Assignment 10

course Math163

03/26/2010 6:53 pm

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.

010.

*********************************************

Question: `q001. Note that this assignment has 10 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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Y=x the graph is a straight line going at a 45 degree angle through the 0 point.

Y=05 the graph is a straight line going at half way between the y=x line and the x axis or about 22.5 degrees

Y=2 the graph is a straight line going above the y=x line

confidence rating #$&* 1

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique rating #$&* Fair

*********************************************

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Y=ax y=05

I don’t really know where to go from here.

confidence rating #$&* 0

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): 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.

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.

------------------------------------------------

Self-critique rating #$&*-1

*********************************************

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Y=x-2 forma a straight line that crosses the y axis at -2 and the x axis at +2. The line is below the 0,0 point

Y=x+3 forms a straight line that crosses the x axis at -3 and the y axis at +3 above the 0,0

confidence rating #$&* 1

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

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

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.

------------------------------------------------

Self-critique rating #$&* Fair

*********************************************

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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Y=2x is 2 points steeper than y=x. If y=x is at a 45 degree angle then y=2x is at an approximate 67 degree angle. It actually creates 2 parallel lines.

the lines are parallel, as you say

they make identical angles with the x axis

the angles are much closer to 67 degrees than to 45 degrees

confidence rating #$&* 1

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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).

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

I really want to have numbers instead of letters, but I think I understand what the solution is.

------------------------------------------------

Self-critique rating #$&* Fair

*********************************************

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Y=2x+(-2

confidence rating #$&* 1

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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).

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): I don’t really know how to work with the greater than and less than

If -2 =< c =< 3, then c would take values like -2, -1.9, -1.8, ..., 2.7, 2.8, 2.9, 3, as well as all numbers in between

------------------------------------------------

Self-critique rating #$&* No OK

*********************************************

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Y=y1 and y=y2 y2-y1

X2-x1

confidence rating #$&* -2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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).

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

I don’t know how to explain the graph that I can create on graph paper

Your graph should show the points (x1, y1) and (x2, y2), at arbitrary points in the plane.

The line segment connecting your two points is the hypotenuse of a right triangle, whose legs are parallel to the x and y axes.

You will see this picture, with which you should be familiar from prerequisite courses (though not all students are), in your upcoming worksheet.

------------------------------------------------

Self-critique rating #$&* Fair

*********************************************

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)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

y-y1

x-x1

slope = rise over run

confidence rating #$&* 1

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

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

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

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

I don’t think I really have an answer.

------------------------------------------------

Self-critique rating #$&* Not OK

*********************************************

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)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

x-x1 and y-y2 will be the same as x1-x2 and y1-y2 because they are only separated by 1. They will create a parallel line

confidence rating #$&* 1

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

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

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique rating #$&* OK

*********************************************

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

x-x1 x1-x2

y-y1= y1-y2

confidence rating #$&* 1

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique rating #$&* OK

*********************************************

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

y-y1 y2-y1

x-x1= x2-x1

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

x2-x1

y= y2-y1 *(x-x1)+y1

x2-x1

confidence rating #$&* 1

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

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.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique rating #$&* Fair

&#Please see my notes and submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end). &#