course Mth 163 Week 5 Quiz 1Version 31 The graph below depicts the demand for a commodity, in thousands, vs. selling price, in thousands of dollars. Construct a good straight-line model for this data and use your model to determine the demand at a selling price of 60 thousand dollars. Determine also the demand at which the selling price will be 23 thousand dollars . The x axis is marked in units of 10 thousand dollars and the y axis in units of 21 thousand units, both starting at (0,0). Interpret both the x and the y intercepts of this graph. I divided the values by 1000 to do away with the zeros for the time being (8000, 137000) = (8,137) (30000, 0 ) = (30,0) 137= 8m+b 0=30m+b subtract ---------- 137 = -22m m=-6.227273 137=8(-6.227273) + b b=186.818182 y=-6.227273x + 186.818182 $60 = -6.227273 + 186.818182 x=20.36496264 * 1000 = 20364.96264 demand $23 = -6.227273x + 186.818182 x=26.30656822 * 1000 = 26306.56822 demand x-int 0=-6227273x + 186.818182 = 29.99999872 * 1000 = (29999.99872,0) y-int 186.818182 * 1000 = (0, 186818.182) ????v?????o??assignment #010 ??????????????Precalculus I 10-08-2006
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14:06:48 query the family of linear functions, Problem 2. Describe the graphs of y = A f(x) for A = -.3 and A = 1.3 and compare; explain the comparison.
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RESPONSE --> Both graphs have a y-int of (0,0) A=-.3 is in the negative direction the point one to the right is (1,-.3) A=1.3 is in the positive direction the point one to the right is (1,1.3) This means that A affects the distance from the x-axis and can make the graph negative if the value of A is negative.
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14:07:00 ** For the basic linear function f(x) = x the A = -.3 graph is obtained by vertically stretching the y = x function by factor -.3, resulting in a straight line thru the origin with slope -.3, basic points (0,0) and (1, -.3), and the A = 1.3 graph is obtained by vertically stretching the y = x function by factor 1.3, resulting in is a straight line thru the origin with slope 1.3, basic points (0,0) and (1, 1.3). **
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14:09:30 describe the graphs of y = f(x) + c for c = .3 and c = -2.7 and compare; explain the comparison.
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RESPONSE --> The graphs have the same slope, making them parallel. When c=.3, the y-int is .3, meaning the graph of y=f(x) was shifted vertically .3. When c=-2.7 the y-int is -2.7 meaning the graph of y=f(x) was shifted vertically -2.7. the value of c causes the graph of f(x) to be shifted vertically, if it is positive, the graph shifts upwards and if the value is negative, the graph shifts downwards.
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14:09:34 ** The graphs will have slopes identical to that of the original function, but their y intercepts will vary from -2.7 to .3. **
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14:10:35 query problem 4. linear function y = f(x) = -1.77 x - 3.87 What are your symbolic expressions, using x1 and x2, for the corresponding y coordinates y1 and y2.
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RESPONSE --> y1= -1.77x1-3.87 y2= -1.77x2-3.87
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14:13:48 ** y1 = f(x1) = -1.77 x1 - 3.87 y2 = f(x2) = -1.77 x2 - 3.87. `dy = y2 - y1 = -1.77 x2 - 3.87 - ( -1.77 x1 - 3.87) = -1.77 x2 + 1.77 x1 - 3.87 + 3.87 = -1.77 ( x2 = x1). Thus slope = `dy / `dx = -1.77 (x2 - x1) / (x2 - x1) = -1.77. This is the slope of the straight line, showing that these symbolic calculations are consistent. **
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RESPONSE --> I didn't include the slope in my answer, but it clearly makes sense.
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14:15:18 query problem 5. graphs of families for y = mx + b. Describe your graph of the family: m = 2, b varies from -3 to 3 by step 1.
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RESPONSE --> The graph is of seven parallel lines with slopes of 1. The y intercepts vary from -3 to 3 by a step of 1 (the b values).
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14:17:05 ** The graphs will all have slope 2 and will pass thru the y axis between y = -3 and y = 3. The family will consist of all such graphs. **
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RESPONSE --> I graphed y=x+b with the variation of b, I did not take into consideration the vertical stretch of 2 brought on by the 2 infront of the x, it got looked over.
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14:18:34 query problem 6. three basic points graph of y = .5 x + 1 what are your three basic points?
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RESPONSE --> y-intercept: (0,1) x-intercept: (-2,0) point 1 unit to right of y-int: (1,1.5)
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14:18:40 ** This is of the form y = mx + b with b= 1. So the y intercept is (0, 1). The point 1 unit to the right is (1, 1.5). The x-intercept occurs when y = 0, which implies .5 x + 1 = 0 with solution x = -2, so the x-intercept is (-2, 0). **
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14:19:34 query problem 6. three basic points graph of y = .5 x + 1 What are your three basic points?
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RESPONSE --> this is the same question that was just asked. y-int: (0,1) x-int: (-2,0) (1,1.5)
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14:19:47 ** The y intercept occurs when x = 0, at which point we have y = .5 * 0 + 1 = 1. So one basic point is (0, 1). The point 1 unit to the right of the y axis occurs at x = 1, where we get y = .5 * 1 + 1 = 1.5 to give us the second basic point (1, 1.5) }The third point, which is not really necessary, is the x intercept, which occurs when y = 0. This gives us the equation 0 = .5 x + 1, with solution x = -2. So the third basic point is (-2, 0). **
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14:22:09 query problem 7. simple pendulum force vs. displacement What are your two points and what line do you get from the two points?
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RESPONSE --> two points: (.6,2.5) and (1,4) Line from these points: y=1.5x+.25
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14:22:31 STUDENT RESPONSE: The two points are (1.1, .21) and (2.0, .54). These points give us the two simultaneous equations .21- m(1.1) + b .54= m(2.0) +b. If we solve for m and b we will get our y = mx + b form. INSTRUCTOR COMMENT: I believe those are data points. I doubt if the best-fit line goes exactly through two data points. In the future you should use points on your sketched line, not data points. However, we'll see how the rest of your solution goes based on these points. **
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14:22:52 what equation do you get from the slope and y-intercept?
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RESPONSE --> y=1.5x+.25
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14:23:10 STUDENT RESPONSE: b= .21 m=.19 INSTRUCTOR COMMENT: ** b would be the y intercept, which is not .21 since y = .21 when x = 1.1 and the slope is nonzero. If you solve the two equations above for m and b you obtain m = .367 and b = -.193. This gives you equation y = mx + b or y = .367 x - .193. **
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14:23:41 what is your linear regression model?
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RESPONSE --> y=4.033576733x+ 0.01061598951
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14:23:56 ** Your linear regression model would be obtained using a graphing calculator or DERIVE. As a distance student you are not required to use these tools but you should be aware that they exist and you may need to use them in other courses. **
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14:25:26 What force would be required to hold the pendulum 47 centimeters from its equilibrium position? what equation did you solve to obtain this result?
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RESPONSE --> 47cm = 4.033577x+ 0.010616 46.989384=4.033577x x=11.649557 lbs. I solved the linear regression equation obtained from derive.
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14:25:37 ** If your model is y = .367 x - .193, with y = force and x= number of cm from equilibrium, then we have x = 47 and we get force = y = .367 * 47 - .193 = 17 approx. The force would be 17 force units. **
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14:28:45 Why would it not make sense to ask what force would be necessary to hold the pendulum 80 meters from its equilibrium position? what equation did you solve to obtain this result?
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RESPONSE --> y is measured in cm not m, but that can be fixed, but 80 meters isn't tangible in a real life situation really. 8000=4.033577x + 0.010616 7999.989384 = 4.033577x x = 1983.348622 lbs
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14:29:09 STUDENT RESPONSE: I used the equation f= .10*47+.21 and got the answer 15.41 which would be to much force to push or pull INSTRUCTOR COMMENT: ** The problem is that you can't hold a pendulum further at a distance greater than its length from its equilibrium point--the string isn't long enough. **
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14:31:01 How far could you hold the pendulum from its equilibrium position using a string with a breaking strength of 25 pounds? what equation did you solve to obtain this result?
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RESPONSE --> y=4.033577(25)+0.010616 = 100.849585 cm or i guess if you wanted to be truely safe you could solve the equation y=4.033577(24)+0.010616 = 96.816464 cm
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14:31:22 ** Using the model y = .367 * x - .193 with y = force = 25 lbs we get the equation 25 = .367 x - .193, which we solve to obtain x = 69 (approx.). Note that this displacement is also unrealistic for this pendulum. **
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14:32:09 What is the average rate of change associated with this model? Explain this average rate in common-sense terms.
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RESPONSE --> average rate of change = 1.025/0.2525 = 4.059406, this means the average change in distance in respect to the average force applied.
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14:33:16 ** Using the model y = .367 * x - .193 with y = force and x = displacement from equilibrium we can use any two (x, y) points to get the rate of change. In all cases we will get rate of change = change in y / change in x = .367. The change in y is the change in the force, while the change in x is the change in position. The rate of change therefore tells us how much the force changes per unit of change in position (e.g., the force increases by 15 pounds for every inch of displacement). **
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RESPONSE --> I had the two values switched (distance/force) which instead of force/distance.
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14:36:22 What is the average slope associated with this model? Explain this average slope in common-sense terms.
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RESPONSE --> I found the average slope has the same value of 4.059406 and the slope shoes the change in distance in respect to the pounds of force applied. I had been trying to figure out why my values were so different from those that are being given, and I allowed the force to be the x value and the distance to be the y value, as opposed to what I believe is being provided here with distance as the x value and force as the y value.
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14:36:29 ** Using the model y = .367 * x - .193 with y = force and x = displacement from equilibrium the average slope is .367. **
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14:38:18 As you gradually pull the pendulum from a point 30 centimeters from its equilibrium position to a point 80 centimeters from its equilibrium position, what average force must you exert?
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RESPONSE --> 30 = 4.033577x + 0.010616 x=7.434935 lbs 80=4.033577x + 0.010616 x=19.830881 lbs 7.434935 = 19.830881 = 27.265816 / 2 = 13.632908 lbs
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14:39:07 ** if it was possible to pull the pendulum back this far and if the model applies you will get Force at 30 cm: y = .367 * 30 - .193 = 10.8 approx. and Force at 80 cm: y = .367 * 80 - .193 = 29 approx. so that ave force between 30 cm and 80 cm is therefore (10.8 + 29) / 2 = 20 approx.. **
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RESPONSE --> I followed the same process, but as before, my x and y values were switched making my model differ.
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14:41:01 query problem 8. flow range What is the linear function range(time)?
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RESPONSE --> (34,80) => 80 = 34m + b (97,20) => 20 = 97m + b subtract to get 60=-63m => m=-.952381 20=97(-.952381)+b b=112.380952 y=-.952381x + 112.380952
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14:44:30 ** STUDENT RESPONSE: I obtained model one by drawing a line through the data points and picking two points on the line and finding the slope between them. I then substituted this value for m and used one of my data points on my line for the x and y value and solved for b. the line I got was range(t) = -.95t + 112.38. y = -16/15x + 98 INSTRUCTOR COMMENT: This looks like a good model. According to the instructions it should however be expressed as range(time) = -16/15 * time + 98. **
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RESPONSE --> I didn't put it back into the range(time) format, but where did the -16/15 come from? the m value was found by dividing 60/-63 which results in about -.952381 where -16/15 gives about 1.0666667
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14:45:21 What is the significance of the average rate of change? Explain this average rate in common-sense terms.
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RESPONSE --> The average rate of change, in this particular scenario tells the average flow range change with respect to clock time, so how many cm change there is per sec.
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14:45:30 ** the average rate of change is change in range / change in clock time. The average rate of change indicates the average rate at which range in cm is changing with respect to clock time in sec, i.e., the average number of cm / sec at which the range changes. Thus the average rate tells us how fast, on the average, the range changes. **
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14:46:41 What is the average slope associated with this model? Explain this average slope in common-sense terms.
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RESPONSE --> The average slope is (20-80)/(97-34) = -60/63 = -.952381. This tells how much the flow range changes per sec.
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14:46:54 ** it's the average rate at which the range of the flow changes--the average rate at which the position of the end of the stream changes. It's the speed with which the point where the stream reaches the ground moves across the ground. **
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14:49:06 query problem 9. If your total wealth at clock time t = 0 hours is $3956, and you earn $8/hour for the next 10 hours, then what is your total wealth function totalWealth( t ), where t is time in hours?
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RESPONSE --> totalWealth(10)=8(10)+3950 =80+3950 =$4030
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14:50:29 ** Total wealth has to be expressed in terms of t. A graph of total wealth vs. t would have y intercept 3956, since that is the t = 0 value, and slope 8, since slope represents change in total wealth / change in t, i.e., the number of dollars per hour. A graph with y-intercept b and slope m has equation y = m t + b. Thus we have totalWealth(t) = 8 * t + 3956 . **
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RESPONSE --> My 3956 morphed into a 3950 is the only setback.
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14:51:10 At what clock time will your total wealth reach $4000? what equation did you solve to obtain this result?
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RESPONSE --> 4000=8x+3956 44=8x x=5.5
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14:51:30 STUDENT RESPONSE: To find the clock time when my total wealth will reach 4000 I solved the equation totalWealth(t) = 4000. The value I got when I solved for t was t = 5.5 hours. 4.4 hours needed to reach 4000 4000 = 10x + 3956 INSTRUCTOR COMMENT: Almost right. You should solve 4000 = 8 x + 3956, obtaining 5.5 hours. This is equivalent to solving totalWealth(t) = 4000 = 8 t + 3956, which is the more meaningful form of the relationship. **
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14:51:50 What is the meaning of the slope of your graph?
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RESPONSE --> The slope is how many dollars are earned per hour.
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14:52:00 GOOD STUDENT RESPONSE: The slope of the graph shows the steady rate at which money is earned on an hourly basis. It shows a steady increase in wealth.
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14:54:02 query problem 10. Experience shows that when a certain widget is sold for $30, a certain store can expect to sell 200 widgets per week, while a selling price of $28 increases the number sold to 300. What linear function numberSold(price) describes this situation?
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RESPONSE --> (30,200) => 200=30m+b (28,300) => 300=28m+b subtract to get -100 = 2m => m=-50 200=30(-50)+b 200=-1500+b 1700 = b Numbersold(price)=-50(price)+1700
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14:55:44 If the store must meet a quota by selling 220 units per week, what price should they set? what equation did you solve to obtain this result?
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RESPONSE --> 220 = -50(price) + 1700 -1480 = -50 (price) (price) = $29.60
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14:55:57 ** If the variables are y and x, you know y so you can solve for x. For the function numberSold(price) = -50 * price + 1700 you substitute 220 for numbersold(price) and solve for price. You get the equation 220 = -50 * price + 1700 which you can solve to get price = 30, approx. **
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15:00:52 If each widget costs the store $25, then how much total profit will be expected from selling prices of $28, $29 and $30? what equation did you solve to obtain this result?
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RESPONSE --> numbersold(28)=-50(28)+1700 => 300 widgets * $25 = $7500; $28*300 widgets = $8400 -$7500 = $900 profit numbersold(29)=-50(29)+1700 => 250 widgets * $25 = $6250; $29*250 widgets = $7250 -$6250 = $1000 profit numbersold(30)=-50(30)+1700 => 200 widgets * $25 = $5000; $30*200 widgets = $6000-$5000 = $1000 profit
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15:01:04 STUDENT RESPONSE: If each widget costs the store $25, then they should expect to earn a profit of 300 dollars from a selling price of $28, 250 dollars from a price of $29 and 200 dollars from a price of $30. To find this I solved the equations numberSold(28); numberSold(29), and numberSold(30). Solving for y after putting the price values in for p. They will sell 300, 250 and 200 widgets, respectively (found by solving the given equation). To get the total profit you have to multiply the number of widgets by the profit per widget. At $28 the profit per widgit is $3 and the total profit is $3 * 300 = $900; at $30 the profit per widgit is $5 and 200 are sold for profit $1000; at $29 what happens? **
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15:04:14 query problem 11. quadratic function depth(t) = .01 t^2 - 2t + 100 representing water depth vs. What is the equation of the straight line connecting the t = 20 point of the graph to the t = 60 point?
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RESPONSE --> depth(20)=.01(20)^2-2(20)+100 = .01(400)-40+100 = 4-40+100 = 64 so, (20,64) depth(60)=.01(60)^2-2(60)+100 = .01(3600)-120+100 = 36-120+100 = 16 so, (60,16) 64=20m+b 16=60m+b subtract to get 48 = -40m => m=-1.2 64=20(-1.2)+b 64=.24+b b=88 y=-1.2x+88
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15:04:51 ** The t = 20 point is (20,64) and the t = 60 point is (60, 16), so the slope is (-48 / 20) = -1.2. This can be plugged into the form y = m t + b to get y = -1.2 t + b. Then plugging in the x and y coordinates of either point you get b = 88. y = -1.2 t + 88 **
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RESPONSE --> I did not put back into depth(t) form. so my y would be depth(t) and my x would be t.
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15:08:54 ** For t = 30 we have y = 49 and for t = 40 we have y = 36. The slope between (30, 49) and (40,36) is (36 - 49) / (40 - 30) = -1.3. This tells you that the depth is changing at an average rate of -1.3 cm / sec. **
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RESPONSE --> Somehow when i was typing it got switched over to the next answer. But anyway, i solved this problem as follows: depth(30)=.01(30)^2-2(30)+100 = 49 depth(40)=.01(40)^2-2(40)+100 = 36 so, 'dy/'dx = (36-49)/(40-30) = -1.3 cm/sec
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15:10:10 what is `dy / `dt based on t = 30 sec and t = 31 sec.
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RESPONSE --> (30,49) depth(31)=.01(31)^2-2(31)+100 = 47.61 'dy/'dx = (49-47.61)/(30-31) = 1.39/-1 = -1.39 cm/sec
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15:10:13 ** Based on t = 30 and t = 31 the value for `dy / `dt is -1.39, following the same steps as before **
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15:11:32 what is `dy / `dt based on t = 30 sec and t = 30.1 sec.
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RESPONSE --> (30,49) depth(30.1) = .01(30.1)^2-2(30.1)+100 = 48.8601 'dy/'dx = (49-48.8601)/(30-30.1) = .1399/-.1 = -1.399 cm/sec
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15:11:36 ** STUDENT RESPONSE: The value for 'dy / `dt based on t = 30 sec and t = 30.1 sec is -1.4 INSTRUCTOR COMMENT: ** Right if you round off the answer. However the answer shouldn't be rounded off. Since you are looking at a progression of numbers (-1.3, -1.39, and this one) and the differences in these numbers get smaller and smaller, you have to use a precision that will always show you the difference. Exact values are feasible here and shoud be used. I believe that this one comes out to -1.399. **
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15:12:04 What do you think you would get for `dy / `dt if you continued this process?
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RESPONSE --> If this process was continued, the closer the t values are the more precise the resulting average rate will be.
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15:12:37 STUDENT RESPONSE: An even more and more accurate slope value. I don't think it would have to continue to decrease. INSTRUCTOR COMMENT **If you look at the sequence -1.3, -1.39, -1.399, ..., what do you think happens? It should be apparent that the limiting value is -1.4 **
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15:14:27 What does the linear function tell you?
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RESPONSE --> The linear function rate(t) = .02t-2 tells you that the slope of the line is .02 and the y-int is -2
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15:15:31 ** The function tells you that at any clock time t the rate of depth change is given by the function .02 t - 2. For t = 30, for example, this gives us .02 * 30 - 2 = -1.4, which is the rate we obtained from the sequence of calculations above. **
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RESPONSE --> Yes, the function gives the depth change for a given clock time, I was thinking more specifically in forms of the resulting graph from the function.
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15:15:55 query problem 14. linear function y = f(x) = .37 x + 8.09
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RESPONSE --> question?
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15:15:58 .
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RESPONSE -->
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15:16:36 what are the first five terms of the basic sequence {f(n), n = 1, 2, 3, ...} for this function.
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RESPONSE --> 8.46, 8.83, 9.20, 9.57, 9.94
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15:16:44 ** The first five terms are 8.46, 8.83, 9.2, 9.57, and 9.94 **
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RESPONSE -->
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15:17:03 What is the pattern of these numbers?
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RESPONSE --> there is a difference of .37 between each of these values.
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15:17:05 ** These numbers increase by .37 at each interval. **
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RESPONSE -->
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15:19:12 If you didn't know the equation for the function, how would you go about finding the 100th member of the sequence? How can you tell your method is valid?
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RESPONSE --> If the function was unknown, multiply .37 * 99 = 36.63 then add the first term 36.63 + 8.46 = 45.09, this method is valid because it comes up with the same answer as what f(100) gives.
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15:19:27 ** You could find the 100th member by noting that you have 99 ?umps?between the first number and the 100 th, each ?ump?being .37. Multiplying 99 times .37 and then adding the result to the 'starting value' (8.46). STUDENT RESPONSE: simply put 100 as the x in the formula .37x +8.09 INSTRUCTOR COMMENT: That's what you do if you have the equation. Given just the numbers you could find the 100th member by multiplying 99 times .37 and then adding the result to the first value 8.46. **
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15:19:56 for quadratic function y = g(x) = .01 x^2 - 2x + 100 what are the first five terms of the basic sequence {g(n), n = 1, 2, 3, ...}?
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RESPONSE --> 98.01, 94.04, 94.09, 92.16, 90.25
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15:20:07 ** We have g(1) = .01 * 1^2 - 2 * 1 + 100 = 98.01 g(2) = .01 * 2^2 - 2 * 2 + 100 = 96.04, etc. The first 5 terms are therefore {98.01, 96.04, 94.09, 92.16, 90.25}
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15:21:10 What is the pattern of these numbers?
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RESPONSE --> the pattern is these values are subtracted by a number increased by .02 each time. the differences in the numbers are -1.97, -1.95, -1.93, -1.91 respectively, all of these numbers have a difference of .02
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15:21:19 ** The changes in these numbers are -1.97, -1.95, -1.93, -1.91. With each interval of x, the change in y is .02 greater than for the previous interval. **
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RESPONSE -->
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15:23:24 If you didn't know the equation for the function, how would you go about finding the next three members of the sequence?
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RESPONSE --> add .02 to -1.91 = -1.89 then add the previous value 90.25 = 88.36 then add .02 to -1.89 = -1.87 + 88.36 = 86.49 add .02 to -1.87 = -1.85 +86.49 = 84.64
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15:23:37 ** According to the pattern established above, the next three changes are -1.89, -1.87, -1.85. This gives us g(6) = g(5) - 1.89, g(7) = g(6) - 1.87, g(7) = g(6) - 1.85. **
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RESPONSE -->
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15:24:09 How can you verify that your method is valid?
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RESPONSE --> adding .02 gives you the next interval to subtract from the g(x) value.
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15:24:28 ** You can verify the result using the original formula; if you evaluate it at 5, 6 and 7 it should confirm your results. That's the best answer that can be given at this point. You should understand, though that even if you verified it for the first million terms, that wouldn't really prove it (who knows what might happen at the ten millionth term, or whatever). It turns out that to prove it would require calculus or the equivalent. **
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15:26:46 query problem 15. The difference equation a(n+1) = a(n) + .4, a(1) = 5 If you substitute n = 1 into a(n+1) = a(n) + .4, how do you determine a(2)?
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RESPONSE --> a(1+1) = a(1) + .4 => a(2) = a(1) + .4, this tells that a(2) = a(1) .4, and it is given that a(1)=5, so substitute in a(2)=5+.4 =>a(2)=5.4
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15:26:49 ** You get a(1+1) = a(1) + .4, or a(2) = a(1) + .4. Knowing a(1) = 5 you get a(2) = 5.4. **
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RESPONSE -->
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15:28:28 If you substitute n = 2 into a(n+1) = a(n) + .4 how do you determine a(3)?
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RESPONSE --> a(2+1) = a(2) + .4 => a(3) = a(2)+.4, substitute the found value of a(2), a(3) = 5.4 + .4 to get a(3) = 5.8
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15:28:30 ** You have to do the substitution. You get a(2+1) = a(2) + .4, or since 2 + 1 = 3, a(3) = a(2) + .4 Then knowing a(2) = 5.4 you get a(3) = 5.4 + .4 = 5.8. **
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RESPONSE -->
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15:31:19 If you substitute n = 3 into a(n+1) = a(n) + .4, how do you determine a(4)?
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RESPONSE --> a(3+1) =a(3)+.4 => a(4) = a(3) + .4, substitute in a(3) value, a(4) = 5.8 + .4 => a(4)=6.2
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15:31:21 ** We get a(4) = a(3) +.4 = 5.8 + .4 = 6.2 **
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RESPONSE -->
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15:32:06 What is a(100)?
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RESPONSE --> .4 * 100 = 40, add a(1) obtaining 45 so a(100) = 45
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15:32:47 ** a(100) would be equal to a(1) plus 99 jumps of .4, or 5 + 99*.4 = 44.6. **
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RESPONSE --> I forgot about taking 1 from 100 for the number of jumps.
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15:34:06 query problem 17. difference equation a(n+1) = a(n) + 2 n, a(1) = 4. What is the pattern of the sequence?
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RESPONSE --> The patten is the values are added by a number increased by 2 each time.
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15:35:20 ** You get a(2) = a(1) + 2 * 1 = 4 + 2 = 6, then a(3) = a(2) + 2 * 3 = 6 + 6 = 12 then a(4) = a(3) + 2 * 4 = 12 + 8 = 20; etc. The sequence is 6, 12, 20, 30, 42, ... . **
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RESPONSE --> I thought the question asked for the pattern not the sequence
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15:35:48 What kind of function do you think a(n) is (e.g., linear, quadratic, exponential, etc.)?
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RESPONSE --> exponential
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15:37:02 ** The differences of the sequence are 6, 8, 10, 12, . . .. The difference change by the same amount each time, which is a property of quadratic functions. **
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RESPONSE -->
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15:40:07 query the slope = slope equation Explain the logic of the slope = slope equation (your may take a little time on this one)
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RESPONSE --> The slope = slope equation is useful to find the equation of a straight line when you are only given the slope and one data point. It takes the given point and an arbitrary point on the line to find the slope (with values (y-y1)/(x-x1) and sets this equal to the known slope. We know that the slope between any two points on a straight line have the same slope, so this allows a linear model to be made for any x and y coordinates.
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15:40:16 ** The slope = slope equation sets the slope between two given points equal to the slope between one of those points and the variable point (x, y). Since all three points lie on the same straight line, the slope between any two of the three points must be equal to the slope between any other pair. **
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RESPONSE -->
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15:40:47 query problem 7. streamRange(t), 50 centimeters at t = 20 seconds, range changes by -10 centimeters over 5 seconds. what is your function?
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RESPONSE --> streamRange(t) = -2(t)+90
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15:40:56 ** The rate at which streamRange changes is change in streamRange / change in t = -10 cm / (5 sec) = -2 cm/s. This will be the slope m of the graph. Since streamRange is 50 cm when t = 20 sec the point (20, 50) lies on the graph. So the graph passes through (20, 50) and has slope -2. The function is therefore of the form y = m t + b with m = -2, and b such that 50 = -2 * 20 + b. Thus b = 90. The function is therefore y = -2 t + 90, or using the meaningful name of the function steamRange(t) = -2t + 90 You need to use function notation. y = f(x) = -2x + 90 would be OK, or just f(x) = -2x + 90. The point is that you need to give the funcion a name. Another idea here is that we can use the 'word' streamRange to stand for the function. If you had 50 different functions and, for example, called them f1, f2, f3, ..., f50 you wouldn't remember which one was which so none of the function names would mean anything. If you call the function streamRange it has a meaning. Of course shorter words are sometimes preferable; just understand that function don't have to be confined to single letters and sometimes it's not a bad idea to make the names easily recognizable. STUDENT RESPONSE: y = -2x + 50 INSTRUCTOR COMMENT: ** At t = 20 sec this would give you y = -2 * 20 + 50 = 10. But y = 50 cm when t = 20 sec. Slope is -10 cm / (5 sec) = -2 cm/s, so you have y = -2 t + b. Plug in y = 50 cm and t = 20 sec and solve for b. You get b = 90 cm. The equation is y = -2 t + 90, or streamRange(t) = -2t + 90. **
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RESPONSE -->
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15:41:33 what is the clock time at which the stream range first falls to 12 centimeters?
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RESPONSE --> 12 cm = -2t+90 -78 = -2t t=39 sec
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15:41:35 10-08-2006 15:41:35 ** Using the correct equation streamRange(t) = -2t + 90, you would set streamRange(t) = 12 and solve 12 = -2t + 90, obtaining t = 39 sec. **
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15:43:39 query problem 9. equation of the straight line through t = 5 sec and the t = 7 sec points of the quadratic function depth(t) = .01 t^2 - 2t + 100 What is the slope and what does it tell you about the depth function?
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RESPONSE --> depth(5) = .01(5)^2-2(5)+100 = .25-10+100 =90.25 depth(7) = .01(7)^2-2(7) +100 =.49-14+100 =86.49 (86.49-90.25)/(7-5) = -3.76/2 = -1.88 The slope says that for every second, the depth decreases by 1.88 cm.
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15:43:55 ** You have to get the whole equation. y = m t + b is now y = -1.88 t + b. You have to solve for b. Plug in the coordinates of the t = 7 point and find b. You get 90.9 = -1.88 * 7 + b so b = 104, approximately. Find the correct value. The equation will end up something like y = depth(t) = -1.88 t + 104. **
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RESPONSE -->
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15:44:02 The slope of the linear function is -1.88. This tells me that the depth is decreasing as the time is increasing at a rate of 1.88 cm per sec.
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RESPONSE -->
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15:46:06 How closely does the linear function approximate the quadratic function at each of the given times?
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RESPONSE --> There is an average deviation of 0.025, it fits well. when t=5 and 7 the deviation is 0 at t=6 it is .01 at t=4 and t=8 it is .03 and at t=3 the quadratic function curves away the most rapidly from the linear function with a deviation of .08
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15:46:12 ** The deviations are for t = 3, 4, 5, 6, 7, & 8 as follows: .08, .03. 0. -.01, 0, .03. **
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RESPONSE -->
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15:46:30 at what t value do we obtain the closest values?
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RESPONSE --> t=.01 gives the closest values
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15:47:01 ** Not counting t= 5 and t = 7, which are 0, the next closest t value is t = 6, the deviation for this is -.01. **
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RESPONSE --> The orginal question in the page said throwing out t=5 and 7 i believe
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15:48:25 On which side of the t = 5 and t = 7 points is the linear approximation closer to the quadratic function? On which side does the quadratic function 'curve away' from the linear most rapidly?
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RESPONSE --> I am not sure what is meant by the first question ""on which side"" and when t=3 the quadratic curves away most rapidly
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15:49:08 ** On the t = 4 side the approximation is closer. The quadratic function curves away on the positive x side. **
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RESPONSE --> I still do not quite understand what this means.
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15:49:12 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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15:49:16 The slope = slope helped me out a lot. Learning that I can solve a linear in different ways was helpful.
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RESPONSE -->
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???????E???zH???? Student Name: assignment #010
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16:04:57 `q001. Note that this assignment has 10 questions Sketch the function y = x and describe your graph. Describe how the graphs of y = .5 x and y = 2 x compare with the graph of y = x.
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RESPONSE --> y=x has a y-int and x-int of (0,0) with the point one right being (1,1) y=.5x has a y and x int of (0,0) with the point one to the right being (1,.5). it is a .5 vertical stretch of the y=x graph y=2x has a y and x int of (0,0) with the point one unit the the right being (1,2). it is a 2 vertical stretch of the y=x graph
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16:05:38 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.
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16:07:42 `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?
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RESPONSE --> All of the lines would run through the origin however, they would all be vertially stretched by different a values. the graph of y=.6x would be .6 times away from the x axis compared to the graph of y=x, it would be the same with all of the a values.
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16:08:13 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.
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16:10:15 `q003. Describe how the graphs of y = x - 2 and y = x + 3 compare with the graph of y = x. If we were to sketch all graphs of the form y = x + c for -2 < x < 3, what would our graph look like?
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RESPONSE --> the graph of y=x-2 is parallel to that of y=x however, it has -2 as a y-int instead of 0. The same with y=x+3, it is parallel to y=x but the y int is 3. If all the graphs were sketched between y=x+c for -2
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16:10:35 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. **
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16:13:00 `q004. Describe how the graph of y = 2 x compares with the graph of y = x. Describe how the graph of y = 2 x - 2 compares with the graph of y = 2 x.
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RESPONSE --> The graph of y=2x would be verically stretched times 2 compared to the graph of y=x, all of the y values in y=x would be multiplied by 2 at the same x value. For example (1,1) would be (1,2). The graph of y=2x-2 is vertically stretched by a factor of 2 so it is parallel to the graph of y=2x but it is shifted two units downwards in comparison
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16:13:07 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).
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16:14:15 `q005. Suppose we graph y = 2 x + c for all values of c for which -2 < c < 3. What with our graph look like?
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RESPONSE --> The graphs will all be vertically stretched by a factor of 2 in comparison to the y=x graph, but they will have varing y ints ranging from -2
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16:14:25 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).
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16:16:10 `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?
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RESPONSE --> slope = (y2-y1)/(x2-x1)
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16:16:17 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).
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16:17:18 `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)?
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RESPONSE --> slope = (y-y1)/(x-x1)
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16:17:23 The slope from (x1, y1) to (x, y) is slope = rise/run = (y - y1) / (x - x1).
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16:17:47 `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)?
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RESPONSE --> equal to because the slope between any two points on a straight line is the same.
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16:17:51 The slope between any two points of a straight line must be the same. The two slopes must therefore be equal.
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RESPONSE -->
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16:21:58 `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?
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RESPONSE --> (y2-y1)/(x2-x1) = (y-y1)/(x-x1)
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16:22:08 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).
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RESPONSE -->
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16:32:28 `q010. Setting the two slopes equal you should have obtained (y - y1)/(x - x1) = (y2 - y1)/(x2 - x1). Solve this equation for y.
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RESPONSE --> (y-y1)/(x-x1) = (y2-y1)/(x2-x1) The value (y2-y1) can be changed to mx2-mx1 which changes to m(x2-x1) so (y-y1)/(x-x1)= [m(x2-x1)]/(x2-x1) (y-y1)/(x-x1)=m
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16:33:23 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.
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RESPONSE --> I was thinking of the slope=slope equation, however this makes sense to me.
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