course Mth 163
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11:20:42 explain the steps in fitting an exponential function to data
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RESPONSE --> 1. Linearize an exponental function in which we replace all our y values with log(y) 2. This new table will be a model for log(y) vs. x 3. Plot points on a graph and use a best-fit straight-line and use them to determine the equation which models the data 4. Form an equation to model our original y vs. x data 5. Compare the values obtained for the given x values to the original data from which the model was obtained.
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11:21:03 04-23-2006 11:21:03 ** If you have two points you can solve the simultaneous equations. If you have a more extensive data set you use transformations. For exponential data you plot log(y) vs. x. If the graph is a straight line then you have a good fit. If the slope is m and the vertical intercept is b then your graph gives you log(y) = m x + b. You invert the transformation using the inverse function 10^x, obtaining 10^log(y) = 10^(mx+b) so that y = 10^(mx) * 10^b, and then rearrange this into the desired form. Note that if a power function fits the data then log y vs. log x will give a straight line so that log y = m log x + b. In this case our solution will be y = 10^b * x^m, a power function rather than an exponential function. **
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NOTES -------> OK
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assignment #019 Ή禈ϏHKޔ Precalculus I 04-23-2006
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11:21:24 explain the steps in fitting an exponential function to data
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RESPONSE -->
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11:21:27 ** If you have two points you can solve the simultaneous equations. If you have a more extensive data set you use transformations. For exponential data you plot log(y) vs. x. If the graph is a straight line then you have a good fit. If the slope is m and the vertical intercept is b then your graph gives you log(y) = m x + b. You invert the transformation using the inverse function 10^x, obtaining 10^log(y) = 10^(mx+b) so that y = 10^(mx) * 10^b, and then rearrange this into the desired form. Note that if a power function fits the data then log y vs. log x will give a straight line so that log y = m log x + b. In this case our solution will be y = 10^b * x^m, a power function rather than an exponential function. **
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RESPONSE -->
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ꫮʇTɳ assignment #020 Ή禈ϏHKޔ Precalculus I 04-23-2006
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17:45:17 What are the zeros of f(x) = 2x - 6 and g(x) = x + 2?
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RESPONSE --> f(x) = 2x - 6 f(x) = 0 x = 3 g(x) = x+2 g(x) = 0 x = -2
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17:45:29 ** f(x) = 2x - 6 is zero when 2x - 6 = 0. This equation is easily solved to yield x = 3. g(x) = x + 2 is zero when x + 2 = 0. This equation is easily solved to yield x = -2. **
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RESPONSE --> OK
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17:53:42 What does the quadratic formula give you for the zeros of the quadratic polynomial q(x)?
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems I'm ""stumped"" on this question, maybe it's the wording, but I can't follow what I'm being ask to do...I'll self-critique
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18:17:29 ** We get q(x) = f(x) * g(x) = (2x - 6) ( x + 2) = 2x ( x + 6) - 6 ( x + 2) = 2 x^2 + 4 x - 6 x - 12 = 2 x^2 - 2 x - 12. This polynomial is zero, by the quadratic formula, when and only when x = [ -(-2) +- sqrt( (-2)^2 - 4(2)(-12) ] / (2 * 2) = [ 2 +- sqrt( 100) ] / 4 = [ 2 +- 10 ] / 4. Simplifying we get x = (2+10) / 4 = 3 or x = (2 - 10) / 4 = -2. This agrees with the fact that f(x) = 0 when and only when x = -3, and g(x) = 0 when and only when x = 2. The only was f(x) * g(x) can be zero is for either f(x) or g(x) to be zero. **
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RESPONSE --> There was no equation to solve on the previous page, but now that I have the problem... using the quadradic formula -b +-sqrt(b^2 - 4ac)/2a x = 2 +-sqrt(100) / 4 x = 12 / 4 = 3 x = -8 / 4 = -2
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18:35:59 2. If z1 and z1 are the zeros of x^2 - x + 6, then what is the evidence that x^2-x + 6=(x - z1) * (x - z2)?
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RESPONSE --> I don't know if I've missed something during the lessons, but I can't recall a practice set like this...the zero's of this equation are 3 and -2. If this is not what you are asking, I will review the anser and self-critique.
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18:38:47 ** z1 and z2 both give zero when plugged into x^2 - x + 6 and also into (x-z1)(x-z2). (x-z1)(x-z2) gives an x^2 term, matching the x^2 term of x^2 - x + 6. Since the zeros and the highest-power term match both functions are obtained from the basic y = x^2 function by the same vertical stretch, both have parabolic graphs and both have the same zeros. They must therefore be identical. **
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RESPONSE --> So, once again, I was incorrect in my answer. I read your explanation, but still have questions on it (explanation). I've made note of the problem, and will research in my notes to see if this is something that I've missed or just something that I need to review.
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18:46:36 3. Explain why, if the quadratic polynomial f(x) = a x^2 + bx + c has no zeros, that polynomial cannot be the product of two linear polynomials.
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RESPONSE --> Honestly, I can't explain this...
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18:48:06 ** If f(x) has linear factors, then if any of these linear factors is zero, multiplying it by the other factors will yield zero. Any linear factor can be set equal to zero and solved for x. Thus if f(x) has linear factors, it has zeros. So if f(x) has no zeros, it cannot have linear factors. **
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RESPONSE --> makes perfect sense.
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18:49:36 4. Explain why no polynomial of degree 2 can be the product of three or more polynomials of degree 1.
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RESPONSE --> Multipling three or more polynomials with a degree of 1 will result in a polynomial with a degree of 3 or more.
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18:49:50 ** If you have 3 polynomials of degree one then each contains a nonzero multiple of x. Multiplying three such factors together will therefore yield a term which is a nonzero multiple of x^3. For example (x-2)(x+3)(x-1) = (x^2 + x - 6)(x+1) = x^3 + 2 x^2 - 5 x - 6. Any polynomial containing a nonzero multiple of x^3 has degree at least 3, and so cannot be of degree 2. Therefore a polynomial of degree 2 cannot be a product of three or more polynomials of degree 1. **
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RESPONSE --> OK
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18:50:42 5. What then would be the zeros and the large-x behavior of y = (x-7)(x+12)
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RESPONSE --> the zero's of this equation are 7 and -12. I don't comprehend the meaning of ""large-x""
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18:51:40 ** y = 0 when x-7 = 0 or x+12 = 0, i.e., when x = 7 or x = -12. If x is a large positive number then both x-7 and x+12 are large positive numbers so that (x-7)(x+12) is a very large positive number. If x is a large negative number then both x-7 and x+12 are large negative numbers so that (x-7)(x+12) is again a very large positive number. So for large positive and negative x the function more and more rapidly approaches infinity. The graph will be decreasing, beginning with very large positive values at large negative x, as it passes through its leftmost zero at x = -12. The rate of decrease will initially be very rapid but will decrease less and less rapidly until the graph reaches a low point between x = 7 and x = -12, at which point it begins increasing at an increasing rate, passing through its rightmost zero at x = 7 and continuing with increasing slope as x becomes large. **
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RESPONSE --> logical
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18:52:40 Describe your graph of this function, describing all intercepts, intervals of increasing or decreasing behavior, concavity, and large-|x| behavior.
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RESPONSE --> what function??? the one on the previous page? if so, I didn't copy it down to reference back to it, I was able to look at it (previously) and give the zero's without using a pencil.
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19:09:27 STUDENT RESPONSE: for large | x | , y gets positive y intercept=-84 parabola opens upward very steeply rising with x intercepts at 7 and -12 INSTRUCTOR COMMENT: Good. Also, you should say that the polynomial is increasing for x > 2.5 and decreasing for x < 2.5
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RESPONSE --> Apparently, I either didn't remember the question correctly or just didn't do it correctly...I didn't get the same response as the ""Student Response"" to this question.
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19:16:06 6. Describe your graph of y = f(x) = (x-3)(x+2)(x+1), describing all intercepts, intervals of increasing or decreasing behavior, concavity, and large-|x| behavior.
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RESPONSE --> The graph of this problem will have x intercepts at 3, -2, -1 with the y intercept being -6. As the line passes thru the -2 x-intercept, it gets more and more negative...as it passes the 3 x-intercept it becomes more and more positive .
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19:17:23 04-23-2006 19:17:23 ** The function has zeros at x = 3, x = -2 and x = -1. For large positive x all three factors will be large positive numbers, so that the product will be a very large positive number. For large negative x all three factors will be large positive numbers, so that the product will be a very large negative number. The graph will be increasing, beginning with very large negative values at large negative x, as it passes through its leftmost zero at x = -2. The rate of increase will initially be very rapid but the graph will increase less and less rapidly until the graph reaches a relative maximum point between x = -2 and x = -1, at which point it begins decreasing. THe function will be decreasing as it passes through its zer0 at x = -1. Somewhere between x = -1 and its next zero at x = 3 the function will reach a relative minimum value after which it will begin to increase more and more rapidly. It will be increasing as it passes through its zero at x = 3 and will continue to increase faster and faster as x becomes larger. **
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NOTES -------> Your explanation describes my graph perfectly...I guess my wording describing a graph needs more detail.
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19:30:34 1. Give the y = (x-x1)(x-x2)(x-x3) form of a degree 3 polynomial with zeros at x = -3, 1 and 2, as well as the y = ax^3 + bx^2 + cx + d form.
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RESPONSE --> The degree 3 form of this equation is y = (x+3)(x-1)(x-2) In the next form of y = ax^3 + bx^2 + cx + d is (x+3)(x^2 + 2x + 3)
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19:32:20 ** The factored form is y=(x+3)(x-1)(x-2) The standard polynomial form is obtained by multiplying these factors to obtain (x+3) ( x^2 - 2x - x + 2) = (x+3)( x^2 - 3x + 2) = (x^3 - 3 x^2 + 2 x) + (3 x^2 - 9 x + 6) = x^3 - 7 x + 6. **
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RESPONSE --> I admit that I didn't do the second part of this question correctly...after reviewing the steps, I see my mistake and made note of such
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19:40:20 2. Describe how the two graphs of y = (x-1)(x+3)(x-4) and y = (1/12) * (x-1)(x+3)(x-4) compare.
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RESPONSE --> The graph of y = (x-1)(x+3)(x-4) gives a parablic graph with the x intercepts at -3, 1, 4 with the y intercept at 12. Starting with a very large negative number, it becomes less negative at it approaches the -3 x-intercept, passes thru it to the y-intercept at 12, getting less positive as it approaches the x-intercept of 1 curving back to the next x-intercept of 4 where it takes off getting more positve. The 1/12 version is just that...taking 1/12 of each of the previous points, the graph follows the same shape, just on a much smaller scale.
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19:40:32 ** The graphs both have zeros when x - 1 = 0, when x + 3 = 0 and when x - 4 = 0. These zeros therefore occur at x = 1, x = -3 and x = 4. The only difference is that the graph of y = 1/12 ( x-1)(x+3)(x-4) is everywhere 12 times closer to the x axis than that of y = (x-1)(x+3)(x-4), with 1/12 the slope at every point. **
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RESPONSE --> OK
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19:45:42 4. What function describes the approximate behavior of the graph of y = p(x) = (x-3)(x-3)(x+4) near the point (3,0)?
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RESPONSE --> This is a polynomial function that graphs as a parabola with the x intercepts at -4 and 3 with a y intercept at 28. The parabola starts with a great negative number becoming less negative as it reaches the x intercept of -4 following to the y intercept of 12 and then becoming less positive as it reaches the x intecept at 3 and then getting more negative.
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19:47:39 ** If x is close to 3 then x + 4 is close to 7 and is not significantly different for various values near x = 3. However the nature of x - 3 depends greatly on just how close x is to 3, and whether x is greater or less than 3. x - 3 = 0 when x = 3, x - 3 < 0 when x > 3 and x - 3 > - when x < 3. (x-3)^2 will be zero when x = 3, and will increase at an increasing rate as x moves away from 3. So the function y = (x-3)(x-3)(x+4) is close to y = 7(x-3)^2. Note that this function describes a parabola with vertex at (3, 0), the 2d-degree zero of the given polynomial, and basic points (3, 0), (4, 7) and (2, 7). So near x = 3 the graph of p(x) = (x-3)(x-3)(x+4) will be very nearly matched by the parabolic graph of the function y = 7 ( x - 3) ^2. As x moves out of the vicinity of x = 3 the graphs will at first gradually, then more and more rapidly move apart. In general near z second-degree 0, like 3 in the present example, the graph of a parabola will look like a parabola whose vertex is at that zero. **
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RESPONSE --> Wow...I didn't see that at all !!!
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19:49:44 Why do we say that near (3,0) the graph of (x-3)(x-3)(x+4) is approximately the same as the graph of 7(x-3)^2?
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RESPONSE --> ?
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19:51:45 with the zero of 3, x+4 will equal 7, so that portion of the graph will appear as a quadratic equation or a parabola
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RESPONSE --> yes, with a zero of 3, x+4 will equal 7 and so does appear that that portion will appear as a parabola.
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20:24:19 Describe the graph of 7(x-3)^2.
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RESPONSE --> It is a parabola with a vertex at 3,0 with the values of -2,25 and 8,25.
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20:27:01 This is a parabola, obtained from the basic y = x^2 parabola by a vertical stretch of 7 and horizontal shift of 3 units. It will be a steep parabola with vertex (3, 0) and basic points at (2, 7) and (4, 7).
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RESPONSE --> this is correct...i graphed the problem and forgot to look back at the original to see the 7. The 7 would show a vertical stretch of 7, that I didn't catch from the original equation.
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20:29:44 How do the graphs made on your calculator or computer compare?
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RESPONSE --> I didn't do derive yet...now that it was done, derive didn't give me the vertical shift that it should have.
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20:29:57 The two graphs should match very closely near (3, 0). To the right the graph of the polynomial will gradually move higher than that of the parabola, and to the left will gradually move lower.
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RESPONSE --> ok
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20:31:55 What does the graph of a polynomial look like near a second-degree zero and why?
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RESPONSE --> Should it not look the same?
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20:33:59 STUDENT ANSWER: parabola, when that portion is factored out it is a quadratic, since that zero is repeated the graph cannot cross the x axis at that point but must touch it sou appearing as a parabola INSTRUCTOR'S ADDITION: Also because the other factors of the polynomial remain nearly constant close to the zero.
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RESPONSE --> if my answer was not clear, I meant a parabola, just as the previous answer. very much correct...since it can't have a zero, therefore it cannot touch the x axis but can be very close
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21:36:41 5. Sketch graphs of y = (x-2)^2 * (x+3)^2 * (x-1) and y = -.5 * (x-3) (x+2)^3, including intercepts, the large-| x | behavior for both positive and negative x, concavity, and intervals of increasing and decreasing behavior.
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RESPONSE --> here goes...I did it on derive and this is what the graph shows me... y = (x-2)^2 * (x+3)^2 * (x-1) is a graph that starts at a very large negative number, comes to the x-axis at -3, curves back down to approximately to the coordinates of -5,70 curves back to the positive and crosses the x-axis at 1 to a point at appox 2.5,1.8 turns back to the negative to the x axis at 2,0 and then takes back off to a very large positive number. y = -.5 * (x-3)(x+2)^3 starts at a very large negative number becoming less negative to the y axis at -2,0, takes back toward positive to appoximately 1.8,32 and takes a dive toward a great negative number crossing the x axis at 3.
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21:38:36 ** The graph of y = (x-2)^2 * (x+3)^2 * (x-1) is nearly parabolic in the vicinity of the zeros at 2 and -3. It only passes through the x axis at x = 1. Near x = 2 we can approximate all factors except (x-2)^2 by substituting x = 2, which gives us y = (x-2)^2 * (x+3)^2 * (x-1) = (x-2)^2 * (2+3)^2 * (2-1) = 25 (x-2)^2, an upward-opening parabola with vertex at x = 2. Near x = -3 we can approximate all factors except (x+3)^2 by substituting x = -3, which gives us y = (x-2)^2 * (x+3)^2 * (x-1) = (-3-2)^2 * (x+3)^2 * (-3-1) = -100 (x+3)^2, a downward-opening parabola with vertex at x = -3. For large positive x the graph is positive and concave up, increasing very rapidly. For large negative x the graph is negative and concave down, decreasing very rapidly. The graph rises from extremely large negative x values to the zero at x = -3, where it touches the x axis and turns back toward negative values without ever passing through the x axis. It reaches a minimum somewhere between x = -3 and x = 1, in the process passing through the y axis at (0, -36). The graph passes through the x axis at x = 1, going from negative to positive. It turns back toward the x axis at some point between x = 1 and x = 2, touches the x axis moving along in which is nearly parabolic in the vicinity of that point, and the turns back upward, increasing with a rapidly increasing slope as x moves to the right. The graph increases at a decreasing rate up to (-3,0), then decreases at an increasing rate until concavity changes from negative to positive sometime before the function reaches its minimum somewhere between (-3,0) and (1,0). Then it decreases at an increasing rate and continues to do so until a point between the local minimum and (1,0), probably close to (1,0), at which concavity again becomes negative. From that point the function increases as a decreasing rate until it reaches a local maximum somewhere between x=1 and x=2, at which point it begins decreasing at an increasing rate, remaining concave down until at some point before (2,0) the concavity becomes upward and the function begins decreasing at a decreasing rate until reaching the local minimum at (2,0). From that point it begins increasing at an increasing rate, maintaining an upward concavity and rapidly increasing to very large y values. ALTERNATIVE DESCRIPTION: The graph of y = -.5 * (x-3) (x+2)^3 passed thru the x axis at x = 3 and at x = -2. Near x = -2 we can approximate all factors except (x-2)^2 by substituting x = 2, which gives us y = -.5 ( -2 - 3) ( x + 2)^3 = 2.5 (x+2)^3. This function gives us a cubic polynomial with zero at x = -2 and basic points (-2, 0), (-3, -2.5) and (3, 2.5). For large positive x the graph is negative and concave down, decreasing very rapidly. For large negative x the graph is negative and concave up, decreasing very rapidly as x moves in the negative direction. The graph rises from extremely large negative x values toward the zero at x = -2, leveling off at (-2, 0) before again beginning to increase at a increasing rate. Somewhere before the zero at x = 3 the graph turns around and begins decreasing, passing downward through (3, 0) as it declines faster and faster into negative values.**
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RESPONSE --> Wow...that was descriptive...I hope that my description will give you an accurate account of what it looks like.
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