Open QA 2

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course Mth 163

002.

Note that there are 12 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 pointSketch 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:

(-4, -1)

(-3, 2)

(-12, -1)

(-13, 2)

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:

If we shifted the point 4 times further from its current position relative to the x axis, it would move the -3 x coordinate 4 times or -12. I suppose you’re referring to moving the point away from the x axis by increasing its vertical value by a multiple of 4. If I had done this portion correctly I could have found the final portion correctly also.

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Self-critique rating: 3

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?

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Your Solution:

(-1,0)

(0,3)

(0,0)

(-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

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Self-critique rating: ok

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:

(0,0)

(-1, 4)

(1, 4)

Then:

(-1,0)

(-2,4)

(0, 4)

Then:

(-1,3)

(-2, 7)

(0, 7)

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

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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.

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Your Solution:

Either sides of the vertex lie one unit to the left or right of the vertex, and then 4 units above the vertex. The first two vertexes are on the origin, the 3rd is 1 unit to the left, and the 4th vertex is 3 units up. The distance between vertex and the points is the same for all 3 parabolas.

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: The shapes are indeed all the same except for the very first one we started with before making our shifts and changes.

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Self-critique rating: ok

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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: Subtracting the 2nd equation from the 3rd we eliminate c and obtain

[200a+10b+c=0] - [60a + 5b + c = 90]

140a+5b=-90

Subtract the 1st from the 3rd and we obtain:

[200a+10b+c=0] - [2a + 3b + c = 128]

198a+7b = -128

Our two equations are:

140a+5b=-90

198a+7b = -128

For the first equation we subtract 5b and divide by 140 to obtain:

A = [-90-5b] / 140

A = -0.64 - 0.04b

198(-0.64 - 0.04b) + 7b = -128

-126.72 - 7.92b+7b = -128

-126.72 - 0.92b = -128

-0.92b = -1.28

B = 1.4

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).

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Self-critique (if necessary):I believe I arranged the equations properly but I went farther in my calculations than was necessary, and they may have been a little off.

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Self-critique rating:

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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: Multiplying the first equation by the negative coefficient of b in the 2nd equation, and multiplying the second equation by the coefficient of the b in the 1st equation gives us

-7[58 a + 2 b = -38] = -406a - 14b = 266

2[198 a + 7 b = -128] = 396a + 14b = -256

Adding the 2nd new equation to the first we get

-10a = 10

A = -1

Substituting this into either equation we get:

58(-1) + 2b = -38

-58 + 2b = -38

2b = 20

B = 10

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.

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Self-critique (if necessary): ok

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Self-critique rating:

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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: Substituting this into either equation we get:

58(-1) + 2b = -38

-58 + 2b = -38

2b = 20

B = 10

If we substitute the value of -1 into either equation we get the same answer for b. -58 + 20 = -38 our solution is valid.

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.

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Self-critique (if necessary): ok

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Self-critique rating: ok

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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

28 + c = 128

C = 100

200(-1) + 10(10) + 100 = 0

Our solution is valid. -200 + 200 = 0

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.

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Self-critique (if necessary): ok

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Self-critique rating: ok

@&

Good. You've been one step ahead on the solution, which is even better.

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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

A+ b + c = -2

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.

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Self-critique (if necessary): ok

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Self-critique rating:

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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

9a + 3b + c = 5

8 = a (7)^2 + b (7) + c

49a + 7b + c = 8

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.

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Self-critique (if necessary): ok

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Self-critique rating: ok

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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:

49 a + 7 b + c = 8.

9 a + 3 b + c = 5.

a + b + c = -2.

Solving the third equation for c we obtain:

C = -2 - a - b

49a + 7b + (-2 - a - b) = 8

@&

To get the next equation you would add 2 to both side, which would give you 48 a + 6 b = 10. This will of course affect the agreement of your solution with the given solution.

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48a + 6b = 6

9a + 3b + (-2 - a - b) = 5

8a + 2b = 3

Our two equations are

48a + 6b = 6

8a + 2b = 3

Multiplying the first equation by 1 and the second by -3 gives us::

[48a + 6b = 6] - [-24a - 6b = -9]

24a = -3

A = -0.125

Substituting this into our 2nd equation gives us:

8(-0.125) + 2b = 3

-1 + 2b = 3

2b = 4

B = 2

Substituting the values for a and b into our equation for c gives us:

C = -2 - (-0.125) - 2

C = -3.875

Substituting these 3 values into a + b + c = -2 validates our answer that all values are correct.

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.

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Self-critique (if necessary): ????If I was able to derive 3 values that all fit the 3 equations then my solution is assuredly correct????

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Self-critique rating:

@&

It's hard to get through these problems without making some sort of arithmetic error. The most common errors involve signs, and you made such an error early in the problem. This is why your solution doesn't match the given solution.

However you clearly understand the process.

*@

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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 = -0.125x^2 + 2x -3.875

Y = -0.125(1)^2 + 2(1) -3.875

Y = -2

Y = -0.125(3)^2 + 2(3) -3.875

Y = -1.125 + 6 -3.875

Y = 1

Y = -0.125(5)^2 + 2(5) -3.875

Y = -3.125 + 10 - 3.875

Y = 3

Y = -0.125(7)^2 + 2(7) -3.875

Y = -6.125 + 14 - 3.875

Y = 4

confidence rating #$&*: Getting all y values as whole numbers that appear to follow a pattern corresponding with the points given makes me think that my solution is correct, although I’m aware the numbers might not always be that pretty.

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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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Self-critique (if necessary):

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Self-critique Rating:

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Self-critique (if necessary):

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Self-critique rating:

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Self-critique (if necessary):

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#*&!

&#Good responses. See my notes and let me know if you have questions. &#