course MTH 163 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `aThe graph goes through the x axis when y = 0 and through the y axis when x = 0. The x-intercept is therefore when 0 = 3x - 4, so 4 = 3x and x = 4/3. The y-intercept is when y = 3 * 0 - 4 = -4. Thus the x intercept is at (4/3, 0) and the y intercept is at (0, -4). Your graph should confirm this. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): same Self-critique Rating: 3 ********************************************* Question: `q002. Does the steepness of the graph in the preceding exercise (of the function y = 3x - 4) change? If so describe how it changes. ********************************************* Your solution: No, the “steepness” aka slope (rise/run) is a constant value of 3. The line is straight. Confidence Assessment: 3
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Given Solution: `aThe graph forms a straight line with no change in steepness. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): same Self-critique Rating: 3 ********************************************* Question: `q003. What is the slope of the graph of the preceding two exercises (the function ia y = 3x - 4;slope is rise / run between two points of the graph)? ********************************************* Your solution: x y -3 -13 -2 -10 -1 -7 0 -4 1 -1 1 1/3 0 2 2 3 5 Slope is easily determined from the graph where the change in x values of one unit (run), is paired with a change in y value of 3 units (rise). It can also be calculated by selecting any two pairs of values and determining the change in the values of the y pairs divided by the change in the values of the x pairs. (-13 - -4) / (-3 – 0) = -9 / -3 = 3 Similarly, using another pair of values, we get (2-5) / (-2-3) = -3/-1 = 3 Confidence Assessment: 3
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Given Solution: `aBetween any two points of the graph rise / run = 3. For example, when x = 2 we have y = 3 * 2 - 4 = 2 and when x = 8 we have y = 3 * 8 - 4 = 20. Between these points the rise is 20 - 2 = 18 and the run is 8 - 2 = 6 so the slope is rise / run = 18 / 6 = 3. Note that 3 is the coefficient of x in y = 3x - 4. Note the following for reference in subsequent problems: The graph of this function is a straight line. The graph increases as we move from left to right. We therefore say that the graph is increasing, and that it is increasing at constant rate because the steepness of a straight line doesn't change. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): same Self-critique Rating:3 ********************************************* Question: `q004. Make a table of y vs. x for y = x^2. Graph y = x^2 between x = 0 and x = 3. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate? ********************************************* Your solution: x y 0 0 1 1 2 4 3 9 y = x^2 This graph is increasing at an increasing rate. This is visually apparent from the graph, and can easily be confirmed by calculating the slope between adjacent points (0-1)/ (0-1) = 1/1 = 1 as beginning slope (1-4)/(1-2) = -3/-1 = 3 as slope of next segment (4-9)/(2-3) = -5/-1 = 5 as slope of next segment Confidence Assessment: 3
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Given Solution: `aGraph points include (0,0), (1,1), (2,4) and (3,9). The y values are 0, 1, 4 and 9, which increase as we move from left to right. The increases between these points are 1, 3 and 5, so the graph not only increases, it increases at an increasing rate STUDENT QUESTION: I understand increasing...im just not sure at what rate...how do you determine increasing at an increasing rate or a constant rate? INSTRUCTOR RESPONSE: Does the y value increase by the same amount, by a greater amount or by a lesser amount every time x increases by 1? In this case the increases get greater and greater. So the graph increases, and at an increasing rate. *&*&. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): same Self-critique Rating: 3 ********************************************* Question: `q005. Make a table of y vs. x for y = x^2. Graph y = x^2 between x = -3 and x = 0. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate? x y -3 9 -2 4 -1 1 0 0 ********************************************* Your solution: y=x^2 This segment is decreasing at a decreasing rate. The change is measured from left to right (for increasing x values) and is clearly visible from the graph. In addition, it can be calculated for each segment by calculating the difference in values between adjacent values. (9-4)/(-3- -2) = 5/-1 = -5 slope (4-1)/(-2 - -1) = 3/-1 = -3 slope (1-0) / (-1-0) = 1/-1 = -1 slope Negative values indicate a decreasing slope, smaller number of units indicates a decreasing rate. Confidence Assessment: 3
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Given Solution: `aFrom left to right the graph is decreasing (points (-3,9), (-2,4), (-1,1), (0,0) show y values 9, 4, 1, 0 as we move from left to right ). The magnitudes of the changes in x from 9 to 4 to 1 to 0 decrease, so the steepness is decreasing. Thus the graph is decreasing, but more and more slowly. We therefore say that the graph is decreasing at a decreasing rate. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: 3 ********************************************* Question: `q006. Make a table of y vs. x for y = `sqrt(x). [note: `sqrt(x) means 'the square root of x']. Graph y = `sqrt(x) between x = 0 and x = 3. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate? ********************************************* Your solution: y=√x x y 0 0 1 1 2 √2 ≈1.414 3 √3≈1.732 The slope is increasing, at a decreasing rate. This can be seen visually or mathematically Graph is now marked in .25 unit grid, with heavier lines marking off single unit blocks. Calculating the *average slope of each segment of the line between the points we get: (0-1)/(0-1) = -1/-1 = 1 Slope of 1 is clearly visible on the graph for this segment. (1-1.414)/(1-2) ≈ -.414/-1≈ .414 Slope of this segment is approximately .414, actual value would be √2 (1.414 – 1.732) / (2-3) ≈ -.318/-1 ≈ .318. This is also an approximation as values for both √2 and √3 are approximated. Comparing the *average slopes for these sections shows a change from 1, to .414, to .318. Clearly the slope is increasing, but at a decreasing rate. *I use *Average slope rather than “slope” as the actual slope between these points cannot be assumed to be a straight line rather than a curve. In fact, a calculation of the values of x and y where x=.25 will clearly NOT be on the line for that segment. (.25,.5). Confidence Assessment: 3
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Given Solution: `aIf you use x values 0, 1, 2, 3, 4 you will obtain graph points (0,0), (1,1), (2,1.414), (3. 1.732), (4,2). The y value changes by less and less for every succeeding x value. Thus the steepness of the graph is decreasing. The graph would be increasing at a decreasing rate. If the graph respresents the profile of a hill, the hill starts out very steep but gets easier and easier to climb. You are still climbing but you go up by less with each step, so the rate of increase is decreasing. If your graph doesn't look like this then you probably are not using a consistent scale for at least one of the axes. If your graph isn't as desribed take another look at your plot and make a note in your response indicating any difficulties. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):same Self-critique Rating: 3 ********************************************* Question: `q007. Make a table of y vs. x for y = 5 * 2^(-x). Graph y = 5 * 2^(-x) between x = 0 and x = 3. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate? ********************************************* Your solution: y=5*2^(-x) = 5 * [1/(2^x)] x y *slope y calculations 0 5.0 5*[1/(2^0)] 5*[1/1] 5*1 1 2.5 -2.5 5*[1/(2^1)] 5*[1/2] 5*.5 2 1.25 -1.25 5*[1/(2^2)] 5*[1/4] 5*.25 3 .625 -.625 5*[1/(2^3)] 5*[1/8] 5*.125 This graph clearly is decreasing, at a decreasing rate, supported by the *slope calculations between specified points added to the chart above (-2.5, then -1.25, then -.625). Once again slope cannot be assumed to follow the straight line use to illustrate the segment being calculated. Actual value would more closely resemble a curve reaching toward but not meeting the x axis. Confidence Assessment: 3
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Given Solution: `a** From basic algebra recall that a^(-b) = 1 / (a^b). So, for example: 2^-2 = 1 / (2^2) = 1/4, so 5 * 2^-2 = 5 * 1/4 = 5/4. 5* 2^-3 = 5 * (1 / 2^3) = 5 * 1/8 = 5/8. Etc. The decimal equivalents of the values for x = 0 to x = 3 will be 5, 2.5, 1.25, .625. These values decrease, but by less and less each time. The graph is therefore decreasing at a decreasing rate. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Same Self-critique Rating: 3 ********************************************* Question: `q008. Suppose you stand still in front of a driveway. A car starts out next to you and moves away from you, traveling faster and faster. If y represents the distance from you to the car and t represents the time in seconds since the car started out, would a graph of y vs. t be increasing or decreasing? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate? ********************************************* Your solution: t y Sample Slope 0 0 1 y2 1 1 .1 .001 2 y3 2 1.5 .11 .002 3 y4 3 2 .111 .003 If it is “traveling faster and faster”, clearly the distance will be increasing at an increasing rate. Speed refers to distance per unit of time (e.g. miles per hour) and it is given that the speed in increasing. We know all distances traveled are positive numbers (or it could not be moving away), and that each distance is greater than the last (or speed cannot continue to increase) Confidence Assessment: 3
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Given Solution: `a** The speed of the car increases so it goes further each second. On a graph of distance vs. clock time there would be a greater change in distance with each second, which would cause a greater slope with each subsequent second. The graph would therefore be increasing at an increasing rate. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): same Self-critique Rating: 3