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18:58:13 What are the zeros of f(x) = 2x - 6 and g(x) = x + 2?
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RESPONSE --> 2x - 6 = 0 2x = 6 x = 3 x + 2 = 0 x = -2 confidence assessment: 3
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18:58:24 ** 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 --> self critique assessment: 3
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19:08:21 ** 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 --> self critique assessment: 3
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19:35:13 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 --> You should be able solve the quadratic formula and get the zeros then substitute them back into the other equation. Solving for the quad. formula: x = [ -(-1) +- sqrt( (-1)^2 - 4(1)(6) ] / (2 * 1) = [ 1 +- sqrt(23) ] / 2 = [ 1 +-4.795831523 ] / 2
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19:38:24 ** 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 --> I did a lot of work, but I have one question.... This doesn't show the quadratic formula answer to prove it. self critique assessment: 2
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19:41:38 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 --> A quadratic polynomial without zeros is irreducible and cannot be factored into two linear factors confidence assessment: 2
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19:41:49 ** 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 --> self critique assessment: 3
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19:51:56 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 --> Three or more polynomials of degree 1 will give you a polynomial of degree of 3, not 2. confidence assessment: 1
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19:52:41 ** 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 --> self critique assessment: 3
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19:53:30 5. What then would be the zeros and the large-x behavior of y = (x-7)(x+12)
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RESPONSE --> 7 or -12 confidence assessment: 2
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20:18:33 ** 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 --> I forgot to add what would happen with the large x behavior would be, but I understand the behavior of the large x graph. self critique assessment: 2
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20:26:18 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 --> The graph will be decreasingfrom the very large positive values at large negative x, as it passes through the x intercept of x = -12. The graph will decrease quickly at first, but but will decrease less and less rapidly until the graph reaches a point between x = 7 and x = -12, where it begins increasing again at an increasing rate, passing through another x intercept of x = 7 and continuing with increasing slope as x becomes large confidence assessment: 1
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20:26:39 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 --> self critique assessment: 3
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21:43:30 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 --> my graph is increasing from very large negative values at large negative x. The rate of increase is initially be very rapid but will decrease less and less rapidly until the graph reaches a high point between x = -2 and x = -1, at which point it begins decreasing at an decreasing rate. Then at (1, -12) the graph begins to increase at an increasing rate, passing through the zero at x = 3 and continuing with increasing slope as x becomes large. confidence assessment: 1
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21:44:03 ** 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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RESPONSE --> self critique assessment: 3
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22:07:36 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 --> y = (x-x1)(x-x2)(x-x3) y = (0-3)(0-1)(0-2) y = (-3)(-1)(-2) y = -6 confidence assessment: 1
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22:12:45 ** 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 misunderstood the question. I do understand where the (x+3)(x-1)(x-2) came from. Then to get the other form, you just multiply these three out: (x^2+2x - 3)(x - 2) x^3 - 7x + 6 self critique assessment: 2
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22:20:55 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 second graph is vertically stretched by 1/2. confidence assessment: 1
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22:26:15 ** 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 --> Wouldn't it be that the graph is 2 times closer to the x axis than the first. I did not mention where the zeros would be. self critique assessment: 1
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22:38:31 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 --> In the following example x1 = x2 = 3 and x3 = -4. Instead of going through the zero on the x axis, the graph just curves down to touch the x axis at that point before curving back up. The y intercept is clearly at y = 3 (just substitute 0). At x = 3, the graph doesn't actually go through the x axist. It just touches the axis then turns and goes up, sort of like a parabola with a vertex at (3,0). confidence assessment: 1
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00:11:04 ** 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 --> self critique assessment: 3
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00:24:35 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 --> To be honest, I am not fully sure where the 7(x-3)^2 came from. The previous explaination lost me, but I hit the button instead of writing that I was unclear. confidence assessment: 0
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00:25:25 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 --> This helps a litttle, but I am still a little confused. self critique assessment: 1
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00:29:01 Describe the graph of 7(x-3)^2.
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RESPONSE --> The graph is a parabola that is vertically stretched by 7. confidence assessment: 1
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00:32:51 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 --> I didn't put down that there was a hrizontal shift of 3, or what the vertex and basic points are. I do understand where that came from. self critique assessment: 1
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00:34:38 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 --> The two graphs are almost perfectly match. self critique assessment: 3
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00:43:01 What does the graph of a polynomial look like near a second-degree zero and why?
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RESPONSE --> It basically looks like a parabola, usually with a vertical stretch. This is because it is derived from the x^2 confidence assessment: 1
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00:46:19 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 --> I did not mention the fact that sicne zero is repeated through the graph, it can't pass the x-axis. self critique assessment: 1
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01:08:50 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 --> For y = (x-2)^2 * (x+3)^2 * (x-1) the zeros are x = 2 x = -3, x = 1 It increases as it approaches x = -3, then decreases to the large x minimum and begins to increase at an increasing rate passing through x=1 with part of a parabola whose vertex is at (1.5, 3) and be geins to decrease again, reaching x = 2 and then begins to increase again. For y = -.5 * (x-3) (x+2)^3 the zeros are x= 3 and x = -2. This graph is increaseing at an increasing rate. It passes through x = -2 adn continues to its large x value then begins to decrease at a decreasing rate again passing through x = 3 confidence assessment: 1
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01:09:52 ** 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 --> This is a lot of info to try to get down. I usually miss sometehing, but I think I mostly hit everything that was asked for. self critique assessment: 1
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