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MATH 2 (Intermediate Algebra Advance Course) NAME : DATE : TEACHER : Mr. E. Ramos jr Handout # : 1 TOPICS :THE RECTANGULAR COORDINATE SYSTEM AND GRAPHING LINEAR EQUATIONS AND THE SLOPE OF A LINE CONCEPTS : Rectangular Coordinate System The following is the rectangular coordinate system: It is made up of two number lines: 1. The horizontal number line is the x- axis. 2. The vertical number line is the y- axis. The origin is where the two intersect. This is where both number lines are 0. It is split into four quadrants which are marked on this graph with Roman numerals. Each point on the graph is associated with an ordered pair. When dealing with an x, y graph, the x coordinate is always first and the y coordinate is always second in the ordered pair (x, y). It is a solution to an equation in two variables. Even though there are two values in the ordered pair, be careful that it associates to ONLY ONE point on the graph, the point lines up with both the x value of the ordered pair (xaxis) and the y value of the ordered pair (y-axis). Example 1: Plot the ordered pairs and name the quadrant or axis in which the point lies. A(2, 3), B(-1, 2), C(-3, -4), D(2, 0), and E(0, 5). Remember that each ordered pair associates with only one point on the graph. Just line up the x value and then the y value to get your location. A(2, 3) lies in quadrant I. B(-1, 2) lies in quadrant II. C(-3, -4) lies in quadrant III. D(2, 0) lies on the x-axis. E(0, 5) lies on the y-axis. Example 2: Find the x- and y- coordinates of the following labeled points Remember that each ordered pair associates with only one point on the graph. Just line up the x value and then the y value to get your ordered pair. Since point A corresponds to 2 on the x-axis and -3 on the y-axis, then A’s ordered pair is (2, -3). Since point B corresponds to 3 on the x-axis and 2 on the y-axis, then B’s ordered pair is (3, 2). Since point C corresponds to -2 on the x-axis and 3 on the y-axis, then C’s ordered pair is (-2, 3). Since point D corresponds to -3 on the x-axis and - 4 on the y-axis, then D’s ordered pair is (-3, - 4). Since point E corresponds to -3 on the x-axis and 0 on the y-axis, then E’s ordered pair is (-3, 0). Since point F corresponds to 0 on the x-axis and 2 on the y-axis, then F’s ordered pair is (0, 2). EXERCISES Practice Problem 1a: Plot each point and name the quadrant or axis in which the point lies. 1a. A(3, 1), B(-2, -1/2), C(2, -2), and D(0,1) Practice Problem 2a: Find the x- and y- coordinates of the following labeled points. 2a. Practice Problems 3a - 3b: Determine if each ordered pair is a solution of the given equation. 3a. y = 4x - 10 ; (0, -10), (1, -14), (-1, -14) 3b. y = -5 ; (2, -5), (-5, 1), (0, -5) GRAPHING LINEAR EQUATIONS AND THE SLOPE OF A LINE Let's draw the graph of this equation. One method we could use is to find the x and y values of two points that satisfy the equation, plot each point, and then draw a line through the points. We can start with any two x values we like, and then find y for each x by substituting the x values into the equation. Let's start with x = 1. Value of x Value of y 1 2.5 2 3 Let's plot these points and draw a line through them. Graphing Using Slope and Y-Intercept There's another way to graph an equation using your knowledge of slope and yintercept. Look at the equation again. We can find the slope and y-intercept of the line just by looking at the equation: m = 1/2 and y intercept = 2. Just by looking at these values, we already know one point on the line! The yintercept gives us the point where the line intersects the y-axis, so we know the coordinates of that point are (0, 2), since the x value of any point that lies on the y axis is zero. To find the second point, we can use the slope of the line. The slope is ½ , which gives us the change in the y value over the change in the x value. The change in the x value, the denominator, is 2, so we move to the right 2 units. The change in the y value, the numerator, is positive one. We move up one unit. This gives us the second point we need. Now we can draw the line through the points. This is the exact same line we found using the first method. Do you see that it's quicker and easier to use the y-intercept and the slope? You can use either method to graph the line, depending on what information you have about the line and its equation. EXERCISES 1. Write the number of the line that matches each equation. 2. Are these statements True False true or false? 1) The coordinates of point I are (5, -3). 2) The slope of line EF is zero. 3) The coordinates of point G are (3, -4). 4) The coordinates of point H are (5, 3). 5) The slope of line JK is -1. 6) The y-intercept of line EF is 6. 7) The equation for line EF is y = 5. 8) The equation for line JK is y = -x. MATH 2 (Intermediate Algebra Advance Course) NAME : DATE : TEACHER : Mr. E. Ramos jr Handout #2 TOPIC : Equations of a Line : Parallel and Perpendicular Lines Equation of a Straight Line The GENERAL equation of a straight line is given in the form: y = mx + b What does it stand for? Gradient y = how far up x = how far along m = Gradient or Slope (how steep the line is) b = the Y Intercept (where the line crosses the Y axis) Knowing this we can work out the equation of a straight line :Example 1 Y Intercept 2 m = = 2 1 b=1 Therefore y = 2x + 1 Example 2 3 = –3 m = -1 b=0 This gives us y = –3x + 0 We do not need the zero! Therefore y = –3x Equations of Straight Lines When working with straight lines, there are many ways to arrive at an equation which represents the line. Information about Slope Slope is always represented by the letter m when writing equations of line. Slope is found by using the formula: Slope is also expressed as rise/run. m= Equation Forms of Straight Lines Slope Intercept Form [if you know the slope and the y-intercept (where the line crosses the y-axis), use this form] Point Slope Form [if you know a point and the slope, use this form] y=mx + b m = slope b = y-intercept (where line crosses the y-axis.) Horizontal Lines y = 3 (or any number) m = slope = any point on the line Vertical Lines x = -2 (or any number) Lines that are horizontal have a slope of Lines that are vertical have no slope (it does zero. They have "run", but no "rise". The not exist). They have "rise", but no "run". rise/run formula for slope always yields zero The rise/run formula for slope always has a since zero denominator and is undefined. rise = 0. y = mx + b These lines are described by what is y = 0x + 3 happening to their x-coordinates. In this y=3 example, the x-coordinates are always equal This equation also describes what is to -2. happening to the y-coordinates on the line. In this case they are always 3. Examples: Examples using Slope-Intercept Form: 1. Find the slope and y-intercept for the equation 2y = -6x + 8. First solve for "y =": y = -3x + 4 Remember the form: y = mx + b Answer: the slope (m) is -3 the y-intercept (b) is 4 2. Find the equation of the line whose slope is 4 and the coordinates of the yintercept are (0,2). In this problem m = 4 and b = 2. Remember the form: y = mx + b Substitute: y = 4x + 2 Examples using Point-Slope Form: 3. Given that the slope of a line is -3 and the line passes through the point (-2,4), write the equation of the line. The slope: m = -3 The point (x1 ,y1) = (-2,4) Remember the form: y - y1 = m ( x - x1) y - 4 = -3 (x - (-2)) y - 4 = -3 ( x + 2) ANS. If asked to express the answer in "y =" form: y -4 = -3x - 6 y = -3x - 2 4. Find the slope of the line that passes through the points (-3,5) and (-5,-8). Find the slope: Use either point: (-3,5) Remember the form: y - y1 = m ( x - x1) Substitute: y - 5 = 6.5 ( x - (-3)) y - 5 = 6.5 (x + 3) Ans. Practice with Equations of Lines Try the following problems: 1. Find the equation of the line that has a slope of 6 and passes through the point (-3,5). 2. Find the equation of the line that has a slope of -2 and a y-intercept of -9. 3. Find the equation of the line that passes through the points (3,4) and (-4,6). 4. Find the slope of the line whose equation is y - 3x = 7. 5. Which equation represents the line whose slope is 1 an whose y-intercept is 0? a) y = x b) x = -y c) y = 1 d) x = 1 Answ er Answ er Answ er Parallel and Perpendicular Lines Mathematics problems often deal with parallel and perpendicular lines. Since these are such popular lines, it is important that we remember some information about their slopes. Parallel Lines: (same slope!) Parallel lines are marked with "feathers" to show that they are parallel. These "feathers" look like "greater than" symbols. Parallel lines have the same slope. y = 3x + 5 y = 3x - 7 y = 3x + 0.5 y = 3x These lines are ALL parallel. They all have the same slope (m). (Remember y = mx + b.) Example: The slope of is and . Find the slope of . Since the lines are parallel, the slopes are the same. The slope of is . ANSWER Perpendicular Lines: (negative reciprocal slopes!) Perpendicular lines have negative reciprocal slopes. To get a negative reciprocal of a number, flip the number over (invert) and negate that value. These lines are perpendicular. Their slopes (m) are negative reciprocals. (Remember y = mx + b.) Example: The slope of is and . Find the slope of . Since the lines are perpendicular, the slopes are negative reciprocals. The slope of is . ANSWER EXERCICES Practice with Parallel and Perpendicular Try the following problems: 1. . If the slope of l1 is and the slope of l2 is find the value of x. 2. , , Is the equation y = 3x + 2 parallel to 2y + 3x = 3? Explain. Answ er Explanation 3. Answ er . If the slope of l1 is and the slope of l2 is , , find the value of x. Explanation 4. Find the slope of a line parallel to a line whose slope is . Explanation 5. Find the slope of the line perpendicular to a line whose slope is 6.. . Find the slope of a line parallel to the line whose equation is 3y + 2x = 6. Answ er 7. Find the slope of a line perpendicular to the line whose equation is 3y + 2x = 6. 8. Find the equation of the line parallel to the line whose equation is y = -3x + 5 and whose y-intercept is -5. 9. Find the equation of the line perpendicular to the line whose equation is 2y - 4x = 7 and whose y-intercept is +5. 10. Find the equation of the line perpendicular to the line whose equation is 2y - 4x = 7 and which passes through the point (1,2). MATH 2 (Intermediate Algebra Advance Course NAME : DATE : TEACHER : Mr. E. Ramos jr Handout # : 3 TOPIC : VARIATIONS : Direct, Inverse, and Joint Direct Variation When two variable quantities have a constant (unchanged) ratio, their relationship is called a direct variation. It is said that one variable "varies directly" as the other. The constant ratio is called the constant of variation. The formula for direct variation is y = kx, where k is the constant of variation. "y varies directly as x" Solving for k: (y = numerator; x = denominator) In a direct variation, the two variables change in the same sense. If one increases, so does the other. Graphically, direct variation y = kx when k > 0. As x increases, y increases. Example # 1: The weekly salary a woman earns, S, varies directly as the number of hours, h, which she works. Express this relation as a formula. Answer: S = hk or (where k is the constant) EXAMPLE 2. If y varies directly as x2, and y = 8 when x = 2, find y when x = 1. Since this is direct variation, the formula is "y = kx2". The reason they've given me the data point (x, y) = (2, 8) is that I have to be able to find the value of "k". So I'll plug in the information they've given me, and solve for k: y = kx2 8 = k(22) 8 = 4k 2=k Now that I have k, I can rewrite the formula completely: y = 2x2. With this, I can answer the question they actually asked: "Find y when x = 1." y = 2x2 y = 2(1)2 y = 2×1 y=2 Then the answer is: y = 2Copyright © Eliza EXAMPLE 3. If y varies directly as x and z, and y = 5 when x = 3 and z = 4, then find y when x = 2 and z = 3. Translating the formula from English to math, I get: y = kxz Plugging in the data point they gave me, and solving for the value of k, I get: 5 = k(3)(4) 5 = 12k 5 /12 = k Now that I have the value of k, I can plug in the new values, and solve for the new value of y: y = ( 5/12)xz y = ( 5/12)(2)(3) y = ( 5/12)(6) y = 5 /2 Then the answer is: y = 5/2 Inverse Variation Inverse Variation (The Opposite of Direct Variation) In an inverse variation, the values of the two variables change in an opposite manner - as one value increases, the other decreases. For instance, a biker traveling at 8 mph can cover 8 miles in 1 hour. If the biker's speed decreases to 4 mph, it will take the biker 2 hours (an increase of one hour), to cover the same distance. Inverse variation: when one variable increases, the other variable decreases. As speed decreases, the time increases. Notice the shape of the graph of inverse variation. If the value of x is increased, then y decreases. If x decreases, the y value increases. We say that y varies inversely as the value of x. An inverse variation between 2 variables, y and x, is a relationship that is expressed as: where the variable k is called the constant of proportionality. As with the direct variation problems, the k value needs to be found using the first set of data. Find the Constant, k: The number of hours, h, it takes for a block of ice to melt varies inversely as the temperature, t. If it takes 2 hours for a square inch of ice to melt at 65º, find the constant of proportionality. Start with the formula: Substitute the values : then solve for k: In a formula, Z varies inversely as p. If Z is 200 when p = 4, find Z when p = 10. Typical Inverse Variation Use the same three process steps that were used in direct variation Problem: problems: 1. Set up the formula. 2. Find the missing constant, k, by using the first set of data given. 3. Using the formula and constant, k, find the missing value in the problem. Inverse Variation Example: In kick boxing, it is found that the force, f, needed to break a board, varies inversely with the length, l, of the board. If it takes 5 lbs of pressure to break a board 2 feet long, how many pounds of pressure will it take to break a board that is 6 feet long? 1. Set up the formula. 2. Find the missing constant, k, using the first set of data given. 3. Using the formula and constant, k, find the missing value in the problem. Combination Variation Example: Variable M varies directly as variable t and inversely as variable s. If M = 24 when t = 3 and s = 2, find M when t = 5 and s = 8. ( In combination problems, there is only one constant value, k, used with the direct and inverse variables.) 1. Set up the formula 2. Find the missing constant of proportionality, k. 3. Using the formula and the constant, k, find the new value in the problem Joint Variation Definition of Joint Variation Joint variation is a variation in which y varies jointly as x or powers of x ( y or powers of z ( ), if there is some nonzero constant k such that where x ≠ 0, z ≠ 0, and n > 0. ) and , Practice problems. Using k as the constant of proportionality, write equations that express the following statements: 1. 2 varies jointly as x and y. 2. S varies jointly as b times the square of r. 3. The length, W, of a radio wave varies jointly as the square root of the inductance, L, and the capacitance, C. Answers: Examples of Joint Variation y = 7xz, here y varies jointly as x and z. , here y varies jointly as and . Solved Example on Joint Variation Assume a varies jointly with b and c. If b = 2 and c = 3, find the value of a. Given that a = 12 when b = 1 and c = 6. Choices: A. 2 B. 3 C. 12 D. 24 Correct Answer: C Solution: Step 1: Given that a = 12 when b = 1 and c = 6. Step 2: As a = kbc, . Step 3: For b = 2 and c = 3, a = k × 2 × 3 = 2 × 2 × 3 = 12. Exercises : 1. The weight of a body varies inversely as the square of its distance from the center of the earth. If the radius of the earth is 4000 miles, how much would a 200-pound man weight 1000 miles above the surface of the earth? 2. Under certain conditions, the thrust T of a propeller varies jointly as the fourth power of it diameter d and the square of the number n of revolutions per second. Show that, if n is doubled and d is halved, the thrust T is decreased by 75%. 3. The number of hours h that it takes m men to assemble x machines varies directly as the number of machines and inversely as the number of men. If four men can assemble 12 machines in four hours, how many men are needed to assemble 36 machines in eight hours? 4. There are about 200 calories in 50 grams of Swiss cheese. Willie ate 70 grams of this cheese. About how many calories were in the cheese that he ate if the number of calories varies directly as the weight of the cheese. 5. The intensity(I) of light varies inversely as the square of the distance(d) between the light source and the observer. Suppose the intensity is measured as 46 units at a distance of 2 m, at what distance would be the intensity of the light source, half of what it was?