Assignment Two-QA

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course Mth 163

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

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_

259.htm

.

Your solution, attempt at solution. If you are unable to attempt a solution, give a

phrase-by-phrase interpretation of the problem along with a statement of what you do

or do not understand about it. This response should be given, based on the work you

did in completing the assignment, before you look at the given solution.

002.

Note that there are 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 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:

-4,-1

-3, 2

-4,-5

-4,-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:

I messed up on a couple of them, and Im not real sure what I did wrong

You need to explain what you were thinking. I can't tell what you did wrong just from a final answer.

&#Your response did not agree with the given solution in all details, and you should therefore have addressed the discrepancy with a full self-critique, detailing the discrepancy and demonstrating exactly what you do and do not understand about the parts of the given solution on which your solution didn't agree, and if necessary asking specific questions (to which I will respond).

&#

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Self-critique rating #$&*1

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:

-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 #$&*3

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), (-1,5), (1,5)

(-1,0), (-2,5), (0,5)

(0,3), (1,8), (-1,8)

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:I did this wrong, but looking at your solution I understand what I did and

will hopefully not do it again.

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Self-critique rating #$&*1

Question `q004: On the coordinate axes you used in the preceding, sketch the

parabola corresponding to the three basic points (0, 0), (-1, 1) and (1, 1).

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

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

Finally sketch the parabola corresponding to the three basic points you indicated with

'+'s'.

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

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Your Solution:

The first one with just a normal dot, kind of looks like a bowl. Then the second one

with the circles has the same center point and is still only two units between the two

end points but its higher up. The third parabola, the one with the x's is the exact same

only the whole thing is moved over one unit. The one with the plus signs is the exact

same only moved to the left one unit and moved up four units.

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

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Self-critique rating #$&*3

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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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

60a+5b+c=90

(-)2a+3b+c=-38

=============

58a+2b=-38

200a+10b+c=0

(-)2a+3b+c=128

=============

198a+7b=-128

confidence rating #$&* 2

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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):

ok

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Self-critique rating #$&*3

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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:

58a+2b=-38

7(58a+2b=-38)

406a+14b=-266

198a+7b=-128

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

- 396a-14b=256

406a+14b=-266

-396a-14b=256

10a=-10

/10=/-10

a=-1

confidence rating #$&* 2

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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 #$&*3

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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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

58(-1)+2b=-38

-58+2b=-38

+58 +58

2b=20

/2 /2

b=10

198(-1)+7b=-128

-198+7b=-128

+198 +198

7b=70

b=10

confidence rating #$&* 3

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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 #$&*3

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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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

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

-2+30+c=128

28+c=128

-28 -28

c=100

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

-200+100+100=0

-100+100=0

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.

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Self-critique (if necessary):

ok

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Self-critique rating #$&*3

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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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

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

confidence rating #$&* 3

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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 #$&*3

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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)

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

(3,5)

y=ax^2+bx+c

5=3a^2+3b+c

(7,8)

y=ax^2+bx+c

8=7a^2+7b+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.

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Self-critique (if necessary):ok

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Self-critique rating #$&*3

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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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

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

5=3a^2+3b+c

8=7a^2+7b+c

3a^2+3b+c=5

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

=================

2a^2+2b=7

7a^2+7b+c=8

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

=================

6a^2+6b+c=10

confidence rating #$&* 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.

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Self-critique (if necessary):

????????????I tried eliminating a so I could slove for b and work from there, but

since a and b had the same number I just ended up cancelling both out, and Im not

sure how not to do that.

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Self-critique rating #$&*0

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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 = a x^2 + b x + c

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

y=(-0.45833)(1)^2+(5.333)(1)+(-6.875)

-0.45833^2+5.333+-6.875

5.5430663889+-6.875

y=-1.3319336111

y=(-0.45833)(3)^2+(5.333)(3)+(-6.875)

1.8905975001+(5.333)(3)+(-6.875)

1.8905975001+15.999+-6.875

y=11.0145975001

y=(-0.45833)(5)^2+(5.333)(5)+(-6.875)

5.2516597225+(5.333)(5)+(-6.875)

5.2516597225+26.665+(-6.875)

25.0416597225

y=(-0.45833)(7)^2+(5.333)(7)+(-6.875)

10.2932530561+30.456

40.7492530561

confidence rating #$&* 2

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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."

&#Your work looks good. See my notes. Let me know if you have any questions. &#

#$&*