course MTH 163 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.
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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 #$&* OK. I am not sure how you got (-4,-1) again?
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Given Solution: Shifting the point -1 units in the horizontal direction we end up at the point (0 + (-1), 0) = (-1, 0). Shifting the point 3 units in the vertical direction we end up at the point (0, 0 + ( 3)) = (0, 3). The point (0, 0) is 0 units from the x axis. If the point is moves 4 times further from the x axis, the y coordinate will be 4 * 0 = 0. The x coordinate will not change. So the coordinates of the new point will be (0, 0). If you then shift the resulting point -1 units in the horizontal direction, it will end up at (-1, 0). If you shift this new point 3 units in the vertical direction, it will end up at (-1, 3) Self-critique: ------------------------------------------------ Self-critique rating #$&* 3 Very similar to problem #1 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 circled points 3 units in the y direction. Put a small '+' through each point. What are the coordinates of your points? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: (0,0) (-1,4) (1,4) (-2,4) (-1,0) (0,4) (-2,7) (-1,3) (0,7) confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Given Solution: Moving each point 4 times further from the x axis: The point (0, 0) is 0 units from the x axis. Multiplying this distance by 4 still gives you 0. So the point (0, 0) will remain where it is. The points (-1, 1) and (1, 1) are both 1 unit above the x axis. Multiplying this distance by 4 gives us 4 * 1 = 4. The x coordinates will not change, so our new points are (-1, 4) and (1, 4). At this stage our three points are (-1, 4) (0, 0) (1, 4) Horizontally shifting each point -1 units, our x coordinates all change by -1. We therefore obtain the points (-1 + (-1), 4) = (-2, 4), (0 + -1, 0) = (-1, 0) and ((1 + (-1), 4) = ( 0, 4), so our points are now (-2, 4) (-1, 0) ( 0, 4) Vertically shifting each point 3 units, our y coordinates all change by 3. We therefore obtain the points (-2, 4 + 3) = (-2, 7) (-1, 0 + 3) = (-1, 3) and ( 0, 4 + 3) = ( 0, 7) Self-critique: ------------------------------------------------ Self-critique rating #$&* OK 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 circle points are narrower than the original parabola. The (x) points look like the same shape as the circled points but are to the left and right of the parabola. The (+) points have the same shape as the (x) points but are below or above the parabola. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Your 'circled-points' parabola will be narrower than the original parabola through (-1, 1), (0, 0) and (1, 1). In fact, each point on the 'circled-points' parabola will lie 4 times further from the x axis than the point on the original parabola. Your 'x'-points parabola will have the same shape as your 'circled-points' parabola, but will lie to the right or left of that parabola, having been shifted -1 units in the horizontal direction. Your '+'-points parabola will have the same shape as the 'x-points' parabola (and the 'circled-point' parabola), but will lie above or below that parabola, having been shifted 3 units in the vertical direction. Self-critique: ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: How long did it take you to complete this exercise? your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv Not too long it took me awhile to make the circles accurate. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q005. Note that this assignment has 8 questions Begin to solve the following system of simultaneous linear equations by first eliminating the variable which is easiest to eliminate. Eliminate the variable from the first and second equations, then from the first and third equations to obtain two equations in the remaining two variables: 2a + 3b + c = 128 60a + 5b + c = 90 200a + 10 b + c = 0. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You first have to eliminate one of the variables. The (c) would be the easiest since all of the cs do not have a coefficient in front of them. 1) Subtract #1 by #2 60a + 10b + c = 90 2a + 3b + c = 128 =58a + 2b = -38 2) Subtract #1 by #3 200a + 10b = 0 2a + 3b + c = 128 =198a + 7b = -128 confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 #$&* 3 OK ********************************************* Question: `q006. Solve the two equations 58 a + 2 b = -38 198 a + 7 b = -128 which can be obtained from the system in the preceding problem, by eliminating the easiest variable. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You must eliminate a variable and in this problem it would be the (b) since the numbers are small. From what I learned in algebra you must multiply each equation to get rid of the (bs). 1) -7(58a + 2b) = -7 (-38) -406a 14 b = 266 2) 2(198a + 7b) = 2* -128 396a + 14b = -256 ------------------------------- -10 = 10 A=-1 confidence rating #$&*: 3 easy ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 #$&* OK. Simple problem. ********************************************* Question: `q007. Having obtained a = -1, use either of the equations 58 a + 2 b = -38 198 a + 7 b = -128 to determine the value of b. Check that a = -1 and the value obtained for b are validated by the other equation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 58(-1) + 2b= -38 -58+ 2b = -38 +58 +58 -------------------- 2b = 20 B=10 confidence rating #$&* 3 Easy problem ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 #$&* 3 easy problem ********************************************* Question: `q008. Having obtained a = -1 and b = 10, determine the value of c by substituting these values for a and b into any of the 3 equations in the original system 2a + 3b + c = 128 60a + 5b + c = 90 200a + 10 b + c = 0. Verify your result by substituting a = -1, b = 10 and the value you obtained for c into another of the original equations. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2 (-1) + 3 (10) + c = 128 -2 + 30 + c = 128 -28 -28 --------------------------- C = 100 60 (-1) + 5 (10) + 100 = 90 90 = 90 confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 #$&* OK simply problem ********************************************* Question: `q009. The graph you sketched in a previous assignment contained the given points (1, -2), (3, 5) and (7, 8). We are going to use simultaneous equations to obtain the equation of that parabola. A graph has a parabolic shape if its the equation of the graph is quadratic. The equation of a graph is quadratic if it has the form y = a x^2 + b x + c. y = a x^2 + b x + c is said to be a quadratic function of x. To find the precise quadratic function that fits our points, we need only determine the values of a, b and c. As we will discover, if we know the coordinates of three points on the graph of a quadratic function, we can use simultaneous equations to find the values of a, b and c. The first step is to obtain an equation using the first known point. What equation do we get if we substitute the x and y values corresponding to the point (1, -2) into the form y = a x^2 + b x + c? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Y = -2 X = 1 -2 = a * 1^2 + b * 1 + c confidence rating #$&* OK ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 #$&* Easy problem ********************************************* Question: `q010. If a graph of y vs. x contains the points (1, -2), (3, 5) and (7, 8), as in the preceding question, then what two equations do we get if we substitute the x and y values corresponding to the point (3, 5), then the point (7, 8) into the form y = a x^2 + b x + c? (each point will give us one equation) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Y=5 X=3 5= a * 3^2 + b * 3 + c Y=8 X=7 8= a * 7^2 + b * 7 + c confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 #$&* Easy problem ********************************************* Question: `q011. If a graph of y vs. x contains the points (1, -2), (3, 5) and (7, 8), as was the case in the preceding question, then we obtain three equations with unknowns a, b and c. You have already done this. Write down the system of equations we got when we substituted the x and y values corresponding to the point (1, -2), (3, 5), and (7, 8), in turn, into the form y = a x^2 + b x + c. Solve the system to find the values of a, b and c. What is the solution of this system? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: a + b + c = -2 9a + 3b + c = 5 49a + 7b + c =8 A = -0.458 B = 5.3 C = -6.875 confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 #$&* Easy problem, its just a different way to insert numbers in the equation. ********************************************* Question: `q012. Substitute the values you obtained in the preceding problem for a, b and c into the form y = a x^2 + b x + c, in order to obtain a specific quadratic function. What is your function? What y values do you get when you substitute x = 1, 3, 5 and 7 into this function? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1) y= -0.45833 x^2 + 5.33333x 6.875 x=1 y= -0.45833 * 1^2 + 5.33333 * 1 -6.875 = -2 x=3 y = -.45833 * 3^2 + 5.33333 * 3 - 6.875 = 5 x=5 y = -.45833 * 5^2 + 5.33333 * 5 - 6.875 = 8.33333. x=7 y = -.45833 * 7^2 + 5.33333 * 7 - 6.875 = 8 . confidence rating #$&*: OK ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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). "