assignment 2

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course Mth 163

If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm . Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution. 002. Note that there are 15 questions in this exercise. We begin with a short exercise intended to introduce you to the 'basic-points' idea, a very important idea which you will use throughout the course to quickly construct graphs of basic functions and their transformations and combinations. This is a main theme which runs throughout the course. We then continue with a series of exercises on solving three simultaneous equations in three unknowns, using the process of elimination. Note on graphs: Graphs in this course aren't something you get from a graphing calculator. The graphing calculator has its uses and its benefits, but is not needed for most of this course. Graphs are generally constructed using transformations and other techniques, applied to a few simple functions. You will soon understand what this means. You aren't expected to take the time to do meticulous artwork in this exercise. You will be asked to sketch points on some graphs, and a few curves (parabolas in this case). These graphs are for your own reference, and only need to be neat enough that you can tell what you're seeing. Nobody else needs to see them. Some you might well be able to imagine in your head, with no need to put anything on paper. You are welcome to use graph paper, but it's probably easier to just sketch a pair of coordinate axes on a piece of paper and mark off an appropriate scale. A set of x and y axes, with each axis running from -8 to 8, will be fine for this exercise. Don't waste time labeling every point on the axes. If you just label the points for -8, -4, 4 and 8 you'll be able to tell what the other coordinates are, and you might not even need to label them (you could just make a larger 'tick mark' at each of these points). You should be able to sketch a usable set of coordinate axes in a minute or less. It is entirely possible that you'll complete at least some of the first five and understand everything in them by simply following the instructions, then reading the given solutions. If this is the case on one or more of the first five exercises, you can simply put 'OK' for the confidence rating and self-critique rating. By doing so you certify that you have done everything correctly and understand the exercise. Question `q001: Sketch a set of coordinate axes, with the x axis horizontal and directed to the right, the y axis vertical and directed upwards. Sketch the point P = (-3, -1) on a set of coordinate axes. Sketch the point you get if you shift this point -1 units in the horizontal direction. What are the coordinates of your point? Sketch the point you get if you shift the original point 3 units in the vertical direction. What are the coordinates of your point? Sketch the point you get if you move the original point 4 times as far from the x axis. What are the coordinates of your point? If you move the original point 4 times as far from the x axis, then shift the resulting point -1 units in the horizontal direction, and finally shift the point 3 units in the vertical direction, what are the coordinates of the final point? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: -1 unit to the hprizontal of the point (-3,-1) gives us (-4,-1) 3 units in the vertical direction of point (-3,-1) gives us (-3,2) 4 times as far from -1 would be -4 from the x axis so the point is (-3,-4) 4 times as far from the x axis makes the point (-3,-4) then shift -1 points horizontally gives you the point (-4,-4). Shifting 3 points in the vertical gives the final point (-4,-1) 3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution: Shifting the point -1 units in the horizontal direction we end up at the point (-3 + (-1), -1) = (-4, -1). Shifting the point 3 units in the vertical direction we end up at the point (-3, -1 + ( 3)) = (-3, 2). The point (-3, -1) is -1 units from the x axis. If the point is moves 4 times further from the x axis, the y coordinate will become 4 * -1 = -4. The x coordinate will not change. So the coordinates of the new point will be (-3, -4). If you then shift the resulting point -1 units in the horizontal direction, it will end up at (-3 + (-1), -4) = (-4, -4). If you shift this new point 3 units in the vertical direction, it will end up at (-4, -4 + 3) = (-4, -1). NOTE: We can express this sequence of transformations in a single step as (-3 + (-1), 4 * -1 + 3) = (-4, -1). Self-critique: ------------------------------------------------ Self-critique rating:

Question `q002: Starting with the point P = (0, 0): Sketch the point you get if you shift this point -1 units in the horizontal direction. What are the coordinates of your point? Sketch the point you get if you shift the original point 3 units in the vertical direction. What are the coordinates of your point? Sketch the point you get if you move the original point 4 times as far from the x axis. What are the coordinates of your point? If you move the original point 4 times as far from the x axis, then shift the resulting point -1 units in the horizontal direction, and finally shift the point 3 units in the vertical direction, what are the coordinates of the final point? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: When starting at 0,0 and shifting -1 in the horizontal direction the new point becomes (-1,0) when starting at 0,0 and shifting 3 points in the vertical direction the new point becomes (0,3) when starting at 0,0 and shifting the point 4 times as far from the x axis the point stays at (0,0) when starting at 0,0 and shifting the point 4 times from the x axis the point becomes (0,0) then shifting -1 in the horizontal and 3 points in the vertical make the final point (-1,3) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3

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Given Solution: Shifting the point -1 units in the horizontal direction we end up at the point (0 + (-1), 0) = (-1, 0). Shifting the point 3 units in the vertical direction we end up at the point (0, 0 + ( 3)) = (0, 3). The point (0, 0) is 0 units from the x axis. If the point is moves 4 times further from the x axis, the y coordinate will be 4 * 0 = 0. The x coordinate will not change. So the coordinates of the new point will be (0, 0). If you then shift the resulting point -1 units in the horizontal direction, it will end up at (-1, 0). If you shift this new point 3 units in the vertical direction, it will end up at (-1, 3) Self-critique: ------------------------------------------------ Self-critique rating:

Question `q003: Plot the points (0, 0), (-1, 1) and (1, 1) on a set of coordinate axes. Now plot the points you get if you move each of these points 4 times further from the x axis, and put a small circle around each point. What are the coordinates of your points? Plot the points that result if you shift each of your three circled points -1 units in the x direction. Put a small 'x' through each point. What are the coordinates of your points? Plot the points that result if you shift each of your three new points (the ones with the x's) 3 units in the y direction. Put a small '+' through each point. What are the coordinates of your points? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: moving each of these 4 times from the x axis gives us the new points of (0,0) (-1,4) and (1,4) moving each of the points -1 in the x direction makes the new points (-1,0) (-2,4) and (0,4) moving each point 3 units in the y diection makes the final points (-1,3) (-2,7) and (0,7) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3

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Given Solution: Moving each point 4 times further from the x axis: The point (0, 0) is 0 units from the x axis. Multiplying this distance by 4 still gives you 0. So the point (0, 0) will remain where it is. The points (-1, 1) and (1, 1) are both 1 unit above the x axis. Multiplying this distance by 4 gives us 4 * 1 = 4. The x coordinates will not change, so our new points are (-1, 4) and (1, 4). At this stage our three points are (-1, 4) (0, 0) (1, 4) Horizontally shifting each point -1 units, our x coordinates all change by -1. We therefore obtain the points (-1 + (-1), 4) = (-2, 4), (0 + -1, 0) = (-1, 0) and ((1 + (-1), 4) = ( 0, 4), so our points are now (-2, 4) (-1, 0) ( 0, 4) Vertically shifting each point 3 units, our y coordinates all change by 3. We therefore obtain the points (-2, 4 + 3) = (-2, 7) (-1, 0 + 3) = (-1, 3) and ( 0, 4 + 3) = ( 0, 7) Self-critique: ------------------------------------------------ Self-critique rating:

Question `q004: On the coordinate axes you used in the preceding, sketch the parabola corresponding to the three basic points (0, 0), (-1, 1) and (1, 1). Then sketch the parabola corresponding to your three circled basic points. Then sketch the parabola corresponding to three basic points you indicated with 'x's'. Finally sketch the parabola corresponding to the three basic points you indicated with '+'s'. Describe how each parabola is related to the one before it. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The first parabola on narrower then the original parabola, the on with x points has the same shape but lies to the left of the circled parabola. The parabola with the + points is the same shape but lies above the parabola. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3

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Given Solution: Your 'circled-points' parabola will be narrower than the original parabola through (-1, 1), (0, 0) and (1, 1). In fact, each point on the 'circled-points' parabola will lie 4 times further from the x axis than the point on the original parabola. Your 'x'-points parabola will have the same shape as your 'circled-points' parabola, but will lie to the right or left of that parabola, having been shifted -1 units in the horizontal direction. Your '+'-points parabola will have the same shape as the 'x-points' parabola (and the 'circled-point' parabola), but will lie above or below that parabola, having been shifted 3 units in the vertical direction. Self-critique: ------------------------------------------------ Self-critique rating:

********************************************* Question: `q005. Begin to solve the following system of simultaneous linear equations by first eliminating the variable which is easiest to eliminate. Eliminate the variable from the first and second equations, then from the first and third equations to obtain two equations in the remaining two variables: 2a + 3b + c = 128 60a + 5b + c = 90 200a + 10 b + c = 0. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: c is the easiest to eliminate since they are all 1. We will subtract the first equation from the second equation and it gives us 58a + 2b = -38 Now we will subtract the first equation from the third equation and we get 198a + 7b = -128 3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution: The variable c is most easily eliminated. We accomplish this if we subtract the first equation from the second, and the first equation from the third, replacing the second and third equations with respective results. Subtracting the first equation from the second, are left-hand side will be the difference of the left-hand sides, which is 2d eqn - 1st eqn left-hand side: (60a + 5b + c )- (2a + 3b + c ) = 58 a + 2 b. The right-hand side will be the difference 90 - 128 = -38, so the second equation will become new' 2d equation: 58 a + 2 b = -38. The 'new' third equation by a similar calculation will be 'new' third equation: 198 a + 7 b = -128. You might well have obtained this system, or one equivalent to it, using a slightly different sequence of calculations. (As one example you might have subtracted the second from the first, and the third from the second). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:

********************************************* Question: `q006. Solve the two equations 58 a + 2 b = -38 198 a + 7 b = -128 which can be obtained from the system in the preceding problem, by eliminating the easiest variable. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: b is the easiest variable to eliminate by multiplying 58a + 2b = -38 by 7 and 198a + 7b = -128 by 2 7(58a + 2b = -38) becomes 406a + 14b = -266 2(198a + 7b = -128) becomes 396a +14b = -256 Now subtract equation 1 from 2 to eliminate b. The new equation becomes 10a = -10 so a=-1 3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution: Neither variable is as easy to eliminate as in the last problem, but the coefficients of b are significantly smaller than those of a. So here we choose eliminate b. It would also have been OK to choose to eliminate a. To eliminate b we will multiply the first equation by -7 and the second by 2, which will make the coefficients of b equal and opposite. The first step is to indicate the multiplications: -7 * ( 58 a + 2 b) = -7 * -38 2 * ( 198 a + 7 b ) = 2 * (-128) Doing the arithmetic we obtain -406 a - 14 b = 266 396 a + 14 b = -256. Adding the two equations we obtain -10 a = 10, so we have a = -1. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:

********************************************* Question: `q007. Having obtained a = -1, use either of the equations 58 a + 2 b = -38 198 a + 7 b = -128 to determine the value of b. Check that a = -1 and the value obtained for b are validated by the other equation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 58(-1) + 2b = -38 -58 + 2b = -38 2b = 20 b = 10 198(-1) + 7(10) = -128 -198 + 70 = -128 This is true 3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution: You might have completed this step in your solution to the preceding problem. Substituting a = -1 into the first equation we have 58 * -1 + 2 b = -38, so 2 b = 20 and b = 10. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:

********************************************* Question: `q008. Having obtained a = -1 and b = 10, determine the value of c by substituting these values for a and b into any of the 3 equations in the original system 2a + 3b + c = 128 60a + 5b + c = 90 200a + 10 b + c = 0. Verify your result by substituting a = -1, b = 10 and the value you obtained for c into another of the original equations. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2(-1) + 3(10) + c = 128 -2 + 30 + c = 128 c=100. 60(-1) + 5(10) + 100 = 90 -60 + 50 + 100 = 90 50 + 100 = 150 3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution: Using first equation 2a + 3b + c = 128 we obtain 2 * -1 + 3 * 10 + c = 128, which we easily solve to get c = 100. Substituting these values into the second equation, in order to check our solution, we obtain 60 * -1 + 5 * 10 + 100 = 90, or -60 + 50 + 100 = 90, or 90 = 90. We could also substitute the values into the third equation, and will again obtain an identity. This would completely validate our solution. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:

********************************************* Question: `q009. The graph you sketched in a previous assignment contained the given points (1, -2), (3, 5) and (7, 8). We are going to use simultaneous equations to obtain the equation of that parabola. A graph has a parabolic shape if its the equation of the graph is quadratic. The equation of a graph is quadratic if it has the form y = a x^2 + b x + c. y = a x^2 + b x + c is said to be a quadratic function of x. To find the precise quadratic function that fits our points, we need only determine the values of a, b and c. As we will discover, if we know the coordinates of three points on the graph of a quadratic function, we can use simultaneous equations to find the values of a, b and c. The first step is to obtain an equation using the first known point. What equation do we get if we substitute the x and y values corresponding to the point (1, -2) into the form y = a x^2 + b x + c? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: -2 = a 1^2 + b(1) + c 3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution: We substitute y = -2 and x = 1 to obtain the equation -2 = a * 1^2 + b * 1 + c, or a + b + c = -2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:

********************************************* Question: `q010. If a graph of y vs. x contains the points (1, -2), (3, 5) and (7, 8), as in the preceding question, then what two equations do we get if we substitute the x and y values corresponding to the point (3, 5), then the point (7, 8) into the form y = a x^2 + b x + c? (each point will give us one equation) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 5 = a 3^2 + b(3) + c which becomes 5 = 9a + 3b + c and 8 = a 7^2 + b(7) + c which becomes 8 = 49a + 7b + c 3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution: Using the second point we substitute y = 5 and x = 3 to obtain the equation 5 = a * 3^2 + b * 3 + c, or 9 a + 3 b + c = 5. Using the third point we substitute y = 8 and x = 7 to obtain the equation 8 = a * 7^2 + b * 7 + c, or 49 a + 7 b + c = 8. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:

********************************************* Question: `q011. If a graph of y vs. x contains the points (1, -2), (3, 5) and (7, 8), as was the case in the preceding question, then we obtain three equations with unknowns a, b and c. You have already done this. Write down the system of equations we got when we substituted the x and y values corresponding to the point (1, -2), (3, 5), and (7, 8), in turn, into the form y = a x^2 + b x + c. Solve the system to find the values of a, b and c. What is the solution of this system? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 8 = 49a + 7b + c 5 = 9a + 3b + c -2 = a + b + c We eliminate c from the second and third equations by subtracting this makes the formula 8a + 2b = 7 we eliminate c from the first and third equations to give 48a + 6b = 10 Now multiply 8a + 2b = 7 by -3 and you get -24a - 6b = -21 Now eliminate the b's -24a - 6b = -21 plus 48a + 6b = 10 gives you 24a = - 11 so a = -11/24 or - 0.458333 now plug in a to find b 8(-0.458333) + 2b = 7 -3.666 + 2b = 7 2b = 10.666 so b = 5.333 now plug in a and b to find c in the first equation -0.458333 + 5.333 + c = -2 c = -6.87 3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution: The system consists of the three equations obtained in the last problem: a + b + c = -2 9 a + 3 b + c = 5 49 a + 7 b + c = 8. This system is solved in the same manner as in the preceding exercise. However in this case the solutions don't come out to be whole numbers. The solution of this system, in decimal form, is approximately a = - 0.45833, b = 5.33333 and c = - 6.875. If you obtained a different solution, you should show your solution. Start by indicating the system of two equations you obtained when you eliminated c, then indicate what multiple of each equation you put together to eliminate either a or b. ADDITIONAL DETAILS ON SOLUTION OF SYSTEM You should have enough practice by now to be able to solve the system; however signs can trip us all up, and I've decided to append the following: The second equation minus the first gives us 8a + 2 b = 7. To avoid a common error in subtracting these questions, note that the right-hand sides of these equations are 5 and -2, and that 5 - (-2) = 5 + 2 = 7. It is very common for students (and the rest of us as well) to get a little careless and calculate the right-hand side as 5 - 2 = 3. The third equation minus the first gives us 48 a + 6 b = 10 (again the right-hand side can trip us up; 8 - (-2) = 10. I often see the incorrect calculation 8 - 2 = 6). Now we solve these two equations, 8 a + 2 b = 7 and 48 a + 6 b = 10: If you subtract 3 times the first from the second you will get 24 a = -11, so that a = -.45833. Substituting this into 8 a + 2 b = 7 and solving for b you get b = 5.33333. Substituting these values of a and b into any of the three original equations you get c = -6.875. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:

********************************************* Question: `q012. Substitute the values you obtained in the preceding problem for a, b and c into the form y = a x^2 + b x + c, in order to obtain a specific quadratic function. What is your function? What y values do you get when you substitute x = 1, 3, 5 and 7 into this function? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = (-0.45833) x^2 + (5.333)x + (-6.875) for x = 1 we get y = (-0.45833) 1^2 + (5.333) + (-6.875) so y = -2 for x = 3 we get y = (-0.45833) 3^2 + (5.333)(3) + (-6.875) so y = 4.999 for x = 5 we get y = (-0.45833) 5^2 + (5.333)(5) + (-6.875) so y = 8.333 for x = 7 we get y = (-0.45833) 7^2 + (5.333)(7) + (-6.875) so y = 8 3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution: Substituting the values of a, b and c into the given form we obtain the equation y = - 0.45833 x^2 + 5.33333 x - 6.875. When we substitute 1 into the equation we obtain y = -.45833 * 1^2 + 5.33333 * 1 - 6.875 = -2. When we substitute 3 into the equation we obtain y = -.45833 * 3^2 + 5.33333 * 3 - 6.875 = 5. When we substitute 5 into the equation we obtain y = -.45833 * 5^2 + 5.33333 * 5 - 6.875 = 8.33333. When we substitute 7 into the equation we obtain y = -.45833 * 7^2 + 5.33333 * 7 - 6.875 = 8. Thus the y values we obtain for our x values yield the points (1, -2), (3, 5) and (7, 8). These are the points we used to obtain the formula. We also get the additional point (5, 8.33333). NOTE THAT ADDITIONAL QUESTIONS RELATED TO THIS EXERCISE CONTINUE IN q_a_ ASSIGNMENT FOR ASSIGNMENT 3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:

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: `q013. Substituting the coordinates of three points into the form y = a x^2 + b x + c of a quadratic function, we get the system a + 2 b - c = -8 4 a - b - c = 3 -2a + b + 3 c = 5 The solution of the system is a = 1, b = -2 and c = 3. What is the corresponding quadratic function? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The funtion is y = (1)x^2 + (-2)x + 3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3 ********************************************* Question: `q014. If you eliminate c from the first two equations of the system a + 2 b - c = -5 4 a - b - c = 3 -2a + b + 3 c = 5 what equation do you get? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 5a + b = -2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3 ********************************************* Question: `q015. If you eliminate c from the last two equations of the system a + 2 b - c = -5 4 a - b - c = 3 -2a + b + 3 c = 5 you get the equation 10 a - 2 b = 14 If you eliminate c from the first and third equations of this system you get the equation a + 7 b = -10 What values of a and b solve this system of two equations? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We have 10 a - 2 b = 14 and a + 7 b = -10. To solve for b we multiply the second equation by -10 and eliminate a -10(a + 7b = -10) becomes -10a - 70b = 100 Now add the two together. -10a - 70b = 100 plus 10 a - 2 b = 14 becomes -72b = 114 so b = -2 Now we plug in -2 to an aquation to find a a + 7(-2) = -10 a + -14 = -10 a=4 So a = 4 and b = -2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3 ------------------------------------------------ Self-critique rating:

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Self-critique (if necessary): ------------------------------------------------ Self-critique rating:

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Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!

`gr31

assignment 2

#$&*

course Mth 163

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Self-critique (if necessary): ------------------------------------------------ Self-critique rating:

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Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!

`gr31

assignment 2

#$&*

course Mth 163

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

002.

Note that there are 15 questions in this exercise.

We begin with a short exercise intended to introduce you to the 'basic-points' idea, a very important idea which you will use throughout the course to quickly construct graphs of basic functions and their transformations and combinations. This is a main theme which runs throughout the course.

We then continue with a series of exercises on solving three simultaneous equations in three unknowns, using the process of elimination.

Note on graphs:

Graphs in this course aren't something you get from a graphing calculator. The graphing calculator has its uses and its benefits, but is not needed for most of this course. Graphs are generally constructed using transformations and other techniques, applied to a few simple functions. You will soon understand what this means.

You aren't expected to take the time to do meticulous artwork in this exercise. You will be asked to sketch points on some graphs, and a few curves (parabolas in this case). These graphs are for your own reference, and only need to be neat enough that you can tell what you're seeing. Nobody else needs to see them. Some you might well be able to imagine in your head, with no need to put anything on paper.

You are welcome to use graph paper, but it's probably easier to just sketch a pair of coordinate axes on a piece of paper and mark off an appropriate scale. A set of x and y axes, with each axis running from -8 to 8, will be fine for this exercise. Don't waste time labeling every point on the axes. If you just label the points for -8, -4, 4 and 8 you'll be able to tell what the other coordinates are, and you might not even need to label them (you could just make a larger 'tick mark' at each of these points). You should be able to sketch a usable set of coordinate axes in a minute or less.

It is entirely possible that you'll complete at least some of the first five and understand everything in them by simply following the instructions, then reading the given solutions. If this is the case on one or more of the first five exercises, you can simply put 'OK' for the confidence rating and self-critique rating. By doing so you certify that you have done everything correctly and understand the exercise.

Question `q001: Sketch a set of coordinate axes, with the x axis horizontal and directed to the right, the y axis vertical and directed upwards.

Sketch the point P = (-3, -1) on a set of coordinate axes.

Sketch the point you get if you shift this point -1 units in the horizontal direction. What are the coordinates of your point?

Sketch the point you get if you shift the original point 3 units in the vertical direction. What are the coordinates of your point?

Sketch the point you get if you move the original point 4 times as far from the x axis. What are the coordinates of your point?

If you move the original point 4 times as far from the x axis, then shift the resulting point -1 units in the horizontal direction, and finally shift the point 3 units in the vertical direction, what are the coordinates of the final point?

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Your solution:

-1 unit to the hprizontal of the point (-3,-1) gives us (-4,-1)

3 units in the vertical direction of point (-3,-1) gives us (-3,2)

4 times as far from -1 would be -4 from the x axis so the point is (-3,-4)

4 times as far from the x axis makes the point (-3,-4) then shift -1 points horizontally gives you the point (-4,-4). Shifting 3 points in the vertical gives the final point (-4,-1)

confidence rating #$&*:

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Given Solution:

Shifting the point -1 units in the horizontal direction we end up at the point (-3 + (-1), -1) = (-4, -1).

Shifting the point 3 units in the vertical direction we end up at the point (-3, -1 + ( 3)) = (-3, 2).

The point (-3, -1) is -1 units from the x axis.

If the point is moves 4 times further from the x axis, the y coordinate will become 4 * -1 = -4.

The x coordinate will not change.

So the coordinates of the new point will be (-3, -4).

If you then shift the resulting point -1 units in the horizontal direction, it will end up at (-3 + (-1), -4) = (-4, -4).

If you shift this new point 3 units in the vertical direction, it will end up at (-4, -4 + 3) = (-4, -1).

NOTE: We can express this sequence of transformations in a single step as

(-3 + (-1), 4 * -1 + 3) = (-4, -1).

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Question `q002: Starting with the point P = (0, 0):

Sketch the point you get if you shift this point -1 units in the horizontal direction. What are the coordinates of your point?

Sketch the point you get if you shift the original point 3 units in the vertical direction. What are the coordinates of your point?

Sketch the point you get if you move the original point 4 times as far from the x axis. What are the coordinates of your point?

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Your Solution:

When starting at 0,0 and shifting -1 in the horizontal direction the new point becomes (-1,0)

when starting at 0,0 and shifting 3 points in the vertical direction the new point becomes (0,3)

when starting at 0,0 and shifting the point 4 times as far from the x axis the point stays at (0,0)

when starting at 0,0 and shifting the point 4 times from the x axis the point becomes (0,0) then shifting -1 in the horizontal and 3 points in the vertical make the final point (-1,3)

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Given Solution:

Shifting the point -1 units in the horizontal direction we end up at the point (0 + (-1), 0) = (-1, 0).

Shifting the point 3 units in the vertical direction we end up at the point (0, 0 + ( 3)) = (0, 3).

The point (0, 0) is 0 units from the x axis.

If the point is moves 4 times further from the x axis, the y coordinate will be 4 * 0 = 0.

The x coordinate will not change.

So the coordinates of the new point will be (0, 0).

If you then shift the resulting point -1 units in the horizontal direction, it will end up at (-1, 0).

If you shift this new point 3 units in the vertical direction, it will end up at (-1, 3)

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Question `q003: Plot the points (0, 0), (-1, 1) and (1, 1) on a set of coordinate axes.

Now plot the points you get if you move each of these points 4 times further from the x axis, and put a small circle around each point. What are the coordinates of your points?

Plot the points that result if you shift each of your three circled points -1 units in the x direction. Put a small 'x' through each point. What are the coordinates of your points?

Plot the points that result if you shift each of your three new points (the ones with the x's) 3 units in the y direction. Put a small '+' through each point. What are the coordinates of your points?

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Your Solution:

moving each of these 4 times from the x axis gives us the new points of (0,0) (-1,4) and (1,4)

moving each of the points -1 in the x direction makes the new points (-1,0) (-2,4) and (0,4)

moving each point 3 units in the y diection makes the final points (-1,3) (-2,7) and (0,7)

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Given Solution:

Moving each point 4 times further from the x axis:

The point (0, 0) is 0 units from the x axis. Multiplying this distance by 4 still gives you 0. So the point (0, 0) will remain where it is.

The points (-1, 1) and (1, 1) are both 1 unit above the x axis. Multiplying this distance by 4 gives us 4 * 1 = 4. The x coordinates will not change, so our new points are (-1, 4) and (1, 4).

At this stage our three points are

(-1, 4)

(0, 0)

(1, 4)

Horizontally shifting each point -1 units, our x coordinates all change by -1. We therefore obtain the points

(-1 + (-1), 4) = (-2, 4),

(0 + -1, 0) = (-1, 0) and

((1 + (-1), 4) = ( 0, 4), so our points are now

(-2, 4)

(-1, 0)

( 0, 4)

Vertically shifting each point 3 units, our y coordinates all change by 3. We therefore obtain the points

(-2, 4 + 3) = (-2, 7)

(-1, 0 + 3) = (-1, 3) and

( 0, 4 + 3) = ( 0, 7)

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Question `q004: On the coordinate axes you used in the preceding, sketch the parabola corresponding to the three basic points (0, 0), (-1, 1) and (1, 1).

Then sketch the parabola corresponding to your three circled basic points.

Then sketch the parabola corresponding to three basic points you indicated with 'x's'.

Finally sketch the parabola corresponding to the three basic points you indicated with '+'s'.

Describe how each parabola is related to the one before it.

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Your Solution:

The first parabola on narrower then the original parabola, the on with x points has the same shape but lies to the left of the circled parabola. The parabola with the + points is the same shape but lies above the parabola.

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Your 'x'-points parabola will have the same shape as your 'circled-points' parabola, but will lie to the right or left of that parabola, having been shifted -1 units in the horizontal direction.

Your '+'-points parabola will have the same shape as the 'x-points' parabola (and the 'circled-point' parabola), but will lie above or below that parabola, having been shifted 3 units in the vertical direction.

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Question: `q005.

Begin to solve the following system of simultaneous linear equations by first eliminating the variable which is easiest to eliminate. Eliminate the variable from the first and second equations, then from the first and third equations to obtain two equations in the remaining two variables:

2a + 3b + c = 128

60a + 5b + c = 90

200a + 10 b + c = 0.

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Your solution:

c is the easiest to eliminate since they are all 1. We will subtract the first equation from the second equation and it gives us 58a + 2b = -38

Now we will subtract the first equation from the third equation and we get 198a + 7b = -128

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Given Solution:

The variable c is most easily eliminated. We accomplish this if we subtract the first equation from the second, and the first equation from the third, replacing the second and third equations with respective results.

Subtracting the first equation from the second, are left-hand side will be the difference of the left-hand sides, which is

2d eqn - 1st eqn left-hand side: (60a + 5b + c )- (2a + 3b + c ) = 58 a + 2 b.

The right-hand side will be the difference 90 - 128 = -38, so the second equation will become

new' 2d equation: 58 a + 2 b = -38.

The 'new' third equation by a similar calculation will be

'new' third equation: 198 a + 7 b = -128.

You might well have obtained this system, or one equivalent to it, using a slightly different sequence of calculations. (As one example you might have subtracted the second from the first, and the third from the second).

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Question: `q006. Solve the two equations

58 a + 2 b = -38

198 a + 7 b = -128

which can be obtained from the system in the preceding problem, by eliminating the easiest variable.

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Your solution:

b is the easiest variable to eliminate by multiplying 58a + 2b = -38 by 7 and 198a + 7b = -128 by 2

7(58a + 2b = -38) becomes 406a + 14b = -266

2(198a + 7b = -128) becomes 396a +14b = -256

Now subtract equation 1 from 2 to eliminate b.

The new equation becomes 10a = -10 so a=-1

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Given Solution:

Neither variable is as easy to eliminate as in the last problem, but the coefficients of b are significantly smaller than those of a. So here we choose eliminate b. It would also have been OK to choose to eliminate a.

To eliminate b we will multiply the first equation by -7 and the second by 2, which will make the coefficients of b equal and opposite. The first step is to indicate the multiplications:

-7 * ( 58 a + 2 b) = -7 * -38

2 * ( 198 a + 7 b ) = 2 * (-128)

Doing the arithmetic we obtain

-406 a - 14 b = 266

396 a + 14 b = -256.

Adding the two equations we obtain

-10 a = 10,

so we have

a = -1.

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Question: `q007. Having obtained a = -1, use either of the equations

58 a + 2 b = -38

198 a + 7 b = -128

to determine the value of b. Check that a = -1 and the value obtained for b are validated by the other equation.

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Your solution:

58(-1) + 2b = -38

-58 + 2b = -38

2b = 20

b = 10

198(-1) + 7(10) = -128

-198 + 70 = -128 This is true

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Given Solution:

You might have completed this step in your solution to the preceding problem.

Substituting a = -1 into the first equation we have

58 * -1 + 2 b = -38, so

2 b = 20 and

b = 10.

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Question: `q008. Having obtained a = -1 and b = 10, determine the value of c by substituting these values for a and b into any of the 3 equations in the original system

2a + 3b + c = 128

60a + 5b + c = 90

200a + 10 b + c = 0.

Verify your result by substituting a = -1, b = 10 and the value you obtained for c into another of the original equations.

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Your solution:

2(-1) + 3(10) + c = 128

-2 + 30 + c = 128

c=100.

60(-1) + 5(10) + 100 = 90

-60 + 50 + 100 = 90

50 + 100 = 150

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Given Solution:

Using first equation 2a + 3b + c = 128 we obtain 2 * -1 + 3 * 10 + c = 128, which we easily solve to get c = 100.

Substituting these values into the second equation, in order to check our solution, we obtain

60 * -1 + 5 * 10 + 100 = 90, or

-60 + 50 + 100 = 90, or

90 = 90.

We could also substitute the values into the third equation, and will again obtain an identity. This would completely validate our solution.

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Question: `q009. The graph you sketched in a previous assignment contained the given points (1, -2), (3, 5) and (7, 8).

We are going to use simultaneous equations to obtain the equation of that parabola.

A graph has a parabolic shape if its the equation of the graph is quadratic.

The equation of a graph is quadratic if it has the form y = a x^2 + b x + c.

y = a x^2 + b x + c is said to be a quadratic function of x.

To find the precise quadratic function that fits our points, we need only determine the values of a, b and c.

As we will discover, if we know the coordinates of three points on the graph of a quadratic function, we can use simultaneous equations to find the values of a, b and c.

The first step is to obtain an equation using the first known point.

What equation do we get if we substitute the x and y values corresponding to the point (1, -2) into the form y = a x^2 + b x + c?

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Your solution:

-2 = a 1^2 + b(1) + c

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Given Solution:

We substitute y = -2 and x = 1 to obtain the equation

-2 = a * 1^2 + b * 1 + c, or

a + b + c = -2.

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Question: `q010. If a graph of y vs. x contains the points (1, -2), (3, 5) and (7, 8), as in the preceding question, then what two equations do we get if we substitute the x and y values corresponding to the point (3, 5), then the point (7, 8) into the form y = a x^2 + b x + c? (each point will give us one equation)

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Your solution:

5 = a 3^2 + b(3) + c which becomes 5 = 9a + 3b + c

and

8 = a 7^2 + b(7) + c which becomes 8 = 49a + 7b + c

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Given Solution:

Using the second point we substitute y = 5 and x = 3 to obtain the equation

5 = a * 3^2 + b * 3 + c, or

9 a + 3 b + c = 5.

Using the third point we substitute y = 8 and x = 7 to obtain the equation

8 = a * 7^2 + b * 7 + c, or

49 a + 7 b + c = 8.

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Question: `q011. If a graph of y vs. x contains the points (1, -2), (3, 5) and (7, 8), as was the case in the preceding question, then we obtain three equations with unknowns a, b and c. You have already done this.

Write down the system of equations we got when we substituted the x and y values corresponding to the point (1, -2), (3, 5), and (7, 8), in turn, into the form y = a x^2 + b x + c.

Solve the system to find the values of a, b and c.

What is the solution of this system?

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Your solution:

8 = 49a + 7b + c

5 = 9a + 3b + c

-2 = a + b + c

We eliminate c from the second and third equations by subtracting this makes the formula 8a + 2b = 7

we eliminate c from the first and third equations to give 48a + 6b = 10

Now multiply 8a + 2b = 7 by -3 and you get -24a - 6b = -21

Now eliminate the b's

-24a - 6b = -21 plus 48a + 6b = 10 gives you 24a = - 11 so a = -11/24 or - 0.458333

now plug in a to find b

8(-0.458333) + 2b = 7

-3.666 + 2b = 7

2b = 10.666 so b = 5.333

now plug in a and b to find c in the first equation

-0.458333 + 5.333 + c = -2

c = -6.87

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Given Solution:

The system consists of the three equations obtained in the last problem:

a + b + c = -2

9 a + 3 b + c = 5

49 a + 7 b + c = 8.

This system is solved in the same manner as in the preceding exercise. However in this case the solutions don't come out to be whole numbers.

The solution of this system, in decimal form, is approximately

a = - 0.45833,

b = 5.33333 and

c = - 6.875.

If you obtained a different solution, you should show your solution. Start by indicating the system of two equations you obtained when you eliminated c, then indicate what multiple of each equation you put together to eliminate either a or b.

ADDITIONAL DETAILS ON SOLUTION OF SYSTEM

You should have enough practice by now to be able to solve the system; however signs can trip us all up, and I've decided to append the following:

The second equation minus the first gives us 8a + 2 b = 7.

To avoid a common error in subtracting these questions, note that the right-hand sides of these equations are 5 and -2, and that 5 - (-2) = 5 + 2 = 7. It is very common for students (and the rest of us as well) to get a little careless and calculate the right-hand side as 5 - 2 = 3.

The third equation minus the first gives us 48 a + 6 b = 10 (again the right-hand side can trip us up; 8 - (-2) = 10. I often see the incorrect calculation 8 - 2 = 6).

Now we solve these two equations, 8 a + 2 b = 7 and 48 a + 6 b = 10:

If you subtract 3 times the first from the second you will get 24 a = -11, so that a = -.45833.

Substituting this into 8 a + 2 b = 7 and solving for b you get b = 5.33333.

Substituting these values of a and b into any of the three original equations you get c = -6.875.

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Question: `q012. Substitute the values you obtained in the preceding problem for a, b and c into the form y = a x^2 + b x + c, in order to obtain a specific quadratic function.

What is your function?

What y values do you get when you substitute x = 1, 3, 5 and 7 into this function?

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Your solution:

y = (-0.45833) x^2 + (5.333)x + (-6.875)

for x = 1 we get y = (-0.45833) 1^2 + (5.333) + (-6.875) so y = -2

for x = 3 we get y = (-0.45833) 3^2 + (5.333)(3) + (-6.875) so y = 4.999

for x = 5 we get y = (-0.45833) 5^2 + (5.333)(5) + (-6.875) so y = 8.333

for x = 7 we get y = (-0.45833) 7^2 + (5.333)(7) + (-6.875) so y = 8

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Given Solution:

Substituting the values of a, b and c into the given form we obtain the equation

y = - 0.45833 x^2 + 5.33333 x - 6.875.

When we substitute 1 into the equation we obtain y = -.45833 * 1^2 + 5.33333 * 1 - 6.875 = -2.

When we substitute 3 into the equation we obtain y = -.45833 * 3^2 + 5.33333 * 3 - 6.875 = 5.

When we substitute 5 into the equation we obtain y = -.45833 * 5^2 + 5.33333 * 5 - 6.875 = 8.33333.

When we substitute 7 into the equation we obtain y = -.45833 * 7^2 + 5.33333 * 7 - 6.875 = 8.

Thus the y values we obtain for our x values yield the points (1, -2), (3, 5) and (7, 8). These are the points we used to obtain the formula. We also get the additional point (5, 8.33333).

NOTE THAT ADDITIONAL QUESTIONS RELATED TO THIS EXERCISE CONTINUE IN q_a_ ASSIGNMENT FOR ASSIGNMENT 3.

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Question: `q013. Substituting the coordinates of three points into the form y = a x^2 + b x + c of a quadratic function, we get the system

a + 2 b - c = -8

4 a - b - c = 3

-2a + b + 3 c = 5

The solution of the system is

a = 1, b = -2 and c = 3.

What is the corresponding quadratic function?

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Your solution:

The funtion is y = (1)x^2 + (-2)x + 3

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Question: `q014. If you eliminate c from the first two equations of the system

a + 2 b - c = -5

4 a - b - c = 3

-2a + b + 3 c = 5

what equation do you get?

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Your solution:

5a + b = -2

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Question: `q015. If you eliminate c from the last two equations of the system

a + 2 b - c = -5

4 a - b - c = 3

-2a + b + 3 c = 5

you get the equation

10 a - 2 b = 14

If you eliminate c from the first and third equations of this system you get the equation

a + 7 b = -10

What values of a and b solve this system of two equations?

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Your solution:

We have 10 a - 2 b = 14 and a + 7 b = -10.

To solve for b we multiply the second equation by -10 and eliminate a

-10(a + 7b = -10) becomes -10a - 70b = 100

Now add the two together.

-10a - 70b = 100 plus 10 a - 2 b = 14 becomes -72b = 114 so b = -2

Now we plug in -2 to an aquation to find a

a + 7(-2) = -10

a + -14 = -10

a=4

So a = 4 and b = -2

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assignment 2

assignment 2

assignment 2

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