Assn 5

Good work. See my notes.

Let me know if anything is unclear, and include specifics about what you do and do not understand.

Did you tell me that you had taken the Major QUiz sometime last week? If so, where and when did you take it?

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23:07:59 query introduction to basic function families problem 1 on basic graphs Why is the graph of y = x a straight line?

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RESPONSE --> There is no variation in the numbers. When y = x the numbers are always the same so that they form a straight line. Changes in the numbers are constant as well as the change in slope.

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23:08:30 ** Since y = x the rise and run between any two points on the graph are equal, which makes the slope 1. A graph with constant slope is a straight line. **

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RESPONSE --> ok

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23:09:52 ** The graph of y = x^2 is symmetric about x = 0 because (-x)^2 = x^2. Thus for any point on the x axis the y values at that point and at the point on the opposite side of the origin are equal, so that the graph on one side of the y axis is a 'reflection' of the graph on the other side. **

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RESPONSE --> Any point put into this equation, whether negative or positive, will be a positive in the end as it is being squared.

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23:13:14 why does y = 2^x keep increasing as x increases, and why does the graph approache the x axis for negative values of x

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RESPONSE --> As you increase the x, you increase the number of times to which the variable 2 is raised. The graph would approach the x axis as negative values because you cannot have a negative exponent, so you must invert the fraction and then use the positive exponent to apply to the fraction.

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23:13:31 ** GOOD STUDENT RESPONSE: y = 2^x will increase as x increases on the positive side because x is the value of the exponent. This will cause the y value to double from its last value when you move one unit in the positive x direction. On the negative side of the y axis y = 2^x will approach the x axis because a negative exponent causes the value to invert into a fractional value of itself--i.e., 2^(-x) = 1 / 2^x. As we move one unit at a time negatively the value will become one half of the previous value so it will never quite reach y = 0. **

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RESPONSE --> ok

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23:17:50 why is y = x^3 antisymmetric about x = 0 (i.e., taking the same values except for the - sign on opposite sides of x = 0)

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RESPONSE --> The equation y = x^3 is antisymmetric about x = 0 because the same values are found with the exception of the positive and negative signs respectively. It is always either on the increase or decrease.

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23:18:08 ** y = x^3 is antisymmetric because if you cube a negative number you get a negative, if you cube a positive number you get a positive, and the magnitude of the cubed number is the cube of the magnitude of the number. So for example (-3)^2 = -27 and 3^3 = 27; the points (-3, -27) and (3, 37) are antisymmetric, one being `down' while the other is `up'. GOOD STUDENT RESPONSE: y = x^3 is antisymmetric about x = 0 because the exponent is an odd number. This will cause negative x values to have a negative y result. The absolute value of the negative y result will be equivalent to its corresponding positive y value. **

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RESPONSE --> ok

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23:19:39 why do y = x^-2 and y = x^-3 rise more and more steeply as x approaches 0, and why do their graphs approach the x axis as we move away from the y axis.

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RESPONSE --> The exponents are negative causing the variable to be inverted into fraction form. Once in fraction form we go ahead and either square or cube the resulting fraction.

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23:20:26 ** And as x approaches 0 the expressions x^-2 and x^-3, which mean 1 / x^2 and 1 / x^3, have smaller and smaller denominators. As the denominators approach zero their reciprocals grow beyond all bound. y = x^-2 and y = x^-3 rise more and more steeply as x approaches zero because they have negative exponents they become fractions of positive expressions x^2 and x^3 respectively which have less and less slope as they approach zero. As x^2 and x^3 approach zero and become fractional, x^-2 and x^-3 begin to increase more and more rapidly because thier functions are then a whole number; (1) being divided by a fraction in which the denominator is increasing at an increasing rate. As y = x^-2 and y = x^-3 move away from the y-axis they approach the x-axis because they have negative exponents. This makes them eqivalent to a fraction of 1 / x^2 or 1 / x^3. As x^2 and x^3 increase in absolute value, the values of y = x^-2 and y = x^-3 constantly close in on the x-axis by becoming a portion of the remaining distance closer, they will never reach x = zero though as this would be division by zero (since it is a fraction) **

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RESPONSE --> ok

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23:23:06 query problem 2. family y = x^2 + c Explain why the family has a series of identical parabolas, each 1 unit higher than the one below it.

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RESPONSE --> The family has a series of identical parabolas, each being shifted on the y axis by the value of ""c"". This would give us the equations y = x^2 +1 and y = x^2 - 1 This would effectively shift the graph by 1 unit up or down the y axis.

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23:23:22 ** GOOD STUDENT RESPONSE: The graph of y = x^2 + c, with c varying from -5 to 4 is a series of identical parabolas each 1 unit higher than the one below it. The c value in the quadratic equation has a direct impact on the vertical shift. The vertex of the graph will be shifted vertically by the amount of the c value, so every time c increases by 1 the graph is raised 1 unit. **

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RESPONSE --> ok

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23:37:29 query problem 4. describe the graph of the exponential family y = A * 2^x for the values A = -3 to 3.

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RESPONSE --> For the equation y = A * 2^x you would get y = -3 * 2^x y = -2 * 2^x y = -1 * 2^x y = 0 * 2^x y = 1 * 2^x y = 2 * 2^x y = 3 * 2^x For the negative exponents you would have values to the negative side of y axis. The x value may be positive or negative. For the positive exponents, you would have values to the positive side of the y axis. This is an exponential equation therefore it should have an increasing line though not a linear equation. The line would simply steadily increase to the positive side of the y axis.

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23:38:02 ** This family includes the functions y = -3 * 2^x, y = -2 * 2^x, y = -1 * 2^x, y = 0 * 2^x, y = 1 * 2^x, y = 2 * 2^2 and y = 3 * 2^x. Each function is obtained by vertically stretching the y = 2^x function. y = -3 * 2^x, y = -2 * 2^x, y = -1 * 2^x all vertically stretch y = 2^x by a negative factor, so the graphs all lie below the x axis, asymptotic to the negative x axis and approaching negative infinity for positive x. They pass thru the y axis as the respective values y = -3, y = -2, y = -1. y = 1 * 2^x, y = 2 * 2^x, y = 3 * 2^x all vertically stretch y = 2^x by a positive factor, so the graphs all lie above the x axis, asymptotic to the negative x axis and approaching positive infinity for positive x. They pass thru the y axis as the respective values y = 1, y = 2, y = 3. y = 0 * 2^x is just y = 0, the x axis. Of course the functions for fractional values are also included (e.g., y = -2.374 * 2^x) but only the integer-valued functions need to be included in order to get a picture of the behavior of the family. ** STUDENT QUESTION: Ok, it was A = -3 to 3. I understand how to substitute these values into y = A * 2^x. I knew that is was an asymptote, but I'm a little confused as to how to graph the asymptote. INSTRUCTOR RESPONSE: For each value of A you have a different function. For A = -3, -2, -1, 0, 1, 2, 3you have seven different functions, so you will get 7 different graphs. Each graph will contain the points for all values of x. For example the A = -3 function is y = -3 * 2^x. This function has basic points (0, -3) and (1, -6). As x takes on the negative values -1, -2, -3, etc., the y values will be -1.5, -.75, -.375, etc.. As x continues through negative values the y values will approach zero. This makes the y axis a horizontal asymptote for the function. You should figure out the x = 0 and x = 1 values for every one of these seven functions, and you should be sure you understand why each function approaches the negative x axis as an asymptote. *&*&

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RESPONSE --> ok

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23:42:20 describe the graph of the exponential family y = 2^x + c for the values c = -3 to 3.

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RESPONSE --> The family would include the equations y = 2^x - 3 y = 2^x - 2 y = 2^x - 1 y = 2^x - 0 y = 2^x + 1 y = 2^x +2 y = 2^x +3 The value of c would shift the line vertically along the x axis. The first 3 equations with a negative variable would shift the graph downward by the appropriate number of units. The positive variable would vertically shift the line upward along the x axis. The 0 would simply leave the line where it originally lies on the graph as there is no vertical shift.

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23:42:38 ** There are 7 graphs, including y = 2^x + 0 or just y = 2^x. The c = 1, 2, 3 functions are y = 2^x + 1, y = 2^x + 2 and y = 2^x + 3, which are shifted by 1, 2 and 3 units upward from the graph of y = 2^x. The c = -1, -2, -3 functions are y = 2^x - 1, y = 2^x - 2 and y = 2^x - 3, which are shifted by 1, 2 and 3 units downward from the graph of y = 2^x. **

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RESPONSE --> ok

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23:50:54 query 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.

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RESPONSE --> For the power function family equation of y = A(x-h)^p +c the equation would be y = 1(x + 3)^-3 + 0 y = 1(x + 2)^-3 + 0 y = 1(x + 1)^-3 + 0 y = 1(x - 1)^-3 + 0 y = 1(x - 2)^-3 + 0 y = 1(x - 3)^-3 + 0 There would be no vertical shift in that the value of c is 0. The horizontal shift would be either -3, -2, -1, 0, 1, 2 or 3 units depending upon the equation used and the value of the h variable.

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23:51:12 ** GOOD STUDENT RESPONSE: I sketched the graph of the power function family y = A (x-h)^p + c for p = -3: A = 1: h = -3 to 3, c = 0. Beginning on the left side of the graph the curve was infinitely close to its asmptote of y = 0. This was determined by the value of c. As we move from left to right the curves decreased at an increasing rate, approaching thier vertical asmptotes which was determined by thier individual values of h. The curves broke at x = c as this value was never possible due to division by zero. The curves resurfaced on the graph high on the right side of thier vertical asymptotes and from there they decreased at a decreasing rate, once again approaching thier horizontal asymptote of y = 0. INSTRUCTOR COMMENTS: Only the h value changes. p=-3, A=1 and c=0, so the functions are y = 1 * (x-h)^-3 or y = (x-h)^-3. For h = -3 to 3 the functions are y = (x - (-3))^-3, y = (x - (-2))^-3, y = (x - (-1))^-3, y = (x - 0)^-3, y = (x - 1)^-3, y = (x - 2)^-3, y = (x - 3)^-3. These graphs march from left to right, moving 1 unit each time. Be sure you see in terms of the tables why this happens. **

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RESPONSE --> ok

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23:51:44 query problem 10 illumination. What function did you evaluate to get your results?

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RESPONSE --> I am not sure what this is asking exactly.

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23:51:52 ** I determined the illumination y from a certain florescent bulb at the distances of 1, 2, 3, and 4 units using the generalized power function for p = -1 with A = 370, h = 0 and c = 0. This power function is y = A (x- h)^p + c = 370 (x - 0)^(-1) + 0, or just y = 370 x^-1. **

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RESPONSE --> ok

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23:52:49 Determine the illumination at distances of 1, 2, 3 and 4 units, and sketch a graph.

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RESPONSE --> Apparently I did not do work on a problem that I have missed.

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23:52:55 ** Student Solution: For x=1 we obtain y=370(-1-0) ^-1=370 For x=2 we obtain y=370(2-0)^-1=185 For x=3 we obtain y=370(3-0)^-1 =123.3 For x=4we obtain y=370(4-0)^-1=92.5**

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RESPONSE --> ok

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23:54:06 Query Add comments on any surprises or insights you experienced as a result of this assignment.

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RESPONSE --> I really have enjoyed this particular assignment. The power function is still giving me some trouble but I am working on it. Is there any easier way to remember all that goes along with these equations?

Here's a summary of what you should know at this point:

There are four types of basic functions at this point, the linear function y = x, the exponential function y = 2^x, the quadratic function y = x^2 and the power functions y = x^p (for any fixed value of p).

You should be able quickly, without a calculator, to evaluate these functions at x = -3, -2, -1, 0, 1, 2, 3 and graph them, and you should thereby understand the basic shapes of these graphs.

You should also be able to similarly evaluate the power functions at x = 1/2 and use this point in the graphing of your power function.

You should understand what points are designated as the 'basic points' of each function (x = 0 and x = 1 point for linear, x = -1, x = 0 and x = 1 points for quadratic, x = 0 and x = 1 points for exponential along with the understanding that the negative x axis is a horizontal asymptote, x = -1, x = 0, x = 1/2, x = 1 and x = 2 points for a power function).

You should be able to make a reasonable graph of each basic function given from only its basic points.

You should understand that y = A f(x - h) + c stretches every point of the y = f(x) graph A times as far from the x axis, shifts every point h units in the x direction, and shifts every point c units in the y direction.

Specifically you should be able to apply these transformations to the basic points and sketch the resulting graph based on only these points and your knowledge of the shape and general behavior of the function.

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23:54:21 ** STUDENT COMMENT: I have never worked with graphs in the power family, and very little in the exponential family. I am always amazed at the patterns that a function produces. It helps me understand the equation so much better than a list of numbers. I do feel that I need the data table with the graph to fully understand it. INSTRUCTOR RESPONSE: The data table is certainly helpful, especially when you see the reasons for the number patterns in the formula as well as you do. **

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RESPONSE --> ok

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21:56:27 `q001. Note that this assignment has 8 questions Evaluate the function y = x^2 for x values -3, -2, -1, 0, 1, 2, and 3. What are your y values?

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RESPONSE --> For function y = x^2 with x values of -3, -2, -1, 0, 1, 2 and 3 the y values would be y = -3^2 = 9 y = -2^2 = 4 y = -1^2 = 1 y = 0^2 = 0 y = 1^2 = 1 y= 2^2 = 4 y = 3^2 = 9 So you have y values of 9, 4, 1, 0, 1, 4, 9.

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21:56:32 You should have obtained y values 9, 4, 1, 0, 1, 4, 9, in that order.

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RESPONSE --> ok

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22:00:50 `q002. Evaluate the function y = 2^x for x values -3, -2, -1, 0, 1, 2, and 3. What are your y values?

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RESPONSE --> For function y = 2^x with x values -3, -2, -1, 0, 1, 2, 3 the y values would be y = 2^-3 = 1/2^3 = 1/8 y = 2^-2 = 1/2^2 = 1/4 y = 2^-1 = 1/2^1 = 1/2 y = 2^0 = 1 y = 2^1 = 2 y = 2^2 = 4 y = 2^3 = 8 So y values are 1/8, 1/4, 1/2, 1, 2, 4, 8

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22:00:58 By velocity exponents, b^-x = 1 / b^x. So for example 2^-2 = 1 / 2^2 = 1/4. Your y values will be 1/8, 1/4, 1/2, 1, 2, 4 and 8. Note that we have used the fact that for any b, b^0 = 1. It is a common error to say that 2^0 is 0. Note that this error would interfere with the pattern or progression of the y values.

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RESPONSE --> ok

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22:07:02 `q003. Evaluate the function y = x^-2 for x values -3, -2, -1, 0, 1, 2, and 3. What are your y values?

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RESPONSE --> For the function y = x^-2 using x values of -3, -2, -1, 0, 1, 2, 3 the y values would be y = -3^-2 = -1/3^2 = 1/9 y = -2^2 = -1/2^2 = 1/4 y = -1^-2 = -1/1^2 = 1 y = 0^-2 = 0

0^(-2) = 1 / 0^2 = 1 / 0, which is undefined.

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22:07:26 By the laws of exponents, x^-p = 1 / x^p. So x^-2 = 1 / x^2, and your x values should be 1/9, 1/4, and 1. Since 1 / 0^2 = 1 / 0 and division by zero is not defined, the x = 0 value is undefined. The last three values will be 1, 1/4, and 1/9.

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RESPONSE --> ok

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22:11:51 `q004. Evaluate the function y = x^3 for x values -3, -2, -1, 0, 1, 2, and 3. What are your y values?

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RESPONSE --> For the function y = x^3 the y values would be y = -3^3 = -27 y = -2^3 = -8 y = -1^3 = -1 y = 0^3 = undefined

0^3 = 0 * 0 * 0, which is just 0. This is very well defined.

1 / 0 is undefined (how many times will 0 go into 1? As many as you like. Hence undefined.

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22:12:52 The y values should be -27, -8, -1, 0, 1, 4, 9.

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RESPONSE --> How is it that the last 2 values for y are 4 and 9 when you were cubing both the 2 and the 3? Wouldn't it be 8 and 27?

Yup. This is an error I've decided not to correct, just to see who's reading. You are, which in your case was not in question in the first place.

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22:33:27 `q005. Sketch graphs for y = x^2, y = 2^x, y = x^-2 and y = x^3, using the values you obtained in the preceding four problems. Describe the graph of each function.

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RESPONSE --> For the graph of y = x^2 you would get a parabola with the points (-3, 9), (-2, 4), (-1, 1), (0,0), (1, 1), (2, 4) and (3, 9). For the graph of y = 2^x you would get a wavy line. It is initially a small slope then once passing the vertex it rises quite rapidly. It would have points (-3, 1/8) , (-2, 1/4), (-1, 1/2), (0, 0), (1, 2), (2, 4) and (3, 8). For the graph of y = x^-2 you would get a wavy line. The line would be fairly uniform in its rise and run on the axis. It would include the points (-3, -1/9), (-2, -1/4), (-1, -1), (0, 0), (1, 1), (2, 1/4) and (3, 1/9). For the graph of y = x^3 you would get a parabola. It would contain the points (-3, -27), (-2, -8), (-1, -1), (0, 0), (1, 1), (2, 8) and (3, 27).

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22:33:55 The graph of y = x^2 is a parabola with its vertex at the origin. It is worth noting that the graph is symmetric with respect to the y-axis. That is, the graph to the left of the y-axis is a mirror image of the graph to the right of the y-axis. The graph of y = 2^x begins at x = -3 with value 1/8, which is relatively close to zero. The graph therefore starts to the left, close to the x-axis. With each succeeding unit of x, with x moving to the right, the y value doubles. This causes the graph to rise more and more quickly as we move from left to right. The graph intercepts the y-axis at y = 1. The graph of y = x^-2 rises more and more rapidly as we approach the y-axis from the left. It might not be clear from the values obtained here that this progression continues, with the y values increasing beyond bound, but this is the case. This behavior is mirrored on the other side of the y-axis, so that the graph rises as we approach the y-axis from either side. In fact the graph rises without bound as we approach the y-axis from either side. The y-axis is therefore a vertical asymptote for this graph. The graph of y = x ^ 3 has negative y values whenever x is negative and positive y values whenever x is positive. As we approach x = 0 from the left, through negative x values, the y values increase toward zero, but the rate of increase slows so that the graph actually levels off for an instant at the point (0,0) before beginning to increase again. To the right of x = 0 the graph increases faster and faster. Be sure to note whether your graph had all these characteristics, and whether your description included these characteristics. Note also any characteristics included in your description that were not included here.

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RESPONSE --> ok

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22:41:18 `q006. Make a table for y = x^2 + 3, using x values -3, -2, -1, 0, 1, 2, 3. How do the y values on the table compare to the y values on the table for y = x^2? How does the graph of y = x^2 + 3 compare to the graph of y = x^2?

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RESPONSE --> x y -3 12 -2 7 -1 4 0 3 1 4 2 7 3 12 The y values for the equation of y = x^2 were 9, 4, 1, 0, 1, 4 and 9. All of the y values for the equation y = x^2 + 3 are 3 units more. The graph is also a parabola, it is simply moved 3 units up on the y axis.

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22:41:30 A list of the y values will include, in order, y = 12, 7, 4, 3, 4, 7, 12. A list for y = x^2 would include, in order, y = 9, 4, 1, 0, 1, 4, 9. The values for y = x^2 + 3 are each 3 units greater than those for the function y = x^2. The graph of y = x^2 + 3 therefore lies 3 units higher at each point than the graph of y = x^2.

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RESPONSE --> ok

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22:52:51 `q007. Make a table for y = (x -1)^3, using x values -3, -2, -1, 0, 1, 2, 3. How did the values on the table compare to the values on the table for y = x^3? Describe the relationship between the graph of y = (x -1)^3 and y = x^3.

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RESPONSE --> Y values for the equation y = (x -1)^3 would be -64, -27, -8, -1, 0, 1, 8. The values of y are changed or shifted -1 unit on the x axis. Instead of the vertex being at (0, 0) it is at (0, -1). The equation of y = (x-1)^3 is simply the y = x^3 shifted by a -1 unit.

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22:53:15 The values you obtained should have been -64, -27, -8, -1, 0, 1, 8. The values for y = x^3 are -27, -8, -1, 0, 1, 8, 27. The values of y = (x-1)^3 are shifted 1 position to the right relative to the values of y = x^3. The graph of y = (x-1)^3 is similarly shifted 1 unit to the right of the graph of y = x^3.

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RESPONSE --> ok

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23:00:37 `q008. Make a table for y = 3 * 2^x, using x values -3, -2, -1, 0, 1, 2, 3. How do the values on the table compare to the values on the table for y = 2^x? Describe the relationship between the graph of y = 3 * 2^x and y = 2^x.

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RESPONSE --> The values of y for the equation y = 3 * 2^x are 3/8, 3/4, 3/2, 3, 6, 12 and 27. All values of y for equation y = 3*2^x are multiplied by 3 from the equation of y = 2^x. The graph is shifted by 3 units upward on the x axis.

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23:01:30 You should have obtained y values 3/8, 3/4, 3/2, 3, 6, 12 and 24. Comparing these with the values 1/8, 1/4, 1/2, 1, 2, 4, 8 of the function y = 2^x we see that the values are each 3 times as great. The graph of y = 3 * 2^x has an overall shape similar to that of y = 2^x, but each point lies 3 times as far from the x-axis. It is also worth noting that at every point the graph of y = 3 * 2^x is three times as the past that of y = 2^x.

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RESPONSE --> ok

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