two points:, and, in the coordinate plane is given by the following equation: d = Midpoint Formula: the midpoint, M (x, y) between two points:, and, is given by the following: M =, Example: find the midpoint of the line segment with end points (2, 7) and ( 6, 5) M =, ( 2, 1) Show that this point represents the midpoint. Find the distance between (2, 7) and ( 2, 1) d = = 52 = 2 13 Find the distance between ( 6, 5) and ( 2, 1) d = = 52 = 2 13 The Graph of a Circle: 1. Identify the center (h, k) and the radius r 2. Plot the center point 3. Plot 4 points on the circle by starting from (h, k) and plot one point r units horizontally left and the 2 nd point r units horizontally right, 3 rd point r units vertically up and the 4 th point r units vertically down. 4. Connect the 4 points with a smooth curve in the shape of a circle 5. Label the center and the radius. Example: graph the circle represented by Center = (5, 2) and radius = 9 = 3 (only consider the positive root) Example: graph the circle x 2 + y 2 = 25 using your graphing calculator. Solve for y and you end up with the following equation y = ± 25. Enter the following equations into your graphing calculator: Y1 = and Y2 =. Page 11

12 Example: find the equation of a circle with center ( 2, 3) and radius = Use completing the square to find the center and radius of a circle Example: Find the center and radius of the following x 2 + y 2 4x + 12y 7 = 0 x 2 4x + y y = 7 (group the x and y terms and place the constant on the other side) Complete the squares for both the x-terms and the y-terms x 2 4x + ( 4/2) 2 ( 4/2) 2 + y y + (12/2) 2 (12/2) 2 = 7 x 2 4x y y = 7 (x 2) 2 + (y + 6) 2 = (x 2) 2 + (y + 6) 2 = 47 center = (2, 6) and the radius = Using Intercepts and Symmetry as Graphing Aids x-intercepts: let y = 0 and solve for x, also called the zeros or roots y-intercepts: let x = 0 and solve for y Example: Find the x and y intercepts of y = x 2 + 2x 8 x-intercepts: set y = 0 x = (x 4)(x + 2) (x 4) = 0 or (x + 2) = 0 x-intercepts are x = 4 and x = 2 y-intercepts: set x = 0 y = (0) 8 y = 8 Symmetry: The graph of equation can be symmetric about the x-axis, y-axis, origin or none of these. Symmetric about the x-axis if you replace y with y the equation is unchanged. For example: y 2 = 3x 4 Symmetric about the y-axis if you replace x with x the equation is unchanged. For example: y = 2x Symmetric about the origin if you replace x with x and y with y the equation is unchanged. For example: x 2 + y 2 = 25 Symmetric about the x-axis Symmetric about the y-axis Symmetric about the origin Page 12

13 Section 0.6: Graphing a Line The general or standard form of a linear equation: ax + by = c where a, b and c are real numbers and a and b cannot both be equal to zero. To graph a line: Find the x-intercept, set y = 0 and solve for x Find the y-intercept, set x = 0 and solve for y Plot the points on a rectangular coordinate system and connect the dots with a straightedge Example: Plot 2x + 3y = 6 Find the x-intercept set y = 0 2x = 6 x = 3 (3, 0) Find the y-intercept set x = 0 3y = 6 y = 2 (0, 2) Note: you only need 2 points to graph any line, in some cases the intercepts are easy points to find. Plot the graph of the same equation using a graphing calculator and show both intercepts. 1) Need to solve the equation for y 2) 2x + 3y = 6 3y = 2x + 6 y = ( 2x + 6)/3 y = + 2 3) Hint: use the intercepts to help determine the viewing window. 4) For example, pick the following window: 4; 4; 4; 4 Slope of a Line: typically represented with the letter m. The slope of a non-vertical line passing through the distinct points (x1, y1) and (x2, y2) is given by: m = = = Page 13

14 Vertical lines, example x = 3, x-intercept at (3, 0) Horizontal lines, example y = 2, y-intercept at (0, 2) Parallel and Perpendicular Lines Two lines are parallel if they lie in the same plane and do not intersect each other. And two lines are perpendicular if they intersect to form right angles (90 o angles). If two distinct lines have the same slope, then the lines are parallel. Conversely, if two lines are parallel, they have the same slope. If two lines have slopes that are negative reciprocals of each other, then the lines are perpendicular, exclusive of vertical and horizontal lines. If l1 and l2 represent two non-vertical lines with slopes m1 and m2, then If l1 and l2 are parallel, then m1 = m2. If l1 and l2 are perpendicular, then m1 = or. m1 m2 = 1 Example: find the slope of the line below: m = = = Slope-intercept form of a line: y = mx + b; where m = slope and b = y-intercept Example: 2x + 3y = 6; solve for y 3y = 2x + 6 y = ( 2x + 6)/3 y = + 2 Slope = m = and y-intercept = b = 2 To graph, you can start with the y-intercept (0, 2) and then using rise over run we can plot another point, for example (3, 0) or ( 3, 4). Graph of the line is provided in the above example. Point-slope form of a line: y y1 = m(x x1) If the slope of line = 3 and one point on the line is ( 1, 4), find the equation of the line in slope-intercept form. Start with the point-slope form: y y1 = m(x x1) y 4 = 3(x ( 1)) y 4 = 3(x + 1) y = 3x y = 3x + 7 Page 14

15 Example: find the equation of the line shown in the graph below in slope-intercept form: y-intercept = (0,1) Slope = m = rise/run = ½ y = x + 1 or y = + 1 Section 0.7: Modeling and Variation Let x and y be two quantities, then 1. If y varies directly with x (or y is directly proportional to x), then there exists a nonzero constant k such that 2. If y varies directly with the nth power of x (or y is directly proportional to the nth power of x), then there exists a nonzero constant k such that 3. If y varies inversely with x (or y is inversely proportional to x), then there exists a nonzero constant k such that. 4. If y varies inversely with the nth power of x (or y is inversely proportional to the nth power of x), then there exists a nonzero constant k such that. Example: The length of a violin string varies inversely as the frequency of its vibrations. A violin string 14 in length vibrates at a frequency of 450 cycles per second. Find the frequency of a 12 violin string. k = Lf = 14(450) = Example: the drag force, F, on a boat varies jointly with the wet surface area, A, of the boat and the square of the speed. A boat with a wet surface area of 72ft 2 traveling at 12.5 MPH experiences a drag force of 675 N. Find the drag force of a boat having a wet surface area of 25 ft 2 and traveling 16 MPH. F = kas = k(72)(12.5) 2 k = 0.06 F = (0.06)(25)(16) Newtons. Page 15

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