#$&* course Mth163
.............................................
Given Solution: The graph of y = x consists of a straight line through the origin, having slope 1. This line has basic points (0,0) and (1,1). The points of the graph of y = .5 x all lie twice as close to the x-axis as the points of the graph of y = x. The point (0,0) of the y = x graph is already on the x-axis, so the corresponding point on the graph of y = .5 x is also (0,0). The point (1,1) of the y = x graph lies 1 unit above the x-axis, so the corresponding point on the graph of y = .5 x will lie twice as close, or .5 units above the x-axis, so that the corresponding point is (1, .5). The graph of y = .5 x Thus passes through the points (0,0) and (1,.5). Of course this result could have been found by simply plugging 0 and 1 into the function y = .5 x, but the point here is to see that we can get the same result if we think of moving all points twice as close. This order thinking will be useful when dealing with more complex functions. Thinking along similar lines we expect the points of the graph of y = 2 x to all lie twice as far from the x-axis as the points of the function y = x. Thus the two basic points (0,0) and (1,1) of the y = x graph will correspond to the points (0,0) and (1,2) on the graph of y = 2 x. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): .5 would be .5 times as close to x axis (which is also twice as close) Y=2x would lie twice as far from x axis ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `q002. If we were to sketch all the graphs of the form y = a x for which .5 < a < 2, what would our sketch look like? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&& . a=1,1.5 . y=1x .y=1.5x &&&& confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: If a =.5 then our function is y = .5 x and the basic points will be (0,0) and (1,.5), as seen in the preceding problem. Similarly if a = 2 then our function is y = 2 x, with basic points (0,0) and (1,2). For .5 < a < 2, our functions will lie between the graphs of y = .5 x and y = 2 x. Since these two functions have slopes .5 and 2, the slopes of all the graphs will lie between .5 and 2. We could represent these functions by sketching dotted-line graphs of y = .5 x and y = 2 x (the dotted lines indicating that these graphs are not included in the family, because the < sign does not include equality). We could then sketch a series of several solid lines through the origin and lying between the two dotted-line graphs. STUDENT COMMENT I don’t know where to go from this point. I graphed the closest thing I could come up with but I don’t know how to explain what it is doing. INSTRUCTOR SUGGESTION You should graph the functions y = .5 x, y = .6 x, y = .8 x, y = 1.1 x, y = 1.5 x and y = 2 x, all on the same graph. Graph each function by plotting its two basic points (the x = 0 point and the x = 1 point), then sketching the straight line through these points. Using your graphs, estimate where the graph of y = .7 x, y = 1.3 x and y = 1.8 x lie. Then insert your description, according to instructions at the end of this document, along with any other work you do in response to other suggestions made below, and resubmit this document. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): &&&&&& “INSTRUCTOR SUGGESTION You should graph the functions y = .5 x, y = .6 x, y = .8 x, y = 1.1 x, y = 1.5 x and y = 2 x, all on the same graph.” Why do we not include every decimal place for the one values? Like 1.2,1.3,1.4,1.6,1.7,1.8,1.9
.............................................
Given Solution: The graph of y = x - 2 lies at every point 2 units below the corresponding point on the graph of y = x, so this graph is parallel to the graph of y = x and 2 units lower. Similarly the graph of y = x + 3 lies parallel to the graph of y = x and 3 units higher. To sketch the family y = x + c for -2 < x < 3, we first can draw dotted-line graphs of y = x - 2 and y = x + 3, then a series of several solid line graphs, all parallel to the graph of y = x, lying between the two dotted-line graphs. STUDENT COMMENT: I got a little confused with y = x + c part, but I understand the first part completely. ** The instructions said to sketch all graphs of the form y = x + c for -2 < x < 3. So for example c could be -1, 0, 1 or 2, giving us the functions y = x - 1, y = x, y = x + 1 andy x+ 2. c could also be -1.9, or .432, or 2.9, giving us functions y = x - 1.9, y = x + .432, y = x + 2.9. c can be any number between -2 and 3. These graphs are as described in the given solution. ** STUDENT COMMENT i didnt really understand how to sketch y=x+c even after reading the intructors comments in the given solution INSTRUCTOR RESPONSE Suppose you were to graph y = x + c for c values -2, -1.9, -1.8, -1.7, ..., 2.8, 2.9, 3.0. This would include 50 graphs. Each of the 50 graphs would lie .1 unit higher than the one before it. The lowest of the graphs would be the c = -2 graph, y = x - 2. The highest of the graphs would be the c = 3 graph, y = x + 3. All the graphs would be parallel. If necessary, you can graph y = x - 2, then y = x - 1.9, then y = x - 1.8. You won't want to graph all 50 lines, but you could then skip to y = x + 2.8, y = x + 2.9 and y = x + 3. STUDENT COMMENT After reading the comments above I agree that I am a little confused. INSTRUCTOR RESPONSE You need to self-critique, giving me a detailed statement of what you do and do not understand about each line and each phrase in the given solution. You should in any case follow the suggestion at the end of the given solution. Graph the indicated graphs, then insert your explanation. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): &&& We would plot all the functions on the same graph and when we do that we apply the previous measurement to the last??? &&& ------------------------------------------------ Self-critique rating:2
.............................................
Given Solution: The graph of y = 2 x lies at every point twice as far the x-axis as that of y = x. This graph passes through the points (0,0) and (1, 2), i.e., passing through the origin with slope 2. The graph of y = 2x - 2 will lie 2 units below the graph of y = 2 x. This graph will therefore have a slope of 2 and will pass-through the y axis at (0, -2). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): &&& The graph of y = 2 x has a slope that increases by 2 (from zero) making the slope 2 The graph of y = 2x - 2 will be underneath the previous graph the slope will be 2 as well because it passes through the y axis at neg two. &&& ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `q005. Suppose we graph y = 2 x + c for all values of c for which -2 < c < 3. What with our graph look like? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&& All graphs will be parallel with differences in there vertical shifts making them thinner and thinner y = 2 x -2 vertical stretch 2 vertical shift down 2 y = 2 x -1vertical stretch 2 vertical shift down 1 y = 2 x +0 vertical stretch 2 y = 2 x +1 vertical stretch 2 vertical shift up 1 y = 2 x +2 vertical stretch 2 vertical shift up 2 y = 2 x +3 vertical stretch 2 vertical shift up 3 &&& confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: Each graph will lie c units vertically from the graph of y = 2 x, therefore having slope 2 the passing through the y-axis at the point (0, c). The family of functions defined by y = 2 x + c will therefore consist of a series of straight lines each with slope 2, passing through the y-axis between (0, -2) and (0, 3). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): &&& Slope is two because your replacing m=slope with 2 &&& ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `q006. Sketch two points, not particularly close to one another, with one point in the second quadrant and the other in the first, with clearly different y values. Label the first point (x1, y1) and the second (x2, y2). Draw a straight line passing through both of these points and extending significantly beyond both. In terms of the symbols x1, x2, y1, and y2, what is slope of this straight line? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& (y2-y1)/(x2-x1) &&& confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: The rise of a line is from y = y1 to y = y2, a rise of y2-y1. The run is similarly found to be x2-x1. The slope is therefore slope = (y2-y1) / (x2-x1). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q007. On the sketch you made for the preceding problem, and add a point (x, y) on your straight line but not between the two points already labeled, and not too close to either. What is the slope from (x1, y1) to (x, y)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& (y-y1)/(x-x1) &&& confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: The slope from (x1, y1) to (x, y) is slope = rise/run = (y - y1) / (x - x1). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q008. Should the slope from (x1, y1) to (x, y) be greater than, equal to or less than the slope from (x1, y1) to (x2, y2)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& It all depends on where you put it the dots for your graph, on mine they are equal to each other. It also depends on what your x and y units are and what your labeled them. I feel like there are a lot of unknowns. &&& confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: The slope between any two points of a straight line must be the same. The two slopes must therefore be equal. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): &&& The slopes between any two points of a straight line must be the same (is this a law??? If not why???) So therefore the slopes must be equal. &&& ------------------------------------------------ Self-critique rating:2 ********************************************* Question: `q009. The slope from (x1, y1) to (x, y) is equal to the slope from (x1, y1) to (x2, y2). If you set the expressions you obtained earlier for the slopes equal to one another, what equation do you get? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& (y-y1)/(x-x1) = (y2-y1)/(x2-x1) &&& confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: The slopes are (y2 - y1) / (x2 - x1) and (y - y1) / (x - x1). Setting the two slopes equal we obtain the equation (y - y1) / (x - x1) = (y2 - y1) / (x2 - x1). STUDENT COMMENT mine is the opposite but i think i would be the same INSTRUCTOR RESPONSE: Your solution was (y2 - y1) / (x2 - x1), if appropriate signs of grouping are inserted to reflect your obvious intent. The signs of both your numerator and denominator would be opposite the signs of the given solution (i.e., y2 - y1 = - (y1 - y2), and x2 - x1 = - (x1 - x2)). When divided the result would therefore be identical (negative / negative is positive). So your solution is completely equivalent to the given solution. However note that you need to group numerator and denominator. y2-y1/x2-x1 means divide y1 by x2, subract that result from y2 then subtract x1 from that. Not what you intended, though I know what you meant. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q010. Setting the two slopes equal you should have obtained (y - y1)/(x - x1) = (y2 - y1)/(x2 - x1). Solve this equation for y. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& (x-x1)(y2-y1)=(y-y1)(x2-x1) . y2(x-x1)-y1(x-x1)=y(x2-x1)-y1(x2-x1) . y2(x)-y2(x1)-y1(x)+y1(x1)=y(x2)-y(x1)-y1(x2)+y1(x1) &&& confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: Starting with (y - y1) / (x - x1) = (y2 - y1) / (x2 - x1), we wish to isolate y on the left-hand side. We therefore begin by multiplying both sides by (x - x1) to obtain (y - y1) = (y2 - y1) / (x2 - x1) * (x - x1). We could then add y1 to both sides to obtain y = (y2 - y1) / (x2 - x1) * (x - x1) + y1. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): &&& I am a little confused when you multiply something over in the dominator to the other side (of equal sign) doesn’t it move to the numerator because you have had to flip it in order to multiply????