course MTH 163 Here are the remaining ten questions:*********************************************
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Given Solution: We easily evaluate the expression: * When x = -2, we get y = 2 x + 3 = 2 * (-2) + 3 = -4 + 3 = -1. * When x = -1, we get y = 2 x + 3 = 2 * (-1) + 3 = -2 + 3 = 1. * When x = 0, we get y = 2 x + 3 = 2 * (0) + 3 = 0 + 3 = 3. * When x = 1, we get y = 2 x + 3 = 2 * (1) + 3 = 2 + 3 = 5. * When x = 2, we get y = 2 x + 3 = 2 * (2) + 3 = 4 + 3 = 7. Filling in the table we have x y -2 -1 -1 1 0 3 1 5 2 7 When we graph these points we find that they lie along a straight line. Only one of the depicted graphs consists of a straight line, and we conclude that the appropriate graph is the one labeled 'linear'. ********************************************* Question: `q008. Let y = x^2 + 3. (Note: Liberal Arts Mathematics students are encouraged to do this problem, but are not required to do it). * Evaluate y for x = -2. What is your result? In your solution explain the steps you took to get this result. * Evaluate y for x values -1, 0, 1 and 2. Write out a copy of the table below. In your solution give the y values you obtained in your table. x y -2 -1 0 1 2 * Sketch a graph of y vs. x on a set of coordinate axes resembling the one shown below. You may of course adjust the scale of the x or the y axis to best depict the shape of your graph. * In your solution, describe your graph in words, and indicate which of the graphs depicted previously your graph most resembles. Explain why you chose the graph you did. ********************************************* Your solution: Evaluating y = x^2 + 3, where x = -2 would equate to y = (-2)^2 + 3. First evaluate -2^2 = 4, then add, y = 4 + 3, solution y = 7. y = (-1)^2 + 3 = 1 + 3 = 4 y = (0)^2 + 3 = 0 + 3 = 3 y = (1)^2 + 3 = 1 + 3 = 4 y = (2)^2 + 3 = 4 + 3 = 7 x y -2 7 -1 4 0 3 1 4 2 7 For the set of coordinates, the graph resembles a symmetric 'V' or 'U' shape, which most closely resembles the quadratic or parabolic illustration. Confidence Assessment: 3 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK
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Given Solution: Evaluating y = x^2 + 3 at the five points: * If x = -2 then we obtain y = x^2 + 3 = (-2)^2 + 3 = 4 + 3 = 7. * If x = -1 then we obtain y = x^2 + 3 = (-1)^2 + 3 = ` + 3 = 4. * If x = 0 then we obtain y = x^2 + 3 = (0)^2 + 3 = 0 + 3 = 3. * If x = 1 then we obtain y = x^2 + 3 = (1)^2 + 3 = 1 + 3 = 4. * If x = 2 then we obtain y = x^2 + 3 = (2)^2 + 3 = 4 + 3 = 7. The table becomes x y -2 7 -1 4 0 3 1 4 2 7 We note that there is a symmetry to the y values. The lowest y value is 3, and whether we move up or down the y column from the value 3, we find the same numbers (i.e., if we move 1 space up from the value 3 the y value is 4, and if we move one space down we again encounter 4; if we move two spaces in either direction from the value 3, we find the value 7). A graph of y vs. x has its lowest point at (0, 3). If we move from this point, 1 unit to the right our graph rises 1 unit, to (1, 4), and if we move 1 unit to the left of our 'low point' the graph rises 1 unit, to (-1, 4). If we move 2 units to the right or the left from our 'low point', the graph rises 4 units, to (2, 7) on the right, and to (-2, 7) on the left. Thus as we move from our 'low point' the graph rises up, becoming increasingly steep, and the behavior is the same whether we move to the left or right of our 'low point'. This reflects the symmetry we observed in the table. So our graph will have a right-left symmetry. Two of the depicted graphs curve upward away from the 'low point'. One is the graph labeled 'quadratic or parabolic'. The other is the graph labeled 'partial graph of degree 3 polynomial'. If we look closely at these graphs, we find that only the first has the right-left symmetry, so the appropriate graph is the 'quadratic or parabolic' graph. ********************************************* Question: `q009. Let y = 2 ^ x + 3. (Note: Liberal Arts Mathematics students are encouraged to do this problem, but are not required to do it). * Evaluate y for x = 1. What is your result? In your solution explain the steps you took to get this result. * Evaluate y for x values 2, 3 and 4. Write out a copy of the table below. In your solution give the y values you obtained in your table. x y 1 2 3 4 * Sketch a graph of y vs. x on a set of coordinate axes resembling the one shown below. You may of course adjust the scale of the x or the y axis to best depict the shape of your graph. * In your solution, describe your graph in words, and indicate which of the graphs depicted previously your graph most resembles. Explain why you chose the graph you did. ********************************************* Your solution: Evaluating y = 2^x + 3, where x = 1 would equate to y = 2^(1) + 3. First evaluate 2^1 = 2, then add, y = 2 + 3, solution y = 5. y = 2^(2) + 3 = 4 + 3 = 7 y = 2^(3) + 3 = 8 + 3 = 11 y = 2^(4) + 3 = 16 + 3 = 19 x y 1 5 2 7 3 11 4 19 For the set of coordinates, the graph resembles a curve increasing from the lower left to the upper right, which most closely resembles the illustration of the exponential graph, which curves upward in a similar fashion. Confidence Assessment: 3 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK
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Given Solution: Recall that the exponentiation in the expression 2^x + 1 must be done before, not after the addition. When x = 1 we obtain y = 2^1 + 3 = 2 + 3 = 5. When x = 2 we obtain y = 2^2 + 3 = 4 + 3 = 7. When x = 3 we obtain y = 2^3 + 3 = 8 + 3 = 11. When x = 4 we obtain y = 2^4 + 3 = 16 + 3 = 19. x y 1 5 2 7 3 11 4 19 Looking at the numbers in the y column we see that they increase as we go down the column, and that the increases get progressively larger. In fact if we look carefully we see that each increase is double the one before it, with increases of 2, then 4, then 8. When we graph these points we find that the graph rises as we go from left to right, and that it rises faster and faster. From our observations on the table we know that the graph in fact that the rise of the graph doubles with each step we take to the right. The only graph that increases from left to right, getting steeper and steeper with each step, is the graph labeled 'exponential'. ********************************************* Question: `q010. If you divide a certain positive number by 1, is the result greater than the original number, less than the original number or equal to the original number, or does the answer to this question depend on the original number? ********************************************* Your solution: Any given positive number divided by 1 will be equal to the original given number. Confidence Assessment: 3 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: `q011. If you divide a certain positive number by a number greater than 1, is the result greater than the original number, less than the original number or equal to the original number, or does the answer to this question depend on the original number? ********************************************* Your solution: If you divide a given positive number by a number greater than one, then the result will be less than the original number, as it is being fractioned into smaller parts. Confidence Assessment: 3 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK
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Given Solution: If you split something up into equal parts, the more parts you have, the less will be in each one. Dividing a positive number by another number is similar. The bigger the number you divide by, the less you get. Now if you divide a positive number by 1, the result is the same as your original number. So if you divide the positive number by a number greater than 1, what you get has to be smaller than the original number. Again it doesn't matter what the original number is, as long as it's positive. Students will often reason from examples. For instance, the following reasoning might be offered: OK, let's say the original number is 36. Let's divide 36 be a few numbers and see what happens: 36/2 = 18. Now 3 is bigger than 2, and 36 / 3 = 12. The quotient got smaller. Now 4 is bigger than 3, and 36 / 4 = 9. The quotient got smaller again. Let's skip 5 because it doesn't divide evenly into 36. 36 / 6 = 4. Again we divided by a larger number and the quotient was smaller. I'm convinced. That is a pretty convincing argument, mainly because it is so consistent with our previous experience. In that sense it's a good argument. It's also useful, giving us a concrete example of how dividing by bigger and bigger numbers gives us smaller and smaller results. However specific examples, however convincing and however useful, don't actually prove anything. The argument given at the beginning of this solution is general, and applies to all positive numbers, not just the specific positive number chosen here. ********************************************* Question: `q012. If you divide a certain positive number by a positive number less than 1, is the result greater than the original number, less than the original number or equal to the original number, or does the answer to this question depend on the original number? ********************************************* Your solution: If you divide a positive number by a number less than 1, you are diving the original number by a decimal or fraction. The result therefore becomes greater than the original number because you splitting the original number up into a smaller fraction, resulting in a larger outcome. This is hard to put into understandable words. Confidence Assessment: 3 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK
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Given Solution: If you split something up into equal parts, the more parts you have, the less will be in each one. Dividing a positive number by some other number is similar. The bigger the number you divide by, the less you get. The smaller the number you divide by, the more you get. Now if you divide a positive number by 1, the result is the same as your original number. So if you divide the positive number by a positive number less than 1, what you get has to be larger than the original number. Again it doesn't matter what the original number is, as long as it's positive.