course mth 163 K|Θf^מrrStudent Name:
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19:11:01 `q001. Note that this assignment has 8 questions Evaluate the function y = x^2 for x values -3, -2, -1, 0, 1, 2, and 3. What are your y values?
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RESPONSE --> for -3, the y value is 9 for -2, the y value is 4 for -1, the y value is 1 for 0, the y value is 0 for 1, the y value is 1 for 2, the y value is 4 for 3, the y value is 9
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19:11:11 You should have obtained y values 9, 4, 1, 0, 1, 4, 9, in that order.
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RESPONSE --> ok
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19:13:07 `q002. Evaluate the function y = 2^x for x values -3, -2, -1, 0, 1, 2, and 3. What are your y values?
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RESPONSE --> the values i got, in order: 1/8, 1/4, 1/2, 1, 2, 4, 8.
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19:13:26 By velocity exponents, b^-x = 1 / b^x. So for example 2^-2 = 1 / 2^2 = 1/4. Your y values will be 1/8, 1/4, 1/2, 1, 2, 4 and 8. Note that we have used the fact that for any b, b^0 = 1. It is a common error to say that 2^0 is 0. Note that this error would interfere with the pattern or progression of the y values.
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RESPONSE --> ok
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19:15:35 `q003. Evaluate the function y = x^-2 for x values -3, -2, -1, 0, 1, 2, and 3. What are your y values?
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RESPONSE --> the values i got, in order: 1/9, 1/4, 1, 0, 1, 1/4, 1/9.
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19:16:13 By the laws of exponents, x^-p = 1 / x^p. So x^-2 = 1 / x^2, and your x values should be 1/9, 1/4, and 1. Since 1 / 0^2 = 1 / 0 and division by zero is not defined, the x = 0 value is undefined. The last three values will be 1, 1/4, and 1/9.
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RESPONSE --> oh, oops. I forgot that would make it division by 0, which would mean it was undefined.
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19:17:38 `q004. Evaluate the function y = x^3 for x values -3, -2, -1, 0, 1, 2, and 3. What are your y values?
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RESPONSE --> I got the values -27, -8, -1, 0, 1, 8, and 27.
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19:18:45 The y values should be -27, -8, -1, 0, 1, 4, 9.
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RESPONSE --> Shouldn't those last two be 8 and 27?
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19:27:48 09-24-2006 19:27:48 `q005. Sketch graphs for y = x^2, y = 2^x, y = x^-2 and y = x^3, using the values you obtained in the preceding four problems. Describe the graph of each function.
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NOTES -------> y = x^2 is a parabola. It is symmetric about the y axis, with its vertex at the origin. Each y value has two corresponding x values, which are identical except for their sign. y = 2^x has an asymptote as y approaches 0 going left, and the y values increase at a greater and greater rate, going right. y = ^-2 is symmetric about the y axis, with a break in the graph at x = 0. Each half of the graph is asymptotic as it approaches y = 0. y = x^3 is antisymmetric about the origin. The upper and lower halves of the graph curve in opposite directions, but are otherwise the same shape.
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19:31:10 The graph of y = x^2 is a parabola with its vertex at the origin. It is worth noting that the graph is symmetric with respect to the y-axis. That is, the graph to the left of the y-axis is a mirror image of the graph to the right of the y-axis. The graph of y = 2^x begins at x = -3 with value 1/8, which is relatively close to zero. The graph therefore starts to the left, close to the x-axis. With each succeeding unit of x, with x moving to the right, the y value doubles. This causes the graph to rise more and more quickly as we move from left to right. The graph intercepts the y-axis at y = 1. The graph of y = x^-2 rises more and more rapidly as we approach the y-axis from the left. It might not be clear from the values obtained here that this progression continues, with the y values increasing beyond bound, but this is the case. This behavior is mirrored on the other side of the y-axis, so that the graph rises as we approach the y-axis from either side. In fact the graph rises without bound as we approach the y-axis from either side. The y-axis is therefore a vertical asymptote for this graph. The graph of y = x ^ 3 has negative y values whenever x is negative and positive y values whenever x is positive. As we approach x = 0 from the left, through negative x values, the y values increase toward zero, but the rate of increase slows so that the graph actually levels off for an instant at the point (0,0) before beginning to increase again. To the right of x = 0 the graph increases faster and faster. Be sure to note whether your graph had all these characteristics, and whether your description included these characteristics. Note also any characteristics included in your description that were not included here.
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RESPONSE --> OK. The graph of y=x^3 has negative Ys for negative Xs, and vice versa, and levels off at (0,0).
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19:33:21 `q006. Make a table for y = x^2 + 3, using x values -3, -2, -1, 0, 1, 2, 3. How do the y values on the table compare to the y values on the table for y = x^2? How does the graph of y = x^2 + 3 compare to the graph of y = x^2?
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RESPONSE --> The y values for each one are 3 greater than the values for y = x^2. The graph is the same, except that it has been shifted up 3 units.
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19:33:31 A list of the y values will include, in order, y = 12, 7, 4, 3, 4, 7, 12. A list for y = x^2 would include, in order, y = 9, 4, 1, 0, 1, 4, 9. The values for y = x^2 + 3 are each 3 units greater than those for the function y = x^2. The graph of y = x^2 + 3 therefore lies 3 units higher at each point than the graph of y = x^2.
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RESPONSE --> ok
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19:38:26 `q007. Make a table for y = (x -1)^3, using x values -3, -2, -1, 0, 1, 2, 3. How did the values on the table compare to the values on the table for y = x^3? Describe the relationship between the graph of y = (x -1)^3 and y = x^3.
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RESPONSE --> The values are -64, -27, -8, -1, 0, 1, and 8. The graph is the same, except that it has been shifted to the right 1 unit.
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19:38:45 The values you obtained should have been -64, -27, -8, -1, 0, 1, 8. The values for y = x^3 are -27, -8, -1, 0, 1, 8, 27. The values of y = (x-1)^3 are shifted 1 position to the right relative to the values of y = x^3. The graph of y = (x-1)^3 is similarly shifted 1 unit to the right of the graph of y = x^3.
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RESPONSE --> ok
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19:44:51 `q008. Make a table for y = 3 * 2^x, using x values -3, -2, -1, 0, 1, 2, 3. How do the values on the table compare to the values on the table for y = 2^x? Describe the relationship between the graph of y = 3 * 2^x and y = 2^x.
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RESPONSE --> The values are 3/8, 3/4, 3/2, 3, 6, 12, and 24. These are each 3 times as high as the values for y = 2^x. The graph is the same shape, but it's steeper as it moves right, and increass faster.
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19:45:00 You should have obtained y values 3/8, 3/4, 3/2, 3, 6, 12 and 24. Comparing these with the values 1/8, 1/4, 1/2, 1, 2, 4, 8 of the function y = 2^x we see that the values are each 3 times as great. The graph of y = 3 * 2^x has an overall shape similar to that of y = 2^x, but each point lies 3 times as far from the x-axis. It is also worth noting that at every point the graph of y = 3 * 2^x is three times as the past that of y = 2^x.
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RESPONSE --> ok
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cܯT] Student Name: assignment #006
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23:06:44 `q001. Note that this assignment has 10 questions Recall that the graph of y = x^2 + 3 was identical to the graph of y = x^2, except that it was raised 3 units. This function is of the form y = x^2 + c. In the case of this specific function, c = 3. What function would this form give us for c = -1? How would the graph of this function compare with the graph of y = x^2?
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RESPONSE --> if c = -1, the function would be y = x^2 - 1. The graph would look the same as y = x^2, except it would be one unit lower.
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23:07:18 If c = -1 the form y = x^2 + c gives us y = x^2 - 1. Every value on a table of this function would be 1 less than the corresponding value on a table of y = x^2, and the graph of y = x^2 - 1 will lie 1 unit lower at each point then the graph of y = x^2.
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RESPONSE --> Yes, the values on the table would be one less, as well.
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23:09:21 `q002. Suppose that we wish to graph the functions y = x^2 + c for c = -3, then for c = -2, then for c = -1, then for c = 0, then for c = 1, then for c = 2, then for c = 3. If all these functions were plotted on the same set of coordinate axes, what would the final graph look like? It is suggested that you actually sketch your graph and describe your sketch.
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RESPONSE --> The final graph is a series of parabolas, each one idential except for their vertical location. They are each one unit apart from the one(s) above or below them-- the lowest three units below y=x^2, and the highest three above it.
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23:09:44 The graph of the c= -3 function y = x^2 - 3 will lie 3 units lower than the graph of y = x^2. The graph of the c= -2 function y = x^2 - 2 will lie 22222 units lower than the graph of y = x^2. The progression should be obvious. The graph of the c= 3 function y = x^2 + 3 will lie 3 units higher than the graph of y = x^2. The final graph will therefore show a series of 7 functions, with the lowest three units below the parabolic graph of y = x^2 and the highest three units above the graph of this function. Each graph will lie one unit higher than its predecessor.
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RESPONSE --> ok
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23:15:45 `q003. The function y = (x -1)^3 is of the form y = (x -k)^3 with k = 1. What function would this form give us for k = 3? How would the graph of this function compare with that of y = x^3?
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RESPONSE --> The function we'd get if k=3 would be y = (x-3)^3. The graph would look like the graph of y=x^3, but shifted three units to the right.
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23:15:56 Recall how the graph of y = (x-1)^3 lies one unit to the right of the graph of y = x^3. The k = 3 function y = (x -3)^3 will lie 3 units to the right of the graph of y = x^3.
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RESPONSE --> ok
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23:18:51 `q004. Suppose we wish to graph the functions y = (x -k)^3 for k values 2, then 3, then 4. If we graph all these functions on the same set of coordinate axes, what will the graph look like? It is suggested that you actually sketch your graph and describe your sketch.
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RESPONSE --> The graph where k = 2 is 2 units right of where the y = x^3 graph would be; where k = 3, it's 3 units right of it, and where k = 4, it's 4 units right of it.
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23:19:22 The k = 2 graph will lie 2 units to the right of the graph of y = x^3, and the k = 4 graph will lie 4 units to the right. The three graphs will all have the same shape as the y = x^3 graph, but will lie 2, 3 and 4 units to the right.
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RESPONSE --> ok
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23:22:03 `q005. The function y = 3 * 2^x is of the form y = A * 2^x for A = 3. What function would this form give us for A = 2? How would the graph of this function compare with that of y = 2^x?
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RESPONSE --> If A = 2, the function we got would be y = 2 * 2^x. Each y value would be twice as large as those of y = 2^x, making the graph the same general shape, but twice as steep.
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23:22:19 As we saw earlier, the graph of y = 3 * 2^x lies 3 times as far from the x-axis as a graph of y = 2^x and is at every point three times as steep. We would therefore expect the A = 2 function y = 2 * 2^x to lie 2 times is far from the x-axis as the graph of y = 2^x.
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RESPONSE --> ok
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23:25:01 `q006. Suppose we wish to graph the functions y = A * 2^x for values A ranging from 2 to 5. If we graph all such functions on the same set of coordinate axes, what will the final graph look like? It is suggested that you actually sketch your graph and describe your sketch.
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RESPONSE --> It will be a series of graphs, each one with y values 2, 3, 4, or 5 times those of the original y = 2^x graph, and each 2, 3, 4, or 5 times as steep.
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23:26:36 These graphs will range from 2 times as far to 5 times as far from the x-axis as the graph of y = 2^x, and will be from 2 to 5 times as steep. The y intercepts of these graphs will be (0,2), (0, 3), (0, 4), (0,5).
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RESPONSE --> OK, yeah, the intercepts are also 2, 3, 4 or 5 times higher as the intercept for y = 2^x, which has a y intercept of 1.
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23:27:42 `q007. What is the slope of a straight line connecting the points (3, 8) and (9, 12)?
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RESPONSE --> The slope is rise/run = 4/6 which reduces to 2/3.
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23:27:56 The rise between the points is from y = 8 to y = 12. This is a rise of 12-8 = 4. The run between these points is from x = 3 to x = 9, a run of 9 - 3 = 6. The slope between these points is therefore rise/run = 4/6 = 2/3, with decimal equivalent .6666....
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RESPONSE --> ok
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23:30:54 `q008. What are the coordinates of the t = 5 and t = 9 points on the graph of y = 2 t^2 + 3? What is the slope of the straight line between these points?
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RESPONSE --> y = 2(5)^2 + 3 = 2(25) + 3 = 53 so (5, 53) y = 2(9)^2 + 3 = 2(81) + 3 = 162 + 3 = 165 so (9, 165) the slope would be (165 - 53) / (9 - 5) = 112/4, or 28.
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23:31:12 The t = 5 value is y = 2 * 5^2 + 3 = 2 * 25 + 3 = 50 + 3 = 53. The t = 9 value is similarly calculated. We obtain y = 165. The rise between these points is therefore 165-53 = 112. The run is from t = 5 to t = 9, a run of 9 - 5 = 4. This slope of a straight line connecting these points is therefore rise/run = 112/4 = 28.
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RESPONSE --> ok
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23:33:29 `q009. Suppose y = 2 t^2 + 3 represents the depth, in cm, of water in a container at clock time t, in seconds. At what average rate does the depth of water change between t = 5 seconds and t = 9 seconds?
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RESPONSE --> `dy/`dx would be (165 - 53) / (9 - 5) = 28 cm/sec.
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23:33:39 The t = 5 second and t = 9 second depths are easily calculated to be y = 53 cm and y = 165 cm. The depth therefore changes from 53 cm to 165 cm, a change of 165 cm - 53 cm = 112 cm, in the 4 seconds between the to clock times. The average rate of depth changes therefore 112 cm/(4 seconds) = 28 cm/second. We note that this calculation is identical to the calculation of the slope of the straight line connecting the t = 5 and t = 9 points of the graph y = 2 t^2 + 3.
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RESPONSE --> ok
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23:34:38 `q010. If we have a graph of depth y vs. clock time t, then how to we know that the slope of a straight line connecting two points of the graph represents the average rate at which depth changes over the corresponding time interval?
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RESPONSE --> It calculates the difference in the depth divided by the difference in the time, which is what a rate is.
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23:35:16 The rise of the graph represents the change in the depth y and the run represents the change clock time t. The slope, which is rise/run, therefore represents change in depth/change in clock time, which is the average rate at which the depth changes.
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RESPONSE --> ok
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y^\J㪴z assignment #005 Ťexzw Precalculus I 09-24-2006
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18:01:52 query introduction to basic function families problem 1 on basic graphs Why is the graph of y = x a straight line?
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RESPONSE --> Because if the x coordinate is 5, then the y coordinate is also 5; If x is 7, then y is 7; if x is 9, y is 9, and so on. This makes the slope of the line 1, meaning that for every unit you move vertically, you move one unit horizontally, as well. This makes it a straight line.
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18:01:59 ** Since y = x the rise and run between any two points on the graph are equal, which makes the slope 1. A graph with constant slope is a straight line. **
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RESPONSE --> ok
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18:03:57 why is y = x^2 symmetric about x = 0 (i.e., taking the same values on either side of x = 0)
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RESPONSE --> The vertex is at (0,0), and a parabola is symmetric about its vertex. This makes it so that there are two x values for every y value on the graph. If y = 1, then both 1 and -1, when squared, will give you 1. If y = 4, then both 2 and -2 will give you 4. Etc.
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18:04:24 ** The graph of y = x^2 is symmetric about x = 0 because (-x)^2 = x^2. Thus for any point on the x axis the y values at that point and at the point on the opposite side of the origin are equal, so that the graph on one side of the y axis is a 'reflection' of the graph on the other side. **
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RESPONSE --> ok
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18:08:06 why does y = 2^x keep increasing as x increases, and why does the graph approache the x axis for negative values of x
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RESPONSE --> As x increases, the y values grow larger and larger, because 2 is being raised to a higher and higher power. 2^2 is 4, 2^3 is 8, 2^4 is 16, and so on. The values increase exponentially. It approaches the x axis for negative values because the values are becoming smaller and smaller-- 2^(-2) is 1/4, 2^(-3) is 1/8, 2^(-4) is 1/16, etc. However, nothing you raise 2 to is ever going to give you 0, so it just approaches the axis.
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18:08:23 ** GOOD STUDENT RESPONSE: y = 2^x will increase as x increases on the positive side because x is the value of the exponent. This will cause the y value to double from its last value when you move one unit in the positive x direction. On the negative side of the y axis y = 2^x will approach the x axis because a negative exponent causes the value to invert into a fractional value of itself--i.e., 2^(-x) = 1 / 2^x. As we move one unit at a time negatively the value will become one half of the previous value so it will never quite reach y = 0. **
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RESPONSE --> ok
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18:09:53 why is y = x^3 antisymmetric about x = 0 (i.e., taking the same values except for the - sign on opposite sides of x = 0)
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RESPONSE --> Raising x to an odd power means that negative x values can give you negative y values, unlike raising x to an even power, which can give only positive y values. x^(-3) is -8, not 8, so it has the same shape of graph as x^3, but the sign is reversed.
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18:10:12 ** y = x^3 is antisymmetric because if you cube a negative number you get a negative, if you cube a positive number you get a positive, and the magnitude of the cubed number is the cube of the magnitude of the number. So for example (-3)^2 = -27 and 3^3 = 27; the points (-3, -27) and (3, 37) are antisymmetric, one being `down' while the other is `up'. GOOD STUDENT RESPONSE: y = x^3 is antisymmetric about x = 0 because the exponent is an odd number. This will cause negative x values to have a negative y result. The absolute value of the negative y result will be equivalent to its corresponding positive y value. **
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RESPONSE --> ok
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18:17:28 why do y = x^-2 and y = x^-3 rise more and more steeply as x approaches 0, and why do their graphs approach the x axis as we move away from the y axis.
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RESPONSE --> They rise as x approaches zero because their values increase. Since raising to a negative exponent makes a number less than one, the closer to zero the x value is, the higher the y will be. Their graphs approach the x axis as we move away from the y axis for the opposite reason: the higher the x values, the lower the y values.
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18:18:15 ** And as x approaches 0 the expressions x^-2 and x^-3, which mean 1 / x^2 and 1 / x^3, have smaller and smaller denominators. As the denominators approach zero their reciprocals grow beyond all bound. y = x^-2 and y = x^-3 rise more and more steeply as x approaches zero because they have negative exponents they become fractions of positive expressions x^2 and x^3 respectively which have less and less slope as they approach zero. As x^2 and x^3 approach zero and become fractional, x^-2 and x^-3 begin to increase more and more rapidly because thier functions are then a whole number; (1) being divided by a fraction in which the denominator is increasing at an increasing rate. As y = x^-2 and y = x^-3 move away from the y-axis they approach the x-axis because they have negative exponents. This makes them eqivalent to a fraction of 1 / x^2 or 1 / x^3. As x^2 and x^3 increase in absolute value, the values of y = x^-2 and y = x^-3 constantly close in on the x-axis by becoming a portion of the remaining distance closer, they will never reach x = zero though as this would be division by zero (since it is a fraction) **
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RESPONSE --> ok
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18:20:49 query problem 2. family y = x^2 + c Explain why the family has a series of identical parabolas, each 1 unit higher than the one below it.
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RESPONSE --> The c value is a vertical shift. Whatever x is, it moves the parabola that many units up or down. Everything else about the parabolas is the same, though, so the vertical location is the only thing changing.
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18:20:58 ** GOOD STUDENT RESPONSE: The graph of y = x^2 + c, with c varying from -5 to 4 is a series of identical parabolas each 1 unit higher than the one below it. The c value in the quadratic equation has a direct impact on the vertical shift. The vertex of the graph will be shifted vertically by the amount of the c value, so every time c increases by 1 the graph is raised 1 unit. **
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RESPONSE --> ok
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18:25:44 query problem 4. describe the graph of the exponential family y = A * 2^x for the values A = -3 to 3.
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RESPONSE --> All the graphs are asymptotic to the x axis s the graph moves left, but vary in how much they are vertically stretched. The graphs of -3 and 3 are the same except for their direction-- -3 curves down, while 3 curves up, and the same is true of -2 and 2, -1 and 1. When A = 0, the graph is just the x axis.
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18:35:04 ** This family includes the functions y = -3 * 2^x, y = -2 * 2^x, y = -1 * 2^x, y = 0 * 2^x, y = 1 * 2^x, y = 2 * 2^2 and y = 3 * 2^x. Each function is obtained by vertically stretching the y = 2^x function. y = -3 * 2^x, y = -2 * 2^x, y = -1 * 2^x all vertically stretch y = 2^x by a negative factor, so the graphs all lie below the x axis, asymptotic to the negative x axis and approaching negative infinity for positive x. They pass thru the y axis as the respective values y = -3, y = -2, y = -1. y = 1 * 2^x, y = 2 * 2^x, y = 3 * 2^x all vertically stretch y = 2^x by a positive factor, so the graphs all lie above the x axis, asymptotic to the negative x axis and approaching positive infinity for positive x. They pass thru the y axis as the respective values y = 1, y = 2, y = 3. y = 0 * 2^x is just y = 0, the x axis. Of course the functions for fractional values are also included (e.g., y = -2.374 * 2^x) but only the integer-valued functions need to be included in order to get a picture of the behavior of the family. ** STUDENT QUESTION: Ok, it was A = -3 to 3. I understand how to substitute these values into y = A * 2^x. I knew that is was an asymptote, but I'm a little confused as to how to graph the asymptote. INSTRUCTOR RESPONSE: For each value of A you have a different function. For A = -3, -2, -1, 0, 1, 2, 3you have seven different functions, so you will get 7 different graphs. Each graph will contain the points for all values of x. For example the A = -3 function is y = -3 * 2^x. This function has basic points (0, -3) and (1, -6). As x takes on the negative values -1, -2, -3, etc., the y values will be -1.5, -.75, -.375, etc.. As x continues through negative values the y values will approach zero. This makes the y axis a horizontal asymptote for the function. You should figure out the x = 0 and x = 1 values for every one of these seven functions, and you should be sure you understand why each function approaches the negative x axis as an asymptote. *&*&
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RESPONSE --> ok
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18:37:13 describe the graph of the exponential family y = 2^x + c for the values c = -3 to 3.
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RESPONSE --> The graphs were different only in their vertical positions. When c = -3, the graph was 3 units lower than y = 2x^2, when it was -1, it was 1 unit lower, etc. When c was 1, the opposite was true-- the graph was 1 unit higher than y = 2x^2, and so on.
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18:40:03 ** There are 7 graphs, including y = 2^x + 0 or just y = 2^x. The c = 1, 2, 3 functions are y = 2^x + 1, y = 2^x + 2 and y = 2^x + 3, which are shifted by 1, 2 and 3 units upward from the graph of y = 2^x. The c = -1, -2, -3 functions are y = 2^x - 1, y = 2^x - 2 and y = 2^x - 3, which are shifted by 1, 2 and 3 units downward from the graph of y = 2^x. **
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RESPONSE --> all right
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18:46:09 query problem 5. power function families Describe the graph of the power function family y = A (x-h) ^ p + c for p = -3: A = 1, h = -3 to 3, c = 0.
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RESPONSE --> The graphs move from left to right one unit each time, depending on the value of h, which is the only difference between them. Each graph was split into two sections, which were not connected. All of them have a horizontal asymptote at the x axis (in the negative direction on the lower section of the graph, and the positive direction on the upper section of the graph), but each one has a different vertical asymptote.
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18:57:44 ** GOOD STUDENT RESPONSE: I sketched the graph of the power function family y = A (x-h)^p + c for p = -3: A = 1: h = -3 to 3, c = 0. Beginning on the left side of the graph the curve was infinitely close to its asmptote of y = 0. This was determined by the value of c. As we move from left to right the curves decreased at an increasing rate, approaching thier vertical asmptotes which was determined by thier individual values of h. The curves broke at x = c as this value was never possible due to division by zero. The curves resurfaced on the graph high on the right side of thier vertical asymptotes and from there they decreased at a decreasing rate, once again approaching thier horizontal asymptote of y = 0. INSTRUCTOR COMMENTS: Only the h value changes. p=-3, A=1 and c=0, so the functions are y = 1 * (x-h)^-3 or y = (x-h)^-3. For h = -3 to 3 the functions are y = (x - (-3))^-3, y = (x - (-2))^-3, y = (x - (-1))^-3, y = (x - 0)^-3, y = (x - 1)^-3, y = (x - 2)^-3, y = (x - 3)^-3. These graphs march from left to right, moving 1 unit each time. Be sure you see in terms of the tables why this happens. **
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RESPONSE --> ok, I forgot to mention that the reason they broke was that x = c would mean you had to divide by 0, which isn't possible.
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19:00:18 query problem 10 illumination. What function did you evaluate to get your results?
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RESPONSE --> To get my results, I used the function 370x^(-1). I got this from y = A(x-h)^p + c. h and c were both zero, which simplified it to y = Ax^p, then i substituted in the A (370) and the p (-1).
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19:00:28 ** I determined the illumination y from a certain florescent bulb at the distances of 1, 2, 3, and 4 units using the generalized power function for p = -1 with A = 370, h = 0 and c = 0. This power function is y = A (x- h)^p + c = 370 (x - 0)^(-1) + 0, or just y = 370 x^-1. **
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RESPONSE --> ok
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19:04:48 Determine the illumination at distances of 1, 2, 3 and 4 units, and sketch a graph.
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RESPONSE --> At 1, the illumination is 370(1)^(-1) = 370. At 2, it's 370(2)^(-1) = 370(1/2) = 185 at 3, it's 370(3)^(-1) = 370(1/3) = 123.3333 at 4, it's 370(4)^(-1) = 370(1/4) = 92.5 that gives you the ordered pairs for the graph: (1, 370), (2, 185), (3, 123.333), (4, 92.5)
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19:04:55 ** Student Solution: For x=1 we obtain y=370(-1-0) ^-1=370 For x=2 we obtain y=370(2-0)^-1=185 For x=3 we obtain y=370(3-0)^-1 =123.3 For x=4we obtain y=370(4-0)^-1=92.5**
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RESPONSE --> ok
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19:05:17 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE -->
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19:05:19 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> It's interesting to connect the graphs to the functions, and see WHY there are asymptotes for certain graphs. When it makes logical sense (not being able to divide by 0, for instance), it makes it easier to remember and understand.
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19:07:50 ** STUDENT COMMENT: I have never worked with graphs in the power family, and very little in the exponential family. I am always amazed at the patterns that a function produces. It helps me understand the equation so much better than a list of numbers. I do feel that I need the data table with the graph to fully understand it. INSTRUCTOR RESPONSE: The data table is certainly helpful, especially when you see the reasons for the number patterns in the formula as well as you do. **
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RESPONSE -->
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