#$&* course Mth 163 1/20 5 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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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: Ok ------------------------------------------------ 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: Moving the original point -1 units in the horizontal direction yields: (-1,0) Shifting the point 3 units vertically would result in the following point: (0,3) Shifting the point 4 times as far from the x axis would not work because 4*0=0, so the point would remain (0,0) If you move the point 4 times away from the x axis, (4*0=0), and shift the horizontal point -1 units in the horizontal direction, (-1,0) and shift the point 3 units upwards, the coordinates are (-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 (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: Ok ------------------------------------------------ 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 the point 4 times further from the x axis results in the following points: 0*4=0, (0,0) 1,1 = 1*4=4 (1,4), (-1, 1*4= (-1,4) Moving the points above in -1 units in the x direction. (-1,0), (0,4), and (-2,4) Increase the previous points 3 units in the y direction. (-2, 7) (0,7) and (-1,3) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Moving each point 4 times further from the x axis: The point (0, 0) is 0 units from the x axis. Multiplying this distance by 4 still gives you 0. So the point (0, 0) will remain where it is. The points (-1, 1) and (1, 1) are both 1 unit above the x axis. Multiplying this distance by 4 gives us 4 * 1 = 4. The x coordinates will not change, so our new points are (-1, 4) and (1, 4). At this stage our three points are (-1, 4) (0, 0) (1, 4) Horizontally shifting each point -1 units, our x coordinates all change by -1. We therefore obtain the points (-1 + (-1), 4) = (-2, 4), (0 + -1, 0) = (-1, 0) and ((1 + (-1), 4) = ( 0, 4), so our points are now (-2, 4) (-1, 0) ( 0, 4) Vertically shifting each point 3 units, our y coordinates all change by 3. We therefore obtain the points (-2, 4 + 3) = (-2, 7) (-1, 0 + 3) = (-1, 3) and ( 0, 4 + 3) = ( 0, 7) Self-critique: ok ------------------------------------------------ Self-critique rating: Question `q004: On the coordinate axes you used in the preceding, sketch the parabola corresponding to the three basic points (0, 0), (-1, 1) and (1, 1). Then sketch the parabola corresponding to your three circled basic points. Then sketch the parabola corresponding to three basic points you indicated with 'x's'. Finally sketch the parabola corresponding to the three basic points you indicated with '+'s'. Describe how each parabola is related to the one before it. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The first parabola is fatter and shorter than the second parabola. The second parabola is shifted to the left by 1 and stretched by 4. The third parabola is shifted up 3 units, so it is the same size as the second, just moved up. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: Ok ------------------------------------------------ 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: I will add the first and second equations together, my goal is to eliminate the c, so I will first multiply the second equation by -1. 2a+3b+c=128 -60a 3b c= -90 Resulting in our newly eliminated equation: -58a 2b= 38 I will now multiply the third equation by -1 and add it to the second equation. 60a + 5b + c= 90 -200a 10 c=0 New equation: -140a 5b = 90 I will now take the first result of the first elimination and multiply it by 5, add it with the second result of the second elimination after it has been multiplied by -2. 5(-58a 2b=38) = -290a -10b= 190 -2(-140a 5b = 90) = 280a + 10b =-180 yielding: -10a = 10 a= -1 I will take this result and plug it back into my first elimination to find B. -58(-1) -2b=38 58 -2b= 38 -2b=-20 b=10 I will take this value along with the A value and plug it into the original form to find c. 2(-1) + 3(10) + c =128 c = 128 28 c=120 confidence rating #$&*: 2.0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): Ok, I think I did the problem correctly, just to much of it. They say bigger is always better but is more better as well? ------------------------------------------------ Self-critique 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): Same answer as before! More has to be better!!!
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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 c= 128-28= c=100 confidence rating #$&*: 3.0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: Putting the point into the y=ax^2 +bx + c form is the following: -2= a(1^2=1)+b(1) + c confidence rating #$&*: 2.0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We substitute y = -2 and x = 1 to obtain the equation -2 = a * 1^2 + b * 1 + c, or a + b + c = -2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok ------------------------------------------------ Self-critique rating: ********************************************* Question: `q010. If a graph of y vs. x contains the points (1, -2), (3, 5) and (7, 8), as in the preceding question, then what two equations do we get if we substitute the x and y values corresponding to the point (3, 5), then the point (7, 8) into the form y = a x^2 + b x + c? (each point will give us one equation) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The system of equations results in repeating the last question for each point which is listed below. 5=a(3^2) + b(3) +c 5=9a + 3b + c 8= a(7^2) + b(7) + c 8= 49a + 7b + c confidence rating #$&*: 2.0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): Ok ------------------------------------------------ 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: Repeating the same steps yields the following system of equations: -2=a + b + c 5= 9a + 3b + c 8= 49a + 7b + c Now, I will eliminate values to yield the values of my variables. First I will eliminate the c variable. I will multiply the second equation by -1 and add to the first equation. a + b + c = -2 -9a 3b c = -5 -8a 2b = -7 Now I will add the second equation by the third equation, but first will multiply the third equation by -1. 9a + 3b + c= 5 -49a -7b c= -8 -40a -4b = -3 I will now the elimination results by each other, first though, I will multiply the first elimination result by -2. 16a + 4b = 14 -40a 4b = -3 -24a = 11 a= -11/24= -.4583333 plugging a into one of the eliminations will give me b. -8 ( -.4583) -2b = -7 -2b = -3.3333 b= 1.667 to find c I will plug a and b into one of the original equations. -11/24 + 1.667 + c = -2 c= -.79 confidence rating #$&*: 2.0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): I messed up on my calculations with step b. However, I do see my mistakes. ------------------------------------------------ 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: The following will all be quadratic functions. y = -.45833x^2 + 5.3333x 6.875 y = -.45833(1)^2 + 5.3333(1) 6.875= -2.03 y = -.45833(3)^2 + 5.3333(3) 6.875= 4.9 y = -.45833(5)^2 + 5.3333(5) 6.875= 8.16 y = -.45833(7)^2 + 5.3333(7) 6.875= 7.9 confidence rating #$&*: 2.0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): Ok ------------------------------------------------ Self-critique 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): Ok ------------------------------------------------ Self-critique Rating: #(*!