#$&* course Mth 163 1/31/13 around 11:00 pm 005.
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Given Solution: You should have obtained y values 9, 4, 1, 0, 1, 4, 9, in that order. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q002. Evaluate the function y = 2^x for x values -3, -2, -1, 0, 1, 2, and 3. What are your y values? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For the function y = 2^x we get the y values of 0.125, 0.25, 0.5, 1, 2, 4, 8. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: By the laws of 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q003. Evaluate the function y = x^-2 for x values -3, -2, -1, 0, 1, 2, and 3. What are your y values? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For the function y = x^-2 we get y values of 0.111, 0.25, 1, undefinied, 1, 0.25, 0.111. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: By the laws of exponents, x^-p = 1 / x^p. So x^-2 = 1 / x^2, and your x values should be 1/9, 1/4, and 1. Since 1 / 0^2 = 1 / 0 and division by zero is not defined, the x = 0 value is undefined. The last three values will be 1, 1/4, and 1/9. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q004. Evaluate the function y = x^3 for x values -3, -2, -1, 0, 1, 2, and 3. What are your y values? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The y values are 27, -8, -1, 0, 1, 8, 27. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The y values should be -27, -8, -1, 0, 1, 8, 27 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The graph for y = x^2 is a parabola with its vertex at 0,0. The graph for y = 2^x looks exponential starting at (-3, 1/8) and increasing as it goes from left to right. The graph of y = x^-2 looks like a power graph with an asymptote on either side of the y axis. The graph of y = x^2 is a power graph that increases from left to right starting negative and ending positively. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This graph is also a parabola and all of the numbers as just 3 more than those for y = x^2. Its vortex is (0,3) so it is 3 points higher on the y axis. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This table included -64, -27, -8, -1, 0, 1, & 8. It was an increase from the other table and really just went up one point higher on the y axis. It still looked very similar to the graph of y = x^3 just more negative. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The values you obtained should have been -64, -27, -8, -1, 0, 1, 8. The values for y = x^3 are -27, -8, -1, 0, 1, 8, 27. The values of y = (x-1)^3 are shifted 1 position to the right relative to the values of y = x^3. The graph of y = (x-1)^3 is similarly shifted 1 unit to the right of the graph of y = x^3. STUDENT QUESTION I assumed the graph was shifted 1 unit down since the graph passes through (0, -1) instead of origin. Then again, it passes through (1, 0), so could it be said that the graph is shifted 1 unit down OR 1 unit to the right? INSTRUCTOR RESPONSE Based on those two points that would be correct. Nowever, for example, (-2, -8) shifts to (-1, -8), a shift to the right, but not to (-2, -9), as would be the case if this was a downward shift. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q008. Make a table for y = 3 * 2^x, using x values -3, -2, -1, 0, 1, 2, 3. How do the values on the table compare to the values on the table for y = 2^x? Describe the relationship between the graph of y = 3 * 2^x and y = 2^x. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The table for y = 3*2^x included the y values of 0.375, 0.75, 1.5, 3, 6, 12, & 24. The values with this table are 3 times more than the values of the y = 2^x table. The graphs look more similar than any of the others but the one for y = 3*2^x gets higher on the x axis toward the end. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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 'high' as the corresponding point of y = 2^x. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: 3 ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The values on this table were 1, 0, 1 4, 9, 16, & 25. Compared to the y = x^2 table these values are larger and do not follow a pattern like y = x^2. There is no symmetry for the graph of y = (x+2)^2. The symmetry for y = x^2 has an axis of symmetry of (0,0). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q010. Explain in terms of the values of y = x^2 for the numbers x = -2, -1, 0, 1, 2 why we expect the graph of y = x^2 to be symmetric about the y axis. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We expect the graph of y = x^2 to be symmetric because we know that it is a parabola and would be in the shape of a U which lets us know that it confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique rating: 3 Exercise 1 Using x = -3, -2, -1, 0, 1, 2, 3, make tables and sketch graphs for the basic linear, quadratic and exponential functions (see table above under 'basic functions'). You should not have to use a calculator to get any of your y values. Using x = -5, -2, -1, -1/2, -1/10, 1/10, 1/2, 1, 2 and 5 and for positive powers and excluding x = 0 for the negative powers, make tables and sketch graphs for the basic power functions y = x ^ 2, y = x ^ 3, y = x ^ -2 and y = x ^ -3. Recall that x ^ -a = 1 / x^a. Use a calculator as little as possible; it really should be unnecessary. Be sure you can construct the table and graph of any of these basic functions in less that 1 minute, anytime you are asked. In terms of your tables explain: why the graph of y = x is a straight line = it increases at the same rate continually why y = x^2 is symmetric about x = 0 (i.e., taking the same values on either side of x = 0) = because it mirrors each other and is the same on both sides. It never crosses in the negative why y = 2^x keeps increasing as x increases, and why the graph approaches the x axis for negative values of x = the exponential rate is increasing, when it is a negative x you would take the reciprocal why y = x^3 is antisymmetric about x = 0 (i.e., taking the same values except for the - sign on either side of x = 0) = antisymmetric because it doesn’t mirror each other, if you flip it then the lines will not match up with each other why y = x^-2 and y = x^-3 rise more and more steeply as x approaches 0, and why their graphs approach the x axis as we move away from the y axis. They get steeper because your x is growing more quickly and x will have a higher increase Constructing Families by Specifying Parameters You are familiar with the y = a x^2 + b x + c form of the family of quadratic functions. This family consists of all functions whose graphs are parabolas with vertical axes of symmetry (i.e., all parabolas which open straight up or down). It would be very difficult to sketch all such parabolas. But we can look at what happens for certain combinations of a, b and c values. For example if b and c are both zero, we are left with y = a x^2. As we saw in our analysis of quadratic functions, y = a x^2 is just a vertical stretch of the basic y = x^2 function and has a graph that fits into the scheme of the graphs shown below: Different families of quadratic functions have different patterns, but they all involve parabolas. A couple of other families are shown below. The first graph is obtained by letting a = 1, b = 0, and c = -5, -4, ..., 4. So we have the functions y = a x^2 + bx + c = 1 x^2 + 0x + c, or y = x^2 + c, with c varying from -5 to 4. The functions are therefore y = x^2 -5 y = x^2 -4 y = x^2 -3 ... y = x^2 +4. The second graph is obtained by letting a=1, c=1 and b = -3, -2, ..., 5. The family is y = ax^2 + bx + c = 1 x^2 + b x + 1, or y = x^2 + bx + 1, with -3 < b < 5. The functions graphed are therefore y = x^2 - 3x + 1 y = x^2 - 2x + 1 y = x^2 - x + 1 ... y = x^2 + 5x + 1. In each of these examples we have specified some of the parameters a, b and c, and we have given a range for others. The exercises that follow apply these ideas to a variety of functions and situations. Exercises 2-10 2. Explain why the family y = x^2 + c sketched above has a series of identical parabolas, each 1 unit higher than the one below it. 3. Using the figure above, determine the approximate x coordinate of the vertex of every graph in the y = x^2 + bx + 1 family Use your knowledge of quadratic functions to obtain the x coordinate of the vertex of each of the functions sketched in this figure. How well does this agree with your estimates? 4. Sketch the graph of the exponential family y = A * 2^x for the values A = -3 to 3. Sketch the graph of the exponential family y = 2^x + c for the values c = -3 to 3. The general form of the exponential function is y = A * 2^(kx) + c. How would you describe the parameters A, k and c for the first of these families? How would you describe the same parameters for the second family? 5. Sketch the graph of the power function family y = A (x-h) ^ p + c for each of the following parameter ranges: for p = -2: A = 1, h = 0, c = -3 to 3. for p = -3: A = 1, h = -3 to 3, c = 0. 6. Make a table of y vs. x for the basic linear function, using x = -3, -2, -1, 0, 1, 2, 3. Sketch the graph. On the same set of axes sketch the graph of the generalized linear function for m = 2, b = 0, and another for m=1, b=2. 7. If f(x) = 2x + 3, then write expressions for the following: f(-2), f(3), f(x+3), f(x) + 3, 3f(x) and f(3x). All these expressions are different, and all involve substituting the expression in parentheses for x in the original form of the function. 8. Make a table for y vs. x for the basic exponential function y = 2^x and plot its graph below. On the same graph plot the generalized exponential function for A = 2, b = 2 and c = 2. Plot also the generalized exponential function for A = 1, b = .5 and c = 1. 9. If f(x) = 2^x, then write expressions for the following: f(-2), f(3), f(x+3), f(x) + 3, 3f(x) and f(3x). All these expressions are different, and all involve substituting the expression in parentheses for x in the original form of the function. 10. The illumination y from a certain florescent bulb is given as a function of distance x by the generalized power function for p = -1 with A = 370, h = 0 and c = 0. Determine the illumination at distances of 1, 2, 3 and 4 units, and sketch a graph. 005. `query 5 ********************************************* Question: `qquery introduction to basic function families problem 1 on basic graphs Why is the graph of y = x a straight line? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: it increases at the same rate continually and the slope is always equal to 1. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Since y = x the rise and run between any two points on the graph are equal, which makes the slope 1. A graph with constant slope is a straight line. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qwhy is y = x^2 symmetric about x = 0 (i.e., taking the same values on either side of x = 0) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: because it mirrors each other and is the same on both sides. It never crosses in the negative confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qwhy does y = 2^x keep increasing as x increases, and why does the graph approache the x axis for negative values of x YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: the exponential rate is increasing, when it is a negative x you would take the reciprocal confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** GOOD STUDENT RESPONSE: y = 2^x will increase as x increases on the positive side because x is the value of the exponent. This will cause the y value to double from its last value when you move one unit in the positive x direction. On the negative side of the y axis y = 2^x will approach the x axis because a negative exponent causes the value to invert into a fractional value of itself--i.e., 2^(-x) = 1 / 2^x. As we move one unit at a time negatively the value will become one half of the previous value so it will never quite reach y = 0. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qwhy is y = x^3 antisymmetric about x = 0 (i.e., taking the same values except for the - sign on opposite sides of x = 0) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: antisymmetric because first cubing a negative number always gives you a negative, second it doesn’t mirror each other, and third if you flip it then the lines will not match up with each other confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qwhy 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: They get steeper because your x is growing more quickly and x will have a higher increase. The expressions would be 1 / x^2 and 1 / x^3 causing them to have smaller and smaller denominators confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** And as x approaches 0 the expressions x^-2 and x^-3, which mean 1 / x^2 and 1 / x^3, have smaller and smaller denominators. As the denominators approach zero their reciprocals grow beyond all bound. y = x^-2 and y = x^-3 rise more and more steeply as x approaches zero because they have negative exponents they become fractions of positive expressions x^2 and x^3 respectively which have less and less slope as they approach zero. As x^2 and x^3 approach zero and become fractional, x^-2 and x^-3 begin to increase more and more rapidly because their functions are then a whole number; (1) being divided by a fraction in which the denominator is increasing at an increasing rate. As y = x^-2 and y = x^-3 move away from the y-axis they approach the x-axis because they have negative exponents. This makes them eqivalent to a fraction of 1 / x^2 or 1 / x^3. As x^2 and x^3 increase in absolute value, the values of y = x^-2 and y = x^-3 constantly close in on the x-axis by becoming a portion of the remaining distance closer, they will never reach x = zero though as this would be division by zero (since it is a fraction) ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qquery problem 2. family y = x^2 + c Explain why the family has a series of identical parabolas, each 1 unit higher than the one below it. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Each number in the series of -5 to 4 is 1 unit higher than the one below it and the value of c changes this change in the vertical direction too. The vertex is also shifted by a unit of 1. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qquery problem 4. describe the graph of the exponential family y = A * 2^x for the values A = -3 to 3. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This espression y = A * 2^x 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. We vertically stretch the y = 2^x function. The positive numbers stretch it positively and the negative, negatively. When it is 0 it is just 0. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. *&*& STUDENT QUESTION: I again am a little confused on this one. I know that vertical shift is being affected and will approach the x axis as the A values approach 0, and will move away from the x axis as the positive A values increase. INSTRUCTOR RESPONSE A vertical shift occurs when the points of one graph are all raised by the same amount. However that is not the case here. The basic points of the y = 2^x function are (-1, 1/2), (0, 1) and (1, 2). Now consider the A = 3 function y = 3 * 2^x. The basic points are (-1, 3/2), (0, 3) and (1, 6). The x = -1 point (-1, 3/2) of y = 3 * 2^x is 1 unit higher than the x = -1 point (-1, 1/2) of the y = 2^x function. The x = 0 point ( 0, 3/2) of y = 3 * 2^x is 2 units higher than the x = 0 point ( 0, 1) of the y = 2^x function. The x = 1 point ( 1, 3/2) of y = 3 * 2^x is 4 units higher than the x = 2 point ( 1, 2) of the y = 2^x function. So this is not a vertical shift. However each point of the y = 3 * 2^x function is 3 times further from the x axis than the corresponding point of the y = 2^x function: (-1,3/2) is 3/2 of a unit from the x axis, 3 times as far as the point (-1, 1/2). (0, 3) is 3 units from the x axis, 3 times as far as the point (0, 1). (1, 6) is 6 units from the x axis, 3 times as far as the point (1, 2). A vertical shift would occur, for example, for the function y = 2^x + 3. The basic points of the y = 2^x + 3 function would be (-1, 7/2), (0, 4) and (1, 6), each point being 3 units higher than the corresponding basic point of the y = 2^x function. As with most descriptions, in order to best understand the explanation you need to sketch it out. In this case you should sketch out these points on a graph and see how they are related. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qdescribe the graph of the exponential family y = 2^x + c for the values c = -3 to 3. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This graph is similar to the previous graph in that you will have a different graph for each solution such as y = 2^x -3 or y = 2^x +3. It is also similar in that the positive numbers will be shifted positively and the negative numbers negatively. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Again this is similar to the above graphs because you will have 7 different graphs to look at. The graph starts on the left side makes like an asymptote of y = 0. The graph curves as you move to the right up to the values of h. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. ** STUDENT COMMENT/QUESTION Lost..... Its not clicking for me. I understand its going to change by 1, but I need to see more information. The book is not helping me and I did a search on the Internet and can only come up with a DNA Biology math reference. Where can I look for more explanation? Thank You INSTRUCTOR RESPONSE: This is easier than you think, once you see it. Your are told that p = -3. So y = A ( x - h)^p + c becomes y = A ( x - h)^(-3) + c. Then you're told that A = 1, so y = 1 ( x - h)^(-3) + c, or just y = (x - h)^(-3) + c. Let's skip the h part for a minute and notice that c = 0. So now we have y = (x - h)^(-3) + 0 or just y = (x - h)^(-3). Now to deal with h, which is said to vary from -3 to 3. There are an infinite number of values between -3 and 3 and of course you're not expected to write a separate function for each of them. You could get the general idea using h values -3, 0 and 3. This gives you the functions y = (x - (-3) ) ^ (-3) y = (x - 0) ^ (-3) y = (x - 3) ^ (-3). Simplifying these you have y = (x + 3) ^ (-3) y = x ^ (-3) y = (x - 3) ^ (-3). The graph of x^(-3) has a vertical asymptote at the y axis (note that the y axis is at x = 0). To the right of the asymptote it decreases at a decreasing rate toward a horizontal asymptote with the positive x axis. As x approaches 0 through the negative numbers, the graph decreases at an increasing rate and forms its asymptote with the negative y axis. The graph of y = (x + 3)^(-3) has its vertical axis at x = -3; i.e., it's shifted 3 units to the left of the graph of y = x^(-3). The graph of y = (x - 3)^(-3) has its vertical axis at x = 3; i.e., it's shifted 3 units to the right of the graph of y = x^(-3). So the graphs 'march' across the x-y plane, from left to right, with vertical asymptotes varying from x = -3 to x = +3. We could fill in the h = -2, -1, 1 and 2 graphs, and as many graphs between these as we might wish, but the pattern should be clear from the three graphs discussed here. STUDENT QUESTION For this equation, I had trouble graphing and seeing how the changes were made. I didn’t understand how to substitute these values and get a shift. For example: y = A(x-h)^p + c then substituting the values you have y= 1 (x - (-3))^-3 + 0 I can see there is no vertical stretch on the y axis and then inside the parenthesis you would have (x + 3) raised to a power of -3 and finally there is no shift on the c value. Does this mean the graph would shift left or right according to the given values? But how do you solve (x + 3)^-3
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Given Solution: ** 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. ** " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: Query Add comments on any surprises or insights you experienced as a result of this assignment. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It was interesting to learn more about the different functions and what they look like. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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. ** " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!