#$&* course MTH 163 October 14th, @ 4:58 PM
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Given Solution: If c = -1 the form y = x^2 + c gives us y = x^2 - 1. Every value on a table of this function would be 1 less than the corresponding value on a table of y = x^2, and the graph of y = x^2 - 1 will lie 1 unit lower at each point then the graph of y = x^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Nailed this one. ------------------------------------------------ Self-critique rating: ********************************************* Question: `q002. Suppose that we wish to graph the functions y = x^2 + c for c = -3, then for c = -2, then for c = -1, then for c = 0, then for c = 1, then for c = 2, then for c = 3. If all these functions were plotted on the same set of coordinate axes, what would the final graph look like? It is suggested that you actually sketch your graph and describe your sketch. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - My graph appeared to be a family of parabolas, resting their vertices along the y axis, from -3 to +3. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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 2 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Are my answers too vague? Or do I need to add more detail? ------------------------------------------------ Self-critique rating:
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - It would look as if three exponential functions stacked on top of each other, with their midpoints running from -2, -3, and -4. Working under the assumption that the sign in the function equation doesn’t change. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The k = 2 graph will lie 2 units to the right of the graph of y = x^3, and the k = 4 graph will lie 4 units to the right. The three graphs will all have the same shape as the y = x^3 graph, but will lie 2, 3 and 4 units to the right. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I have noted that I mistook my graph reading to have lowered the graph instead of move it horizontally. I can completely see where my mistake was now; it will run 2,3,and 4 units to the right (which I originally though would be toward the negative side of the y-axis). ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q005. The function y = 3 * 2^x is of the form y = A * 2^x for A = 3. What function would this form give us for A = 2? How would the graph of this function compare with that of y = 2^x? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - Without checking with a calculator, I could see that perhaps the change in A relates to raising the graph up three units from the original function, y = 2^x. Perhaps this refers to being three times as far from the x axis. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. STUDENT QUESTION In this case, is A stretching or shifting? I said shifting in my response, but as I'm recalling from the previous assignment I suspect it may actually be stretching the graph by 2 units instead of shifting. INSTRUCTOR RESPONSE Your suspicion is correct. The graph is stretching. When a graph shifts, every point moves by the same amount. When a graph stretches, every point moves to a multiple of its original distance from some axis. Points further from that axis move more, points closer to the axis move less. For example the graph below depicts the x any y axes, and the graphs of the two functions. Vertical lines are drawn at x = 1 and x = 2. You should sketch a good copy of this graph and actually trace out the properties discussed below, and annotate your graph accordingly: Look first at the vertical line corresponding to x = 1. It should be clear that, along this line, the graph of the second function is about twice as far from the x axis as the graph of the first. Now look at the vertical line corresponding to x = 2. It should be clear that, along this line, the graph of the second function is about twice as far from the x axis as the graph of the first. You should also see that along the x = 1 line the second graph lies at a certain distance above the first, while at the x = 2 line the second graph lies at a greater distance above the first. At the origin, the graphs meet. Then if we move from left to right, starting at the origin, the vertical distance between the graphs keeps increasing. The graph below includes 'heavier' vertical line segments representing the increasing vertical distance between the graphs: This graph represents a vertical stretch, in every point of the second graph lies at double the vertical distance from the horizontal axis as the corresponding point of the first. By contrast, consider the graph shown below, in which the second graph is shifted in the vertical direction relative to the first. On this graph the vertical distance is the same on every vertical line. It probably doesn't look like this is the case. There's an optical illusion at work here, which is due to the fact that the upper graph gets closer and closer to the lower graph. However, despite appearances, this isn't the case if the distances are measured along the vertical lines. In the first figure below we show the line segments which represent these vertical distances. They are all of the same length. In the second figure below, these line segments are depicted without the graph. In this example, all points of the second graph lie at the same vertical distance above the first. This graph represents a vertical shift. In a vertical shift all points point of one graph lie at the same vertical displacement relative to the first. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Just to clarify the difference between a shift and a stretch, the shifting will result in all points moving the same exact number of units, whereas a stretch will do the same, except the farther away from the x-axis (in this case) will cause it to ‘stretch’ three times as much? Or maybe a better way to put it is that the higher on the original graph, the more it will ultimately move. ------------------------------------------------ Self-critique rating:
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Given Solution: 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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q007. What is the slope of a straight line connecting the points (3, 8) and (9, 12)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - I did this purely with using just the graph. My points had a midline, (6,10), which I used to estimate that the slope was 2/3. Basically, my mind tells me to think of slope being Rise/Run, so I count up two points and over three points to meet the ‘midpoint’. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The rise between the points is from y = 8 to y = 12. This is a rise of 12-8 = 4. The run between these points is from x = 3 to x = 9, a run of 9 - 3 = 6. The slope between these points is therefore rise/run = 4/6 = 2/3, with decimal equivalent .6666.... &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Is the way I figured this fine? Or was it too simple and subject to inaccuracy? This was of course using a protractor and graphing paper, so I would assume using numbers and calculations would provide more accuracy when the aforementioned tools aren’t available. ------------------------------------------------ Self-critique rating:
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Given Solution: The t = 5 value is y = 2 * 5^2 + 3 = 2 * 25 + 3 = 50 + 3 = 53. The t = 9 value is similarly calculated. We obtain y = 165. The rise between these points is therefore 165-53 = 112. The run is from t = 5 to t = 9, a run of 9 - 5 = 4. This slope of a straight line connecting these points is therefore rise/run = 112/4 = 28. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Nailed this one. ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q009. Suppose y = 2 t^2 + 3 represents the depth, in cm, of water in a container at clock time t, in seconds. At what average rate does the depth of water change between t = 5 seconds and t = 9 seconds? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - These types of problems always throw me for a bit of a loop and I think about it too hard. - When I see “average” or “mean” I think to add the numbers and divide by how many numbers in question. - Thus, the rate changes at 7 cm per second. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): When given a word problem, I mess up. But, in the exact same circumstance, I got it right. Hm. Regardless, I think I can see the connection here, but I would ask for a deeper explanation. ------------------------------------------------ Self-critique rating: