#$&* course Mth 163 1/17/13 around 7:00p.m. 000. As with other documents you have seen so far, you will copy this document into a text editor to complete it before submitting it. However in a text editor the graphs won't show. So you might have to look back at this document to see the graphs.
.............................................
Given Solution: The vertex of the 'blue' parabola is at the point (2, -1), the 'lowest' point on the parabola. The vertex of the 'purple' parabola is at the point (-2, 3), the 'highest' point on the parabola. Self-critique: ok ------------------------------------------------ Self-critique rating: ok Question `q002: What are the coordinates of the other two 'basic points' of each parabola? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The coordinates of the points on the blue graph are (0, 3) & (4, 3). The coordinates of the points on the purple graph are close to (-3, 1) & (-1,1). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: ok ------------------------------------------------ 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: You would move to the right or left 1 and vertically up 2. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: For the parabola with vertex (2, 1), if we move 1 unit to the right or left we will be at the point (3, 1) or (1, 1), putting us directly below one of the other two basic points. If we then move 2 units upward, we will be at the point (3, 3) or (1, 3). So if we move 1 unit to the right or left, we need to move 2 units upward to get to another basic point. Self-critique: ok ------------------------------------------------ Self-critique rating: ok Question `q004: For each of the other two parabolas, how far would we have to move to the right or the left, starting from the vertex, in order to be directly above or below another of its 'basic points'? How far would we then have to move in the vertical direction to reach that point? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: For the blue graph you would move to the left or right 1 unit and up 4 units. For the purple graph you would move left or right 1 unit and down 4 units. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: I dont understand why you would just move 2 units and 1 unit. That would not put you at the vertex???
.............................................
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): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q006. Solve the following system of simultaneous linear equations using the method of elimination: 4a + 5b = 18 6a + 9b = 30. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: In this equation we have to multiply both equations by a number in order to get something to cancel. I chose to multiply the top equation by 6 giving us 24a + 30b = 108. Then I multiplied the bottom equation by -4 giving -24a -36b = -120. Cancel out the 24a & -24a and subtract the other numbers. This leaves us with -6b = -12. Divide 12 by 6 and the answer is b = 2. To find a we must substitute b=2 into one of the equations. I chose the first equation where substitution gives you 4a + 5*2 = 18. If you multiply 5*2 then subtract it from 18 you are left with 4a = 8. Then dividing 8 by 4 we are left with a = 2. So the answer is a=2 and b = 2. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: In the system 4a + 5b = 18 6a + 9b = 30 we see that the coefficients of b are relatively prime; they therefore have a least common multiple equal to 5 * 9. The coefficients 4 and 6 of a have a least common multiple of 12. common multiple of 12. We have a choice of which variable to eliminate. We could 'match' the b by multiplying the first equation by 9 and the second by -5, or we could match the coefficients of a by multiplying the first equation by 3 and the second by -2. Either choice would work. The numbers required to 'match' the coefficients of a are smaller, but the numbers required to 'match the coefficients of b would otherwise work equally well. Choosing to 'match' the coefficient of a, we obtain 3 * [4a + 5b ] = 3 * 18 -2 * [ 6a + 9b ] = -2 * 30, so the system becomes 12 a + 15 b = 54 -12 a - 18 b = -60. Adding the equations we get -3 b = -6, so b = 2. Substituting this value of b into the first equation we obtain 4 a + 5 * 2 = 18, or 4 a + 10 = 18, which we easily solve to obtain a = 2. Substituting this value of a into the second equation we obtain 6 * 2 + 9 * 2 = 30, which verifies our solution. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q007. If y = 5x + 8, then for what value of x will we have y = 13? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To solve we must substitute 13 in for y giving us 13 = 5x + 8. Then subtract 8 from 13 leaving 5 = 5x. Divide 5 by 5 and x = 1. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): ok ------------------------------------------------ Self-critique Rating: 3 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. ********************************************* Question: `q001. 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: I did not fully get the answer but here is as far as I got: I wrote the first and second equation down and eliminated c by multiplying the second equation by -1. This gave me -60a - 5b -c = -90. When I eliminated the c I was left with -58a -2b = 38. I then took the first and third equation down and eliminated c again. This left me with the equation -198a -7b = 128. I then took the equations I was left with and eliminated b by multiplying the first equation by 7 and the second equation by -2. I was left with -406a - 14b = 266 and 396a + 14b = -256. When I eliminated b I was left with -10a = 10. Then divide 10 by 10 leaving me with a = 1. I was not sure what to do after this. I thought about possibly taking those equations I was left with and eliminate the a variable and solve but I couldnt find a number to correlate so I am stuck. confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): I got all of those answers but I thought the problem was to solve for a, b and c??? I guess that is why you said begin to solve. I should read the instructions a little better. ------------------------------------------------ Self-critique Rating: 3
.............................................
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): I got the signs wrong, but everything else looked right. I am not sure why you did not follow through and find out the answer for b as well??? ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q003. 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: I substituted -1 in for a and added that to the -38 leaving me with 2b = 20. When you divide 20 by 2 we are left with b = 10. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q004. 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: I substituted a = -1 and b = 10 into the first equation giving me 2 (-1) + 3 (10) + c = 128. I then multiplied 2 by -1 and 3 by 10 giving me -2 + 30 + c = 128. I then added -2 and 30 together and subtracted that from 128 giving me c = 100. I verified the result by substituting all three answers into the second equation. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: Using first equation 2a + 3b + c = 128 we obtain 2 * -1 + 3 * 10 + c = 128, which we easily solve to get c = 100. Substituting these values into the second equation, in order to check our solution, we obtain 60 * -1 + 5 * 10 + 100 = 90, or -60 + 50 + 100 = 90, or 90 = 90. We could also substitute the values into the third equation, and will again obtain an identity. This would completely validate our solution. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q005. The graph you sketched in a previous assignment contained the given points (1, -2), (3, 5) and (7, 8). We are going to use simultaneous equations to obtain the equation of that parabola. A graph has a parabolic shape if its the equation of the graph is quadratic. The equation of a graph is quadratic if it has the form y = a x^2 + b x + c. y = a x^2 + b x + c is said to be a quadratic function of x. To find the precise quadratic function that fits our points, we need only determine the values of a, b and c. As we will discover, if we know the coordinates of three points on the graph of a quadratic function, we can use simultaneous equations to find the values of a, b and c. The first step is to obtain an equation using the first known point. What equation do we get if we substitute the x and y values corresponding to the point (1, -2) into the form y = a x^2 + b x + c? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If we substituted (1, -2) into the equation given we will get -2 = a 1^2 + b *1 + c. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): I didnt think about simplifying it all, but I understand why you did and how you obtained this. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q006. 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: If you substitute the values in for (3, 5) will get 5 = a 3^2 + b *3 + c. Simplifying this equation would give you 5 = 9a + 3b + c If you substitute the values in for (7,8) you will get 8 = a 7^2 + b *7 + c. Simplifying this equation would give you 8 = 49a + 7b + c confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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 = 7. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I dont understand why the last equation is equal to 7??? ------------------------------------------------ Self-critique Rating: 1
.............................................
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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I am not sure why I got a different solution for c, but I would like to see it. ------------------------------------------------ Self-critique Rating: 2
.............................................
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). Self Critique: I do not understand how you got these answers????? 002. ********************************************* Question: `q001. 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: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: `q002. 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: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: Neither variable is as easy to eliminate as in the last problem, but the coefficients of b are significantly smaller than those of a. So here we choose eliminate b.It would also have been OK to choose to eliminate a. To eliminate b we will multiply the first equation by -7 and the second by 2, which will make the coefficients of b equal and opposite. The first step is to indicate the multiplications: -7 * ( 58 a + 2 b) = -7 * -38 2 * ( 198 a + 7 b ) = 2 * (-128) Doing the arithmetic we obtain -406 a - 14 b = 266 396 a + 14 b = -256. Adding the two equations we obtain -10 a = 10, so we have a = -1. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q003. Having obtained a = -1, use either of the equations 58 a + 2 b = -38 198 a + 7 b = -128 to determine the value of b. Check that a = -1 and the value obtained for b are validated by the other equation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: `q004. 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: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: Using first equation 2a + 3b + c = 128 we obtain 2 * -1 + 3 * 10 + c = 128, which we easily solve to get c = 100. Substituting these values into the second equation, in order to check our solution, we obtain 60 * -1 + 5 * 10 + 100 = 90, or -60 + 50 + 100 = 90, or 90 = 90. We could also substitute the values into the third equation, and will again obtain an identity. This would completely validate our solution. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q005. The graph you sketched in a previous assignment contained the given points (1, -2), (3, 5) and (7, 8). We are going to use simultaneous equations to obtain the equation of that parabola. A graph has a parabolic shape if its the equation of the graph is quadratic. The equation of a graph is quadratic if it has the form y = a x^2 + b x + c. y = a x^2 + b x + c is said to be a quadratic function of x. To find the precise quadratic function that fits our points, we need only determine the values of a, b and c. As we will discover, if we know the coordinates of three points on the graph of a quadratic function, we can use simultaneous equations to find the values of a, b and c. The first step is to obtain an equation using the first known point. What equation do we get if we substitute the x and y values corresponding to the point (1, -2) into the form y = a x^2 + b x + c? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: `q006. 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: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: Using the second point we substitute y = 5 and x = 3 to obtain the equation 5 = a * 3^2 + b * 3 + c, or 9 a + 3 b + c = 5. Using the third point we substitute y = 8 and x = 7 to obtain the equation 8 = a * 7^2 + b * 7 + c, or 49 a + 7 b + c = 7. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q007. If a graph of y vs. x contains the points (1, -2), (3, 5) and (7, 8), as was the case in the preceding question, then we obtain three equations with unknowns a, b and c. You have already done this. Write down the system of equations we got when we substituted the x and y values corresponding to the point (1, -2), (3, 5), and (7, 8), in turn, into the form y = a x^2 + b x + c. Solve the system to find the values of a, b and c. What is the solution of this system? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q008. 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 aspecific 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: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: Substituting the values of a, b and c into the given form we obtain the equation y = - 0.45833 x^2 + 5.33333 x - 6.875. When we substitute 1 into the equation we obtain y = -.45833 * 1^2 + 5.33333 * 1 - 6.875 = -2. When we substitute 3 into the equation we obtain y = -.45833 * 3^2 + 5.33333 * 3 - 6.875 = 5. When we substitute 5 into the equation we obtain y = -.45833 * 5^2 + 5.33333 * 5 - 6.875 = 8.33333. When we substitute 7 into the equation we obtain y = -.45833 * 7^2 + 5.33333 * 7 - 6.875 = 8. Thus the y values we obtain for our x values yield the points (1, -2), (3, 5) and (7, 8). These are the points we used to obtain the formula. We also get the additional point (5, 8.33333). " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q008. Substitute the values you obtained in the preceding problem for a, b and c into the form y = a x^2 + b x + c, in order to obtain aspecific quadratic function. What is your function? What y values do you get when you substitute x = 1, 3, 5 and 7 into this function? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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). " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!