#$&* course mth 163 1/17 10:44 p.m. If your solution to stated problem does not match the given solution, you should self-critique per instructions athttp://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
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Given Solution: The vertex of the 'blue' parabola is at the point (2, -1), the 'lowest' point on the parabola. The vertex of the 'purple' parabola is at the point (-2, 3), the 'highest' point on the parabola. Self-critique: ------------------------------------------------ Self-critique rating:ok Question `q002: What are the coordinates of the other two 'basic points' of each parabola? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: For the blue parabola it would be (1, 0) and (3, 0). For the parabola on the left it would be (-1, 1) and (-3, 1) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: For the 'blue' parabola, the points (1, 0) and (3, 0) are two units apart, and lie on the same horizontal line. The horizontal line is the x axis. For the 'purple' parabola, the points (-3, 1) and (-1, 1) are two units apart. These points lie on the horizontal line where y = 2. Self-critique: ------------------------------------------------ Self-critique rating:ok Question `q003: For the first parabola, the one whose vertex is (2, 1), how far would we have to move to the right or the left, starting from the vertex, in order to be directly above or below another of its 'basic points'? How far would we then have to move in the vertical direction to reach that point? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: We would have to move to the right or left 1 point and up 2 points confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: For the parabola with vertex (2, 1), if we move 1 unit to the right or left we will be at the point (3, 1) or (1, 1), putting us directly below one of the other two basic points. If we then move 2 units upward, we will be at the point (3, 3) or (1, 3). So if we move 1 unit to the right or left, we need to move 2 units upward to get to another basic point. Self-critique: ------------------------------------------------ Self-critique rating:ok Question `q004: For each of the other two parabolas, how far would we have to move to the right or the left, starting from the vertex, in order to be directly above or below another of its 'basic points'? How far would we then have to move in the vertical direction to reach that point? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: For the blue parabola, we would need to move over to the right or left one point and up one point to reach the next data point. For the parabola on the left, we would have to move over in either direction 1 point and down two points. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: For either of the other two parabolas, if we move 1 unit to the right or left we will be directly above or below one of its basic points (above in the case of the 'third' parabola, whose vertex is (-2, 3), below in the case of the 'second' parabola, whose vertex is (2, -1).) To get to the basic points of the 'third' parabola we will need to move 2 units downward. To get to the basic points of the 'second' parabola we will need to move 1 unit upward. Self-critique: ------------------------------------------------ Self-critique rating:ok Question `q005. Solve the following system of simultaneous linear equations: 3a + 3b = 9 6a + 5b = 16. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Start by setting up the problem to subtract it. I multiplied the top equation by -2 to give me -6a-6b=-18 and added it to the second equation which gave me b=2, then I substituted 2 in for b in equation 1 which gave me a=1. Then I checked the answer by substituting the numbers in for the a and b in equation 2 which gave me 6(1)+5(2)=16. 6+10=16, 16=16, so the answer checks. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The system 3a + 3b = 9 6a + 5b = 16 can be solved by adding an appropriate multiple of one equation in order to eliminate one of the variables. Since the coefficient of a in the second equation (the coefficient of a in the second equation is 6)) is double that in the first (the coefficient of a in the first equation is 3), we can multiply the first equation by -2 in order to make the coefficients of a equal and opposite: -2 * [ 3a + 3b ] = -2 [ 9 ] 6a + 5b = 16 gives us -6a - 6 b = -18 6a + 5b = 16 . Adding the two equations together we obtain -b = -2, or just b = 2. Substituting b = 2 into the first equation we obtain 3 a + 3(2) = 9, or 3 a + 6 = 9 so that 3 a = 3 and a = 1. Our solution is therefore a = 1, b = 2. We used the first equation in our last step, so we verigy this solution is by substituting these values into the second equation, where we get 6 * 1 + 5 * 2 = 6 + 10 = 16. STUDENT QUESTION I got my answer in a very different way than the solution given. I have been trying to remember things from the classes I took a long time ago and came up with this answer. Is it alright to use this method? INSTRUCTOR RESPONSE Here is a synopsis of your solution: I'll first solve the first equation for a: 3a+3b=9 so a+b=3 so a=3-b. Now I'll substitute this expression for a into the second equation 6 a + 5 b = 16 Replacing a with 3 - b: 6(3-b)+5b=16 18-6b+5b=16 -b=-2 b=2 a = 3 - b so a=3 - 2 = 1 Substituting a = 1 and b = 2 into the two equations we get 3(1)+3(2)=9 so 9 = 9 6(1)+5(2)=16 so 16 = 16. The solution checks with the two equations. You have an excellent solution. The method you have used is performed correctly and is equally valid with the method used in the solutions. It is called the 'substitution method'. For these first few problems in this course the substitution method and the elimination method are equally efficient. However the elimination method is also important, and since elimination works better on most of the problems we'll be encountering in the near future, it is the method I use in the given solutions. You can use either method, as long as you know both. However you might find the given solutions easier to understand if you use the elimination method. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q006. Solve the following system of simultaneous linear equations using the method of elimination: 4a + 5b = 18 6a + 9b = 30. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I multiplied the first equation by 6 and the second equation by -4 so that I could eliminate the a’s. This gave me 24a+30b=108 -24a-36b=-120 After subtracting I was left with -6b= -12, so I divided by -6 to get b=2. Then I substituted 2 in for b in the first equation which gave me a=2. Then I checked it by inputting both solutions into the second equation and got 12+18=30 and finally 30=30 so I knew my solution worked. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: In the system 4a + 5b = 18 6a + 9b = 30 we see that the coefficients of b are relatively prime; they therefore have a least common multiple equal to 5 * 9. The coefficients 4 and 6 of a have a least common multiple of 12. • We have a choice of which variable to eliminate. We could 'match' the b by multiplying the first equation by 9 and the second by -5, or we could match the coefficients of a by multiplying the first equation by 3 and the second by -2. • Either choice would work. The numbers required to 'match' the coefficients of a are smaller, but the numbers required to 'match the coefficients of b would otherwise work equally well. Choosing to 'match' the coefficient of a, we obtain 3 * [4a + 5b ] = 3 * 18 -2 * [ 6a + 9b ] = -2 * 30, so the system becomes 12 a + 15 b = 54 -12 a - 18 b = -60. Adding the equations we get -3 b = -6, so b = 2. Substituting this value of b into the first equation we obtain 4 a + 5 * 2 = 18, or 4 a + 10 = 18, which we easily solve to obtain a = 2. Substituting this value of a into the second equation we obtain 6 * 2 + 9 * 2 = 30, which verifies our solution. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q007. If y = 5x + 8, then for what value of x will we have y = 13? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I began by substituting 13 in for y. Then I subtracted 8 from both sides to get 5=5x, then divide by 5 to get x=1 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We first substitute y = 13 into the equation y = 5 x + 8 to obtain 13 = 5 x + 8. Subtracting 8 from both equations and reversing the equality we obtain 5 x = 5, which we easily solve to obtain x = 1. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `q008. Sketch a set of coordinate axes representing y vs. x, with y on the vertical axis and x on the horizontal axis. • Plot the points (1, -2), (3, 5) and (7, 8). • Sketch a smooth curve passing through these three points. On your curve, what are the y coordinates corresponding to x coordinates 1, 3, 5 and 7? Estimate these coordinates as accurately as you can from your graph. Retain your sketch for use in future questions. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The y coordinates are -2, 5, 7.5, 8. The coordinates for 1, 3, and 7 where already given and I estimated the coordinate for the x value of 5 to be a little more than 7.5. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The x coordinates 1, 3 and 7 match the x coordinates of the three given points, the y coordinates will be the y coordinates -2, 5 and 8, respectively, of those points. • At x = 5 the precise value of x, for a perfect parabola, would be 8 1/3, or about 8.333. • You are unlikely to have drawn a perfect parabola and your estimate will almost certainly differ from this value. Any estimate with .5 or so of this value would be a good estimate. Drawn with complete accuracy a parabola through these points will peak between x = 3 in and x = 7. • The peak of the actual parabola will occur close to x = 6. Any estimate between x = 5 and x = 7 is pretty good, and even an estimate between x = 7 and x = 8 is not unreasonable. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `q009. Using your sketch from the preceding exercise, estimate the x coordinates corresponding to y coordinates 1, 3, 5 and 7. Also estimate the x values at which y is 0. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When y is zero, I estimate that x is 1.3. When y is 1, x is about 1.7, When y is 3, x is about 2.2, when y is 5, x is 3(given in original points), when y is 7, x is approx. 4 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The easiest way to estimate your points would be to make horizontal lines on your graph at y = 1, 3, 5 and 7. You would easily locate the points were these lines intersect your graph, then estimate the x coordinates of these points. For the actual parabola passing through the given points, y will be 1 when x = 1.7 (and also, if your graph extended that far, near x = 10). • y = 3 near x = 2.3 (and near x = 9.3). • y = 5 at the given point (3, 5), where x = 3. • y = 7 near x = 4 (and also near x = 7.7). Any set of estimates in the vicinity of these values is good. STUDENT COMMENT: I’m still not clear on this. I didn’t get the same as you and my graph went to 12. I know when we estimate we will not get the same answer, but did I do this right? If not what do I do to correct it? INSTRUCTOR RESPONSE: The following commentary need not be self-critiqued, though you may certainly ask questions if you wish. However I have two reasons for including it: • The estimates here were given by a good student whose graph for some reason didn't match the points on this particular problem. If this student missed his/her estimates, then most students will need some clarification at this point. • The discussion and the graphs shown here should be helpful to you. You should be able to sketch these graphs, label the coordinate axes based on the coordinates of the three points, and make reasonable verification of the values given here. • This situation is closely related to the work we will be doing in the first few assignments, where we will actually learn to do parabolic or quadratic curve fits. Instructor response to student: Some of your estimates were reasonable, but others weren't. Your solution was if y = 0 then x = 0 if y = 1 then x = approx. 1.7 if y = 3 then x = approx. 4.6 if y = 5 then x = approx. 8.1 if y = 7 then x = approx. 11.5 If the x coordinates 1,3 7 match the y coordinates -2,5,8, then y = 1 at x = 1.7 is a very reasonable estimate y = 3 at x = 4.6 is not a reasonable estimate. By the time x = 3, y already has a value of 5, and is increasing toward value y = 8 at x = 7. For an x value between 3 and 5 we would expect y to be greater than 5; we would not expect a y value of 3 to occur on this x interval. y = 5 at x = 8.1 is a reasonable estimate. When x = 7 we have y = 8, and it's possible to draw a reasonable graph on which y decreases to 5 by the time x = 8. y = 7 at x = 11.5 isn't consistent with your previous estimates. You already have y down to value 5 by the time x = 8.1, and there's no reason to expect the graph to turn around and go back up to 7 by the time x reaches 11.5. The figure below depicts the three data points. Your graph should have similar form. The graph below shows the parabola that actually passes through these points. Your graph won't necessarily look the same, since you probably aren't practiced at drawing an accurate parabola through three given points, and besides nobody told you to use a parabola. So it's no problem if your graph differs from this one, but this 'ideal' graph is the basis for the numbers in the given solution. For this graph, y = 0 somewhere between x = 1 and x = 3, closer to x = 1; we might estimate that this occurs at x = 1.5 (a fairly precise value for the actual parabola is about x = 1.476). You will learn a lot more about parabolas within the next couple of weeks. In fact, by Assignment 4 or 5, you will have learned how to create and use a parabolic model for any three points, and you'll see that while it's not really what you'd call easy, it's not all that difficult either. The graph below isn't parabolic, but it's still a smooth curve and passes through the given points, and your graph could well look like this. The y value at x = 5 appears to be pretty close to the x = 7 value, which is y = 8. The y = 0 value again occurs between x = 1 and x = 3; perhaps a little to the left of the point in the above graph, maybe at x = 1.4. Another reasonable graph is shown below. At x = 5 the y value is probably around 7. The graph below agrees with the estimates you gave. While it fits the overall trend of the data by 'staying in the middle', it doesn't fit any of the three given points very well. This graph has been included so you can verify that it agrees reasonably with your given values if y = 0 then x = 0 if y = 1 then x = approx. 1.7 if y = 3 then x = approx. 4.6 if y = 5 then x = approx. 8.1 if y = 7 then x = approx. 11.5 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I did not extend my graph all the way around but I was actually quite surprised to find that I was relatively close on my graphing. ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `q010. Suppose the graph you used in the preceding two exercises represents the profit y on an item, with profit given in cents, when the selling price is x, with selling price in dollars. According to your graph: • What would be the profit if the item is sold for 4 dollars? • What selling price would result in a profit of 7 cents? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If the item sold for 4 dollars, I would have a profit of approx. 7 cents. So the selling price of the item would have to be about 4 dollars for me to have a profit of approx. 7 cents. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: To find the profit for a selling price of x = 4 dollars, we would look at the x = 4 point on the graph. • This point is easily located by sketching a vertical line through x = 4. Projecting over to the y-axis from this point, you should have obtained an x value somewhere around 6 of 7, depending on how you drew the graph. This represents a profit of about 6 or 7 cents. The profit is the y value, so to obtain the selling price x corresponding to a profit of y = 7 we sketch the horizontal line at y = 7, which as in a preceding exercise will give us x values of about 4 or 5 (y = 7 would also occur at x = 7.7, approx., if the entire parabola was drawn). STUDENT COMMENT: I think I'm doing my graphs wrong. Can you help? INSTRUCTOR RESPONSE: See also my notes on the preceding problems, especially the ones with the graphs. The graph depicted below doesn't fit the original points very well, but we'll use it to illustrate the situation. As before you should review the coordinates of the three given points and make sense of them on this graph. The 'red' line is a vertical line through x = 4. This line is drawn first. The 'green' horizontal line is then drawn through the point where the x = 4 line intersects the graph. You should verify for yourself that the 'green' line intersects the y axis around y = 6 (the actual coordinate is about y = 6.2, but you aren't expected to be able to verify the value that accurately; you should however be able to tell that the coordinate is closer to 6 than to 5 or 7, based on the coordinates of the three given points). This would tell us that if the selling price is around 4 units, the profit is around 6 units. To find the selling price that will bring a profit of 7 units, we first draw the horizontal 'blue' line through y = 7. Then we draw the vertical 'purple' line, through the point where our horizontal line intersects the graph. This line intersects the x axis somewhere between x = 4 and x = 5. It's hard to tell whether the x value is closer to 4 than to 5, so we might estimate that this occurs around x = 4.5 (the actual value as depicted is about x = 4.43). This would tell us that to achieve a profit of 7 units, the selling price should be about 4.5 units. Your estimates will of course differ from these, depending on how you drew your graph. For example, if the parabola is taken as the 'perfect' model for the data, the results might vary from these by as much as a unit. Worth noting, even though you probably don't understand it yet: For a number of reasons you will soon understand, it turns out that the parabola is a very reasonable model for estimating how price and profit are related. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `q011. On another set of coordinate axes, plot the points (-3, 4) and (5, -2). Sketch a straight line through these points. We will proceed step-by-step obtain an approximate equation for this line: First substitute the x and y coordinates of the first point into the form y = m x + b. • What equation do you obtain when you make this substitution? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 4=m(-3)+b confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Substituting x = -3 and y = 4 into the form y = m x + b, we obtain the equation • 4 = -3 m + b. We can reverse the right- and left-hand sides to get • -3 m + b = 4. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `q012. Substitute the coordinates of the point (5, -2) into the form y = m x + b. What equation do you get? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: -2=m(5) + b confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Substituting x = 5 and y = -2 into the form y = m x + b, we obtain the equation • -2 = 5 m + b. Reversing the sides we have • 5 m + b = -2 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `q013. You have obtained the equations -3 m + b = 4 and 5 m + b = -2. Use the method of elimination to solve these simultaneous equations for m and b. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I started by subtracting the two equations to eliminate the b’s. I got -8m=6 and then divided by -8 to get m= -3/4. Then I substituted that in to the first equation for m and got, b= 7/4. I then input both amounts into the second equation to check it. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Starting with the system -3 m + b = 4 5 m + b = -2 we can easily eliminate b by subtracting the equations. If we subtract the first equation from the second we obtain -8 m = 6, with solution m = -3/4. Substituting this value into the first equation we obtain (-3/4) * -3 + b = 4, which we easily solve to obtain b = 7/4. To check our solution we substitute m = -3/4 and b = 7/4 into the second equation, obtaining 5 ( -3/4) + 7/4 = -2, which gives us -8/4 = -2 or -2 = -2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `q014. Substitute your solutions b = 7/4 and m = -3/4 into the original form y = m x + b. • What equation do you obtain? • What is the significance of this equation? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y= - 3/4x+ 7/4 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Substituting b = 7/4 and m = -3/4 into the form y = m x + b, we obtain the equation y = -3/4 x + 7/4. This is the equation of the straight line through the given points (-3, 4) and (5, -2). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `q014. Substitute your solutions b = 7/4 and m = -3/4 into the original form y = m x + b. • What equation do you obtain? • What is the significance of this equation? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y= - 3/4x+ 7/4 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Substituting b = 7/4 and m = -3/4 into the form y = m x + b, we obtain the equation y = -3/4 x + 7/4. This is the equation of the straight line through the given points (-3, 4) and (5, -2). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ok #*&! ********************************************* Question: `q014. Substitute your solutions b = 7/4 and m = -3/4 into the original form y = m x + b. • What equation do you obtain? • What is the significance of this equation? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y= - 3/4x+ 7/4 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Substituting b = 7/4 and m = -3/4 into the form y = m x + b, we obtain the equation y = -3/4 x + 7/4. This is the equation of the straight line through the given points (-3, 4) and (5, -2). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ok #*&!#*&!