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Mth163

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graph 27june2013

** **

Question: `q009. How do the values on a table for y = (x + 2)^2 compare to those for y = x^2? Use x values -3, -2, -1, 0, 1, 2, 3 to construct each table. What is the axis of symmetry for this function?

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Your solution:

&&&

Your y values will move 2 units horizontally to the left

y = (x + 2)^2 (y values)

1

0

1

4

9

16

25

y = x^2 (y values)

-9

-4

-1

0

1

4

9

The AOS is -b/2(a)

The vertex is the point of symmetry, so the vertex of y = x^2 is (0,0) the vertex of y = (x + 2)^2 is (0,0)

@&

You didn't include x values in your table.

If you had you would see clearly that (0, 0) is not a point of the graph of y = (x+2)^2.

So that point can't be the vertex.

Sketch that parabola based on your table and see if you can locate the correct vertex.

** **

using this

y = (x + 2)^2 (y values)

1

0

1

4

9

16

25

I graphed it on graph paper

** **

so my vertex would be at (-1,1)

@&

(-1, 1) is a point on the graph.

However the vertex lies on the axis of symmetry for the graph. There is no line through (-1, 1) for which the graph is symmetric about that line.

There is only one possible line of symmetry for this graph. Where is that axis of symmetry, and at what point does it intersect your graph?

*@

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Mth163

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y=x^3

** **

Precalculus I Class Notes 9/08/98

Introduction to Function Families

Analyzing a Quadratic Function

We analyze a quadratic function y = a t^2 + b t + c by determining the locations of its zeros, if any, the location of its vertex, the points 1 unit to the right and left of the vertex, and the y intercept. We can find the value of y corresponding to a given value of t by substitution; we can fine the value(s) of t corresponding to a given y using the quadratic formula.

Basic Function Families

The first three of the four basic functions are y = x, y = x^2 and y = 2^x. We graph these functions by first making a table for each. We see that y = x yields a straight-line or linear graph, y = x^2 yields a parabolic graph with vertex at the origin, and y = 2^x yields a graph which is asymptotic to the negative x axis and which increases more and more rapidly for increasing values of x.

Stretching and Shifting the Basic Functions

When a graph is stretched vertically by a given factor a, every point on the graph moves in the vertical direction until it is a time as far from the horizontal axis as it was before. If | a | is less than 1, every point moves closer to the horizontal axis and the graph appears to be compressed; if | a | is greater than 1, every point moves further from the horizontal axis and the graph appears stretched. If a < 0, then positive values of the basic function are transformed into negative values and negative values into positive, and the graph appears inverted. When the linear function y = x is vertically stretched by factor a the its slope is multiplied by a.

When a graph is shifted either horizontally or vertically, every point moves the corresponding horizontal or vertical distance, and the graph simply shifts left or right, or up or down, as the case may be.

When applying a series of stretches and shifts to a graph, we always apply the stretches first. Applying the shifts first would give a different result.

A variety of function families can be generated by applying one or more fixed transformations, and by also applying a transformation whose parameter varies over a given range.

The Number of Transformations Required to obtain the General Function for each Basic Function

A linear function is characterized by its slope and y intercept. It requires only a vertical stretch of the basic y = x function to match the slope of any desired linear function, and only a vertical shift to match the y intercept.

A quadratic function y = a t^2 + b t + c is characterized by the location of its vertex and by the value of a. It requires only three transformations to transform y = t^2 into a given quadratic. We first stretched the function vertically by factor a, then shift it horizontally and vertically to reposition the vertex.

A general exponential function y = A * 2^(kt) + c can be obtained from y = 2^t by a vertical stretch by factor A, a horizontal stretch by factor 1/k (or the horizontal compression by factor k), and a vertical shift c.

Analyzing a Quadratic Function

Given the graph of a quadratic function, as shown below, we wish to indicate

the zeros of the function

the point or points where f(t) = 0

the vertex of the parabola

the point corresponding to the value f(0)

the value y = f(-3)

the value of t for which y = 20.

The zeros are the points where the graph crosses the t axis.

We estimate these points to be (-.6, 0) and (4.6, 0) [note: not (3.6, 0) as incorrectly labeled in the figure).

The expression f(t) = 5 means that f(t), whose value is plotted on the vertical axis, has the value 5.

Thus we look for the point on the graph where the vertical coordinate is 5.

This point is approximately -1.5, 5; its projection lines to the horizontal vertical axes are indicated in red.

The vertex of the parabola is the extreme point of the parabola and lies on the axis of symmetry.

We estimate this point to be approximately (2, -7).

When we refer to f(0), we are talking about the value of f(t) when t = 0.

We therefore look for the graph point corresponding to t = 0.

Since t = 0 on the vertical axis, we see that we are looking for the point at which the graph crosses the vertical axis.

This point is approximately (0, -3), and is indicated as (0,f(0)) on the graph.

In the expression y = f(-3), we see that the -3 replaces t in the expression y = f(t).

We thus look for the graph point corresponding to t = -3.

This point is found by following the blue dotted line up from the t axis to the graph; the line continues over to the vertical axis, where we label the value f(-3). This value is a bit less than 20.

To find the t value of the function for which y = 20, we locate the y = 20 point on the vertical axis and proceed straight across to the graph to the y = 20 point.

We then project straight down to the t axis to find the t value, which is a bit less than -3.

pc02.jpg

video clip #01

If we wish to find the values of a, b and c for the quadratic function y = a t^2 + b t + c, we note that the graph gives us the following information:

The vertex is at x = 2, so -b / (2a) = -2.

If we go over one unit from the vertex we go up about one unit (it's a bit hard to tell for sure be to the scale of the graph, but 1 vertical unit seems about right). So we figure that a = 1.

Since a = 1 and -b / (2a) = -2, we must have -b / (2 * 1) = -2, so b = -4.

Since for t = 0, y = f(t) = a t^2 + b t + c takes value f(0) = c, c = -3 (again estimating the y coordinate corresponding to t = 0 from the graph).

Our function is thus approximately y = 1 t^2 - 4 t - 3.

Basic Function Families

Before introducing the idea of basic function families, we review of couple of essential facts about exponents.

In the first place, b^x * b ^ y = b ^ (x + y).

This isn't a rule made up to confuse algebra students, but is simple common sense about multiplying numbers:

b^x means b multiplied by itself x times, and b^y means b multiplied by itself y times.

So b^x * b^y will consist of b multiplied by itself a total of x + y times. Again, just common sense.

It follows from this that, for example, 2 ^ -2 * 2 ^ 2 = 2 ^ (-2 + 2) = 2 ^ 0.

People often make a mistake of thinking that 2 ^ 0 = 0, but this seemingly commonsense assumption actually violates common sense.

This can easily be seen as follows:

2 ^ -2 = 1 / (2^2) = 1/4 (recall that raising a number to a negative exponent results in an inversion of sort e.g., 2 ^ -2 = 1 / (2^2) = 1/4).

So when we multiply 2 ^ -2 by 2 ^ 2 we get 1/4 * 4 = 1.

It follows that 2 ^ 0 = 1.

We could duplicate this argument for any number b, coming to the conclusion that b ^ 0 = 1 for any number b.

pc04.jpg

video clip #02

We now introduce the first three of our four most basic function families.

These are the families of the functions

y = x,

y = x^2 and

y = 2 ^ x.

We begin by making a table for each of the functions, using x values from -3 to 3.

We observe the different behaviors of the four functions. We note:

the constant changes in y values for the y = x function,

the symmetry of the y = equal x ^ 2 function, and

the fact that the y = 2^x function doubles with every successive x value on the table.

pc07.jpg

video clip #03

We sketch the graphs of these functions.

Not surprisingly the y = x function has a graph which is a straight line through the origin.

The y = x ^ 2 function has a graph in the shape of a parabola, symmetric about the y axis.

The y = 2 ^ x function approaches the negative x axis more and more closely as x becomes more and more negative, while the graph gets steeper and steeper for increasing positive values of x.

The y = 2 ^ x function is called an exponential function; exponential functions all have the same basic shape, though some might be upside down and some inverted from left to right as compared to the function shown here.

On one side or another an exponential function always approaches a horizontal line, coming closer and closer without limit but never quite reaching the line.

This line is called a horizontal asymptote of the graph.

pc05.jpg

video clip #04

Stretching and Shifting the Basic Functions

Starting with a basic function, we can think of stretching its graph either in the vertical or the horizontal direction, or both, then shifting the graph to the right or left (horizontally) or up or down (vertically).

As we will see, such transformations leave straight lines straight, leave parabolas parabolas, and transform graphs of exponential functions into graphs of other exponential functions.

If we consider what happens to a parabola when we stretch it vertically, we will gain valuable insight into how this process works.

Begin with the basic parabola y = x ^ 2. This graph is shown in blue in the figure below.

To stretch the graph vertically by a factor of 2, we take each point of the original (blue) graph and move it twice as far from the x axis to obtain the red graph.

This has the effect of doubling the value of y that corresponds to each value of x.

As a result the function y = x^2 is transformed into the 'doubled' function y = 2 x ^ 2.

We could also have stretched the y = x ^ 2 graph by a factor of 1/2, moving each point of the y = x ^ 2 graph to 1/2 its distance from the x axis.

Since each y value is halved, we see that the new function is y = 1/2 x^2.

We can even extend the idea of stretching to negative factors.

If we stretch the function y = x ^ 2 by factor of -1, we will maintain the distance of every point from the x axis but will reverse its direction from the x axis.

This says the effect of inverting the graph, as shown below.

pc08.jpg

pc09.jpg

video clip #05

If we apply the same stretching idea to a linear graph, we see that stretching by a factor of 2, as it moves every point twice as far away from the x axis, has the effect of doubling the slope of the graph.

pc10.jpg

Suppose we wish to match the green dotted line (supposed to be straight but not quite looking that way), using just stretching and shifting transformations. How would we go about it?

video clip #06

If we wish to match the green dotted line using just stretching and shifting transformations, we might begin by applying a stretching transformation.

Note that we always apply stretching transformations before shifting transformations.

We begin by stretching the graph by a factor of approximately 1/3 (this is just an estimate based on the graph; 1/4 might also be close for this graph).

This gives us the graph of the function y = 1/3 x.

Then to match the green graph, we would have to raise the graph an estimated 6 units.

This is indicated by the purple vertical arrows 'lifting' selected points of the 'red' graph to the target 'green' graph.

This process increases the values of y = 1/3 x by 6, resulting in the graph of the function y = 1/3 x + 6.

We note that this shift raises the point (0,0) to (0,6). The latter point is called the y-intercept of the graph of y = 1/3 x + 6.

More generally, if we stretch the y = x graph by factor m, then raise it b units vertically, we obtain the graph of the general linear function y = mx + b.

The factor m is the 'vertical stretch' factor and gives us the slope of the graph of the transformed function.

The vertical shift b raises the origin to (0,b), which is called the y intercept of the graph.

Thus the function y = mx + b has a graph whose slope is m and its y intercept is at (0,b).

pc11.jpg

We can see the variety of graphs possible for this general linear function by imagining that the parameter b is held fixed, which results in a fixed y intercept, while the parameter m is permitted to vary through all possible real numbers.

The first figure below gives representative members of this family.

We might on the other hand wish to keep m constant and let b vary.

This would result in a series of graphs with the same slope, and hence parallel to one another, though with different y intercepts.

Representatives of this family are sketched in the second graph below.

pc12.jpg

video clip #07

The Number of Transformations Required to obtain the General Function for each Basic Function

Any linear function can be constructed by two geometric transformations: a vertical stretch and a vertical shift.

These two transformations are related in the way discussed above to the two parameters m and b in the general form y = mx + b of a linear function.

Each independent transformation we apply to a function will be related to a parameter in the general form of the function.

To obtain a given parabola by geometric transformations we must usually employ three transformations, a vertical stretch to get the 'width' of the parabola correct, then a horizontal and a vertical shift.

We know that a parabolic (or quadratic) function has the general form y = a x^2 + b x + c, with three parameters.

These parameters are determined by the transformations, but each transformation is not directly related to exactly one of the parameters (as we will see, one of the transformations directly determines the value of a, but the other two have a combined influence on b and c).

Another way of expressing the basic form of a quadratic function is y = a (x - h) ^ 2 + k.

This form is completely equivalent of the other, but a is the vertical stretch, the h is the horizontal and k is the vertical shift.

pc15.jpg

When we stretch and shift our basic exponential function y = 2 ^ x, we will think of a vertical stretch, a horizontal compression, and a vertical shift.

Again we use three geometric transformations and we obtain a function form with three parameters.

One form is y = A b ^ x + c, with parameters A, b and c; another is y = A * 2 ^ (kx) + c, with parameters A, k and c.

The second of these forms reveals the vertical stretch A, the horizontal compression k and the vertical shift c.

Just as a we were able to fit a quadratic function to any three data points, we can fit an exponential function to any three data points.

We substitute the coordinates of the three points into one of the forms above, or a third equivalent form to be discussed later, and obtain three simultaneous equations.

The equations we got for the parameters of the quadratic model were linear so we could solve them by adding multiples of equations to one another and multiplying equations by constants; the equations we get for the exponential model are not linear and must be solved by other means. But the principal is the same in that we obtain three simultaneous equations and solve them for the parameters of the model.

pc16.jpg

video clip #08

Question: `q010. Explain in terms of the values of y = x^2 for the numbers x = -2, -1, 0, 1, 2 why we expect the graph of y = x^2 to be symmetric about the y axis.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&

0^2 = 0 =y, and x=0 so the x and y are both zero is is the midpoint for the reflection of these coordinates

@&

You can't explain this just in terms of the value at x = 0.

0^2 = 0, but your statements don't apply to the graph of y = x^2.

*@

&&&

confidence rating #$&*:

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Self-critique rating:2

@&

See if you can revise your answers to thos last two questions according to my notes.

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

Be sure to include the entire document, including my notes.

&#

*@

** **

You can't explain this just in terms of the value at x = 0.

0^2 = 0, but your statements don't apply to the graph of y = x^2.

** **

I thought this was my equation would would I be trying to find y=x^3, when the problem asks for y=x^2???

@&

I have corrected the ^3 references in the above. All should have indicated ^2.

I'm sorry for the confusion.

The point of my notes is that the symmetry of a graph is not determined by one of its points. What is it about the y values that does determine the symmetry?

*@

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Mth163

Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.

** Question Form_labelMessages **

A = 1, h = -3 to 3, c = 0

** **

Question: `qquery problem 5. power function families

Describe the graph of the power function family y = A (x-h) ^ p + c for p = -3: A = 1, h = -3 to 3, c = 0.

** **

** **

y=1(x--3)^p+0

y=1(x--2)^p+0

y=1(x--1)^p+0

y=1(x-0)^p+0

y=1(x-1)^p+0

y=1(x-2)^p+0

y=1(x-3)^p+0

@&

Good so far, but you are given the value of p.

You will need to then sketch the graphs and describe the family. There is a clear pattern to the graphs, which is determined by the changing values of h. The key to understanding why the pattern is as it is will lie in way the values shift the tables of values for these functions.

*@

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Mth163

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** Question Form_labelMessages **

June 27,2013 14:54

** **

Question: `q004. Suppose we wish to graph the functions y = (x -k)^3 for k values 2, then 3, then 4. If we graph all these functions on the same set of coordinate axes, what will the graph look like? It is suggested that you actually sketch your graph and describe your sketch.

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Your solution:

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My y coordinates are labeled by twenties and x coordinates by ones.

2=k

-125

-64

-27

-8

-1

0

1

My curve starts at (-3,-125) touches zero at -1 then ends at coordinates (3,1)

3=k

-216

-125

-64

-27

-8

-1

Starts at (-3,-216) goes through zero at y=3 and ends at (0,3). Also it touches the other graph at (1,-8)

4=k

-343

-216

-125

-64

-27

-8

-1

Each are one unit to the right of the other

&&&

confidence rating #$&*:

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Given Solution:

The k = 2 graph will lie 2 units to the right of the graph of y = x^3, and the k = 4 graph will lie 4 units to the right. The three graphs will all have the same shape as the y = x^3 graph, but will lie 2, 3 and 4 units to the right.

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Self-critique (if necessary):

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

Self-critique rating: OK

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Question: `q005. The function y = 3 * 2^x is of the form y = A * 2^x for A = 3. What function would this form give us for A = 2? How would the graph of this function compare with that of y = 2^x?

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Your solution:

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confidence rating #$&*:

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Given Solution:

As we saw earlier, the graph of y = 3 * 2^x lies 3 times as far from the x-axis as a graph of y = 2^x and is at every point three times as steep. We would therefore expect the A = 2 function y = 2 * 2^x to lie 2 times is far from the x-axis as the graph of y = 2^x.

STUDENT QUESTION

In this case, is A stretching or shifting? I said shifting in my response, but as I'm recalling from the previous assignment I suspect it may actually be stretching the graph by 2 units instead of shifting.

INSTRUCTOR RESPONSE

Your suspicion is correct. The graph is stretching.

When a graph shifts, every point moves by the same amount.

When a graph stretches, every point moves to a multiple of its original distance from some axis. Points further from that axis move more, points closer to the axis move less.

For example the graph below depicts the x any y axes, and the graphs of the two functions. Vertical lines are drawn at x = 1 and x = 2. You should sketch a good copy of this graph and actually trace out the properties discussed below, and annotate your graph accordingly:

• Look first at the vertical line corresponding to x = 1. It should be clear that, along this line, the graph of the second function is about twice as far from the x axis as the graph of the first.

• Now look at the vertical line corresponding to x = 2. It should be clear that, along this line, the graph of the second function is about twice as far from the x axis as the graph of the first.

• You should also see that along the x = 1 line the second graph lies at a certain distance above the first, while at the x = 2 line the second graph lies at a greater distance above the first.

• At the origin, the graphs meet. Then if we move from left to right, starting at the origin, the vertical distance between the graphs keeps increasing.

The graph below includes 'heavier' vertical line segments representing the increasing vertical distance between the graphs:

This graph represents a vertical stretch, in every point of the second graph lies at double the vertical distance from the horizontal axis as the corresponding point of the first.

By contrast, consider the graph shown below, in which the second graph is shifted in the vertical direction relative to the first.

• On this graph the vertical distance is the same on every vertical line.

• It probably doesn't look like this is the case. There's an optical illusion at work here, which is due to the fact that the upper graph gets closer and closer to the lower graph. However, despite appearances, this isn't the case if the distances are measured along the vertical lines.

• In the first figure below we show the line segments which represent these vertical distances. They are all of the same length. In the second figure below, these line segments are depicted without the graph.

In this example, all points of the second graph lie at the same vertical distance above the first.

This graph represents a vertical shift. In a vertical shift all points point of one graph lie at the same vertical displacement relative to the first.

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Self-critique (if necessary):

@&

&#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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** **

I am pretty sure i skipped this on (probability didn't notice)

** **

My solution

&&&&

y=2*2^x

It would be more vertically stretched by 2 units upward

&&&&

@&

A graph on which every point moves 2 units higher is said to have vertically shifted by 2 units.

This graph is a vertical stretch, as you say; but a vertical stretch will not move each point 2 units higher. Instead, every point moves twice as far from the x axis.

*@

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#$&*

Mth163

Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.

** Question Form_labelMessages **

june27 15:06

** **

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

Question: `q011. Consider the family of functions of form y = 2 ( x - h ) ^ 2.

What is the vertex of the graph of the h = 0 function?

What is the vertex of the graph of the h = -1 function?

What is the vertex of the graph of the h = 1 function?

What are the other two basic points of each of these functions?

Graph these three functions and describe your graph.

Make a sketch representing the family y = 2 ( x - h ) ^ 2 for -4 <= h <= 2 and describe your sketch.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

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I’m going to say x=2 for these functions.

. h=0 vertex @ (2,8) other two basic points (3,18)(1,2)

. h=-1 vertex @ (2,9) other two basic points at (3,16)(1,4)

. h=1 vertex @ (2,9) other two basic points at (3,16)(1,4)

@&

These aren't correct.

I'll need you to explain how you got your results. Then I should be able to help you clarify the situation.

*@

&&&

confidence rating #$&*:

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

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

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** **

What i originally did was plug in x for 2 and the other h values to find my answers, but can i use h=b to find AOS???

@&

To find the vertices of these graphs, and understand why the graphs are related as they are, you need to plug in a series of x values for each function.

What is the table of values for the h = 2 function, and where do you conclude the vertex of the corresponding graph will lie?

What is the table of values for the h = 1 function, and where do you conclude the vertex of the corresponding graph will lie?

What is the table of values for the h = 0 function, and where do you conclude the vertex of the corresponding graph will lie?

What happens to the graph and the location of its vertex as h decreases by 1 unit? Does this pattern continue for h values -1, then -2, etc..?

*@

@&

Remember that the vertex of a parabola lies on its axis of symmetry. So you need a table of values for each graph, and you need to recognize the symmetry in each table and how it is related to the symmetry, and hence the vertex, of each graph.

See if you can answer the questions I've inserted above.

*@

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#$&*

Mth163

Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.

** Question Form_labelMessages **

y vs x june27

** **

Question: `q012. The depth function for a certain flow is depth = y(t) = .01 (t - 20)^2, for 0 <= t <= 20. What is the average slope of the graph of depth vs. clock time between the t = 5 point and the t = 10 point? What is the average rate of change of depth with respect to clock time for the interval 12 <= t <= 16?

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Your solution:

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Average slope is -.25

respect to clock time for the interval 12 <= t <= 16 , slope is -.12

found these by change in y/change in x

@&

You need to show the details of your reasoning. You don't say what the two points are, what the rise is, or the run, on all of which you base your slope.

*@

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confidence rating #$&*:

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

Self-critique rating:1

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** **

&&&&

y(t) = .01 (t - 20)^2

y(t) = .01 (5 - 20)^2

y(t) = .01 (10 - 20)^2

using my two new coordinates i find slope by rise/run

(10,1)(5,2.25)

[(10-5)/(1-2.25)]

slope equals -4.0cm/sec

which would be your rate of change

&&&&

self-critique #$&*

#$&* self-critique

self-critique rating

rating #$&*:

@&

The points (10,1) and (5,2.25) are correct.

What are the units of those quantities?

What is the rise between the points?

What is the run between the points?

What therefore is the slope between the points, and in terms of its units, what does the slope mean?

*@

@&

I've inserted some more detailed questions. You're on the right track, but you're still missing some details.

*@

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Mth163

Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.

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slopes june27

** **

Question: `qdescribe the pattern to the depth change rates

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Your solution:

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The coordinates are (125,-162.5)(215,-.5)(35,-.5) usually when you have two of the same y values while the x values are on different sides of the vertex and not on the same line you have a parabola. So this pattern indicates at parabola.

@&

This is a good observation. Note that the axis of symmetry of the parabola must be halfway between these two points, at x coordinate (215 + 35) / 2 = 125. Your point (125, -162.5) must therefore be the vertex.

*@

&&&

confidence rating #$&*:1

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Given Solution:

** Rates are -3.8, -3 and -2.2 for the three intervals (20,40), (40,60) and (60,80).

For each interval of `dt = 20 the rate changes by +.8. **

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Self-critique (if necessary):

You meant for me to use the questions above, using (20, 40), (40, 60) and (60, 80)

I do not understand how you found your rates.

Can you please go through the process???

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Self-critique Rating:

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t = 125, 215, 35 values are not relevant to this question.

The coordinates of the graph at t = 20, 40, 60 and 80 are

(20, 58)

(40, -18)

(60, -78)

(80, -122]

What are the slopes between each consecutive pair of points?

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(20, 58)

(40, -18)

40-20/-18-58=20/-76=-5/19=-.556

that is my answer for the 1st two points

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You appear to be dividing the run by the rise rather than the rise by the run.

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(40, -18)

(60, -78)

60-40/-78--18=20/-60=-1/3=-.333

(60, -78)

(80, -122]

80-60/-122--78=20/-44

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You're on the right track but check my notes.

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question form

#$&*

Mth163

Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.

** Question Form_labelMessages **

cups of water revision 27june

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Question: `qcomment on how the actual graph of the data compared with your prediction

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Your solution:

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My graph represents exponential decay, more cups of water added to the gallon makes the gallon drop farther.

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More water does cause a greater drop.

If the 'run' required to fall by some fixed percent is constant, then the behavior is exponential. For example, if the 'run' required for the length to drop to half is always the same, the behavior would be exponential.

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When you document your solution, you need to provide the information required to evaluate it.

You would need to justify your answer by providing the data.

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confidence rating #$&*: 3

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I constantly used the same unit of water, the the bottom of a mountain dew from end of label to bottom of plastic I started at one inch the amount of length traveled was zero. inch mark on yard stick=x,length traveled=y. (1,0)(3,2)(6,3)(10.5,4.5)(15,5.5)(21.5,6.5)(26.5,5)(31.5,5)(38.9,4.4)(39.9,1) counting by 5's on my x axis and by 1's on my y axis It kinda makes an upside down half circle, semi-circle

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This graph is relevant to some very interesting behavior of the system, but it isn't the graph of length vs. amount of water, which I believe was what was requested.

Length traveled would not be direct data, but would be calculated from previous length and new length.

How much water was present, and what was the corresponding length of the chain, for each observation?

Can you report this as a table, then make the required graph?

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