Assignment 10 QA

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course Mth 163

10/13/2014 11:14 am

010.

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Question: `q001. Note that this assignment has 11 questions

Sketch the function y = x and describe your graph. Describe how the graphs of y = .5 x and y = 2 x compare with the graph of y = x.

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Your solution:

Note: using “…….” To indicate these points keep traveling in this pattern since x is not specified.

This graph is linear laying diagonally with points …(-3, -3), (-2, -2), (-1, -1), (0, 0), (1, 1), (2, 2). (3, 3)….

When we graph y = .5x

These points are similar and lay diagonally but the y values are half as much (except the point at (0, 0)) as the previous graph. These points are:

….(-3, -1.5), (-2, -1), (-1, -.5), (0, 0), (1, .5), (2, 1), (3, 1.5)

When y = 2x

These points also lie diagonally but the y values are 2 times the amount if the original graph (except the point at (0, 0)):

(-2, -4), (-1, -2), (0, 0), (1, 2), (2, 4), (3, 6)….

confidence rating #$&*:

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3

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Given Solution:

The graph of y = x consists of a straight line through the origin, having slope 1. This line has basic points (0,0) and (1,1).

The points of the graph of y = .5 x all lie twice as close to the x-axis as the points of the graph of y = x. The point (0,0) of the y = x graph is already on the x-axis, so the corresponding point on the graph of y = .5 x is also (0,0). The point (1,1) of the y = x graph lies 1 unit above the x-axis, so the corresponding point on the graph of y = .5 x will lie twice as close, or .5 units above the x-axis, so that the corresponding point is (1, .5). The graph of y = .5 x Thus passes through the points (0,0) and (1,.5).

Of course this result could have been found by simply plugging 0 and 1 into the function y = .5 x, but the point here is to see that we can get the same result if we think of moving all points twice as close. This order thinking will be useful when dealing with more complex functions.

Thinking along similar lines we expect the points of the graph of y = 2 x to all lie twice as far from the x-axis as the points of the function y = x. Thus the two basic points (0,0) and (1,1) of the y = x graph will correspond to the points (0,0) and (1,2) on the graph of y = 2 x.

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Self-critique (if necessary):

I did not include the slope or the effect of the slope or the basic points or the effects on them. But my graphs and the effects were the same.

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Self-critique rating:

3

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Question: `q002. If we were to sketch all the graphs of the form y = a x for which .5 < a < 2, what would our sketch look like?

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Your solution:

These points would all lie between the 2 follow-up graphs we made in the previous exercise. In other words between, y = .5x and y = 2x these points would also make straight lines.

confidence rating #$&*:

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3

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Given Solution:

If a =.5 then our function is y = .5 x and the basic points will be (0,0) and (1,.5), as seen in the preceding problem.

Similarly if a = 2 then our function is y = 2 x, with basic points (0,0) and (1,2).

For .5 < a < 2, our functions will lie between the graphs of y = .5 x and y = 2 x. Since these two functions have slopes .5 and 2, the slopes of all the graphs will lie between .5 and 2.

We could represent these functions by sketching dotted-line graphs of y = .5 x and y = 2 x (the dotted lines indicating that these graphs are not included in the family, because the < sign does not include equality). We could then sketch a series of several solid lines through the origin and lying between the two dotted-line graphs.

STUDENT COMMENT

I don’t know where to go from this point. I graphed the closest thing I could come up with

but I don’t know how to explain what it is doing.

INSTRUCTOR SUGGESTION

You should graph the functions y = .5 x, y = .6 x, y = .8 x, y = 1.1 x, y = 1.5 x and y = 2 x, all on the same graph.

Graph each function by plotting its two basic points (the x = 0 point and the x = 1 point), then sketching the straight line through these points.

Using your graphs, estimate where the graph of y = .7 x, y = 1.3 x and y = 1.8 x lie.

Then insert your description, according to instructions at the end of this document, along with any other work you do in response to other suggestions made below, and resubmit this document.

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Self-critique (if necessary):

OK

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Self-critique rating:

OK

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Question: `q003. Describe how the graphs of y = x - 2 and y = x + 3 compare with the graph of y = x. If we were to sketch all graphs of the form y = x + c for -2 < x < 3, what would our graph look like?

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Your solution:

Note: using “…….” To indicate these points keep traveling in this pattern since x is not specified.

When y = x -2 will have y values that are - 2 from the original graphs y values. These values will be:

…..(-3, -5), (-2, -4), (-1, -3), (0, -2), (1, -1), (2, 0), (3, 1)…..

When y = x + 3 These points will have y values that shift + 3 from the original graph:

….(-3, 0), (-2, 1), (-1, 2), (0, 3), (1, 4)…….

Once again the graphs contained in the for y = x + c when -2 < x < 3 will all lie between the 2 previous graphs.

confidence rating #$&*:

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Given Solution:

The graph of y = x - 2 lies at every point 2 units below the corresponding point on the graph of y = x, so this graph is parallel to the graph of y = x and 2 units lower. Similarly the graph of y = x + 3 lies parallel to the graph of y = x and 3 units higher.

To sketch the family y = x + c for -2 < x < 3, we first can draw dotted-line graphs of y = x - 2 and y = x + 3, then a series of several solid line graphs, all parallel to the graph of y = x, lying between the two dotted-line graphs.

STUDENT COMMENT: I got a little confused with y = x + c part, but I understand the first part completely.

** The instructions said to sketch all graphs of the form y = x + c for -2 < x < 3. So for example c could be -1, 0, 1 or 2, giving us the functions y = x - 1, y = x, y = x + 1 andy x+ 2. c could also be -1.9, or .432, or 2.9, giving us functions y = x - 1.9, y = x + .432, y = x + 2.9. c can be any number between -2 and 3.

These graphs are as described in the given solution. **

STUDENT COMMENT

i didnt really understand how to sketch y=x+c even after reading the

intructors comments in the given solution

INSTRUCTOR RESPONSE

Suppose you were to graph y = x + c for c values -2, -1.9, -1.8, -1.7, ..., 2.8, 2.9, 3.0. This would include 50 graphs.

Each of the 50 graphs would lie .1 unit higher than the one before it.

The lowest of the graphs would be the c = -2 graph, y = x - 2.

The highest of the graphs would be the c = 3 graph, y = x + 3.

All the graphs would be parallel.

If necessary, you can graph y = x - 2, then y = x - 1.9, then y = x - 1.8. You won't want to graph all 50 lines, but you could then skip to y = x + 2.8, y = x + 2.9 and y = x + 3.

STUDENT COMMENT

After reading the comments above I agree that I am a little confused.

INSTRUCTOR RESPONSE

You need to self-critique, giving me a detailed statement of what you do and do not understand about each line and each phrase in the given solution.

You should in any case follow the suggestion at the end of the given solution. Graph the indicated graphs, then insert your explanation.

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Self-critique (if necessary):

OK

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Self-critique rating:

OK

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Question: `q004. Describe how the graph of y = 2 x compares with the graph of y = x.

Describe how the graph of y = 2 x - 2 compares with the graph of y = 2 x.

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Your solution:

The y = x graph is linear laying diagonally with points …(-3, -3), (-2, -2), (-1, -1), (0, 0), (1, 1), (2, 2). (3, 3)….

This has a slope of 1.

When y = 2x

These points also lie diagonally but the y values are 2 times the amount if the original graph (except the point at (0, 0)):

(-2, -4), (-1, -2), (0, 0), (1, 2), (2, 4), (3, 6)….

This has a slope of 2.

Then the graph y = 2x -2 contains y values that are 2 times the original values for y =x and then shifted down for -2, meaning this graph lies 2 units below y = 2x.

The points would follow the pattern:

….(-3, -8), (-2, -6), (-1, -4), (0, -2), (1, 0), (2, 2)….

This has a slope of 2.

confidence rating #$&*:

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3

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Given Solution:

The graph of y = 2 x lies at every point twice as far the x-axis as that of y = x. This graph passes through the points (0,0) and (1, 2), i.e., passing through the origin with slope 2.

The graph of y = 2x - 2 will lie 2 units below the graph of y = 2 x. This graph will therefore have a slope of 2 and will pass-through the y axis at (0, -2).

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Self-critique (if necessary):

Did not include the b variable for the y intercept.

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Self-critique rating:

3

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Question: `q005. Suppose we graph y = 2 x + c for all values of c for which -2 < c < 3. What with our graph look like?

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Your solution:

All of the graphs would lie between the graph y = 2x -2 and y = 2x + 3 and all of these graphs would have the slope of 2 being linear.

confidence rating #$&*:

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Given Solution:

Each graph will lie c units vertically from the graph of y = 2 x, therefore having slope 2 the passing through the y-axis at the point (0, c). The family of functions defined by y = 2 x + c will therefore consist of a series of straight lines each with slope 2, passing through the y-axis between (0, -2) and (0, 3).

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Self-critique (if necessary):

Did not state that the graphs would be vertically displaced from y = 2x according to their c value.

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Self-critique rating:

3

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Question: `q006. Sketch two points, not particularly close to one another, with one point in the second quadrant and the other in the first, with clearly different y values. Label the first point (x1, y1) and the second (x2, y2). Draw a straight line passing through both of these points and extending significantly beyond both. In terms of the symbols x1, x2, y1, and y2, what is slope of this straight line?

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Your solution:

For the points (x1, y1) and (x2, y2)

The slope is as follows:

Rise / run

(y2 - y1) / (x2 - x1)

confidence rating #$&*:

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3

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Given Solution:

The rise of a line is from y = y1 to y = y2, a rise of y2-y1. The run is similarly found to be x2-x1. The slope is therefore

slope = (y2-y1) / (x2-x1).

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Self-critique (if necessary):

OK

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Self-critique rating:

OK

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Question: `q007. On the sketch you made for the preceding problem, and add a point (x, y) on your straight line but not between the two points already labeled, and not too close to either. What is the slope from (x1, y1) to (x, y)?

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Your solution:

(y - y1) for the rise

(x - x1) for the run

Making the slope:

(y - y1) / (x - x1)

confidence rating #$&*:

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OK

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Given Solution:

The slope from (x1, y1) to (x, y) is

slope = rise/run = (y - y1) / (x - x1).

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Self-critique (if necessary):

OK

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Self-critique rating:

OK

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Question: `q008. Should the slope from (x1, y1) to (x, y) be greater than, equal to or less than the slope from (x1, y1) to (x2, y2)?

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Your solution:

These slopes should be equal since a line should have a consistent slope between all of its points.

confidence rating #$&*:

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3

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Given Solution:

The slope between any two points of a straight line must be the same. The two slopes must therefore be equal.

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Self-critique (if necessary):

OK

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Self-critique rating:

OK’

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Question: `q009. The slope from (x1, y1) to (x, y) is equal to the slope from (x1, y1) to (x2, y2). If you set the expressions you obtained earlier for the slopes equal to one another, what equation do you get?

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Your solution:

When these equations equal one another the equation obtained is:

(y2 - y1) / (x2 - x1) = (y - y1) / (x - x1).

confidence rating #$&*:

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Given Solution:

The slopes are (y2 - y1) / (x2 - x1) and (y - y1) / (x - x1). Setting the two slopes equal we obtain the equation

(y - y1) / (x - x1) = (y2 - y1) / (x2 - x1).

STUDENT COMMENT

mine is the opposite but i think i would be the same

INSTRUCTOR RESPONSE:

Your solution was (y2 - y1) / (x2 - x1), if appropriate signs of grouping are inserted to reflect your obvious intent.

The signs of both your numerator and denominator would be opposite the signs of the given solution (i.e., y2 - y1 = - (y1 - y2), and x2 - x1 = - (x1 - x2)). When divided the result would therefore be identical (negative / negative is positive). So your solution is completely equivalent to the given solution.

However note that you need to group numerator and denominator. y2-y1/x2-x1 means divide y1 by x2, subract that result from y2 then subtract x1 from that. Not what you intended, though I know what you meant.

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Self-critique (if necessary):

I believe mine is correct if I’m understanding the above explanation correctly.

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Self-critique rating:

@&

The only difference between your result and mine is that the sides of the equation are switched, which is in fact not a difference at all.

*@

3

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Question: `q010. Setting the two slopes equal you should have obtained (y - y1)/(x - x1) = (y2 - y1)/(x2 - x1). Solve this equation for y.

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Your solution:

(y - y1)/(x - x1) = (y2 - y1)/(x2 - x1)

* (x - x1)

= y - y1 = (y2 - y1)/(x2 - x1) * (x - x1)

-y1

Y = (y2 - y1)/(x2 - x1) * (x - x1) + y1

confidence rating #$&*:

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1

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Given Solution:

Starting with (y - y1) / (x - x1) = (y2 - y1) / (x2 - x1), we wish to isolate y on the left-hand side. We therefore begin by multiplying both sides by (x - x1) to obtain

(y - y1) = (y2 - y1) / (x2 - x1) * (x - x1).

We could then add y1 to both sides to obtain

y = (y2 - y1) / (x2 - x1) * (x - x1) + y1.

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Self-critique (if necessary):

OK

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Self-critique rating:

OK

If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties.

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Question: `q011. What is the slope of the straight line defined by the points (2, 5) and (6, 8)?

In terms of x and y, what is the slope of the straight line from (2, 5) to the unspecified point (x, y)?

It (x, y) lies on the straight line from (2, 5) to (6, 8), then the slope from (2, 5) to (x, y) must be equal to the slope from (2, 5) to (6, 8). Your answers to the preceding two questions therefore constitute two expressions for the slope of the line, one expression being a simple fraction and the other a symbolic expression in terms of x and y.

Set these two expressions equal to get an equation involving x and y. What is your equation?

Solve your equation for y. What is your result?

Do the coordinates (6, 8) satisfy your equation? Should they?

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Your solution:

The slope of the line through the points (2, 5) and (6, 8) would be:

(8-5) / (6-2) = Ύ or .75

The slope for the line from (2, 5) to the unspecified point (x, y) would be defined as:

(Y - 5) / (x - 2)

These expressions as an equation:

(y - 5) / (x - 2) = (8-5) / (6-2)

Solving for y:

(y - 5) / (x - 2) = (8-5) / (6-2)

*(x- 2)

Y - 5 = Ύ * (x -2)

+5

Y = Ύ * (x -2) + 5

Y = [(3x-2) / 4] + 5

@&

If you simplify the right-hand side you get

y = 3/4 x + 7/2.

*@

Plugging in the point (6, 8)

[(3*6 - 2) / 4] +5 = y

16 / 4 + 5 = y

@&

You would get

(y-8) / (x-6) = 3/4

Both equations yield

y = 3/4 x + 7/2

when solved for y.

*@

4 + 5 = 9

This does not result in the same coordinates. I’m not sure if the equation is wrong or if this is correct.

confidence rating #$&*:

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2

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Self-critique rating:

""OK

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Self-critique (if necessary):

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Self-critique rating:

""OK

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Self-critique (if necessary):

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Self-critique rating:

#*&!

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