#$&* course Mth163 Question: `q001. Note that this assignment has 12 questions
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Given Solution: If c = -1 the form y = x^2 + c gives us y = x^2 - 1. Every value on a table of this function would be 1 less than the corresponding value on a table of y = x^2, and the graph of y = x^2 - 1 will lie 1 unit lower at each point then the graph of y = x^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q002. Suppose that we wish to graph the functions y = x^2 + c for c = -3, then for c = -2, then for c = -1, then for c = 0, then for c = 1, then for c = 2, then for c = 3. If all these functions were plotted on the same set of coordinate axes, what would the final graph look like? It is suggested that you actually sketch your graph and describe your sketch. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& c = -3, shifted down three units vertically c = -2, shifted down two units vertically c = -1, shifted down one unit vertically c = 0, due to horizontal and vertical asymptotes zero is undefined because it never touches x axis or y c = 1, shifts one unit vertically upward c = 2, shifts two units vertically upward c = 3 shifts three units vertically upward confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): The big picture will have a series of 7 functions with the lowest units below the horizontal asymptote and the highest three units about the original (or horizontal asymptote). Each graph will lie higher than the last. ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& The overall function would move to the right k units. The graph would have shifted 3 units to the right compared to the original. confidence rating #$&*:: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Recall how the graph of y = (x-1)^3 lies one unit to the right of the graph of y = x^3. The k = 3 function y = (x -3)^3 will lie 3 units to the right of the graph of y = x^3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q004. Suppose we wish to graph the functions y = (x -k)^3 for k values 2, then 3, then 4. If we graph all these functions on the same set of coordinate axes, what will the graph look like? It is suggested that you actually sketch your graph and describe your sketch. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& My y coordinates are labeled by twenties and x coordinates by ones. 2=k -125 -64 -27 -8 -1 0 1 My curve starts at (-3,-125) touches zero at -1 then ends at coordinates (3,1) 3=k -216 -125 -64 -27 -8 -1 Starts at (-3,-216) goes through zero at y=3 and ends at (0,3). Also it touches the other graph at (1,-8) 4=k -343 -216 -125 -64 -27 -8 -1 Each are one unit to the right of the other &&& confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique rating: OK ********************************************* 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: &&& confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):
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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): A controls steepness (as in slope as well??) ------------------------------------------------ Self-critique rating:2
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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): Self-critique rating:OK ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& (5,53) (9,165) Your slope is 28 &&& confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique rating:OK ********************************************* 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: &&& The average rate of change in water depth is 28 cm/sec &&& confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The t = 5 second and t = 9 second depths are easily calculated to be y = 53 cm and y = 165 cm. The depth therefore changes from 53 cm to 165 cm, a change of 165 cm - 53 cm = 112 cm, in the 4 seconds between the to clock times. The average rate of depth changes therefore 112 cm/(4 seconds) = 28 cm/second. We note that this calculation is identical to the calculation of the slope of the straight line connecting the t = 5 and t = 9 points of the graph y = 2 t^2 + 3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q010. If we have a graph of depth y vs. clock time t, then how to we know that the slope of a straight line connecting two points of the graph represents the average rate at which depth changes over the corresponding time interval? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& Change in y / change in x &&& confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:OK If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties. ********************************************* Question: `q011. Consider the family of functions of form y = 2 ( x - h ) ^ 2. What is the vertex of the graph of the h = 0 function? What is the vertex of the graph of the h = -1 function? What is the vertex of the graph of the h = 1 function? What are the other two basic points of each of these functions? Graph these three functions and describe your graph. Make a sketch representing the family y = 2 ( x - h ) ^ 2 for -4 <= h <= 2 and describe your sketch. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& I’m going to say x=2 for these functions. . h=0 vertex @ (2,8) other two basic points (3,18)(1,2) . h=-1 vertex @ (2,9) other two basic points at (3,16)(1,4) . h=1 vertex @ (2,9) other two basic points at (3,16)(1,4)
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