course Mth163 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): need to say x = 3 and -2 Self-critique rating: ********************************************* Question: `qWhat does the quadratic formula give you for the zeros of the quadratic polynomial q(x)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: it gives you the three places on the graph where it crosses the x axis and it also gives you the y intercept if x = 0 Confidence rating: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok, we have f(x) * g(x)..which is ..(2x-6)(x+2) .worked out as .2x^2 + 4x -6x -12 ..which is 2x^2 -2x -12, the polynomial is zero, by quadratic formula, only when: (2 + 10) / 4 =3 or (2 10) / 4 = -2, this agrees that f(x) = 0 when x = -3 and g(x) = 0 when when x = -2. The only for f(x) or g(x) to be zero is if one of them = 0. Self-critique rating: ********************************************* Question: `q2. 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)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I know that x^2 x + 6 = 0 when (x-z1) * (x-z2) = 0 , which is equivalent to x^2 x + 6 Confidence rating: unsure ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): the zeros and the highest power term match both functions and are obtained from the basic y = x^2 function which is the function for a parabola graph with the same vertical stretch and they both have the same zeros, therefore they must be identical. Self-critique rating: ********************************************* Question: `q3. 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If f(x) cannot be further factored into two linear factors it has no zeros and is called irreducible. Only a quadratic polynomial that can be reduced or further factored into two linear factors can have zeros. Confidence rating: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Any linear factor can be set equal to zero and solved for x Self-critique rating: ********************************************* Question: `q4. Explain why no polynomial of degree 2 can be the product of three or more polynomials of degree 1. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: if you have 3 polynomials like (x -3)(x+2)(x 5) and work it out you have: (x^2-x -6)(x-5) = (x^3 5x^2 3x +30) Showing you that a three degree polynomial is the product of three or more one degree polynomials and not a 2 degree polynomial. Confidence rating: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Each degree contains a nonzero multiple of x together yielding a nonzero multiple of x^3. Self-critique rating: ********************************************* Question: `q5. What then would be the zeros and the large-x behavior of y = (x-7)(x+12) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The y = 0 when x = 7 and x = -12 because: y = (x-7)(x+12) y = (7-7)(-12+12) = 0 * 0 ..which gives you y = 0 If x is a large neg. # the product of the polynomial will give us large positive #. If x is a large positive #, the product of the polynomial will give us a large positive #. The graph will increase rapidly in the positive direction and decrease more and more rapidly in the negative direction. Confidence rating: mostly ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): approaches more and more towards infinity. Graph decreases starting at large positive passes thro zero at x = -12, then decreases rapidly slowing down until it reaches between x = 7 and x = -12, where it begins to increase at increasing rate passing through zero at x = 7, and continuing with increasing slope as x becomes large. Self-critique rating: ********************************************* Question: `qDescribe your graph of this function, describing all intercepts, intervals of increasing or decreasing behavior, concavity, and large-|x| behavior. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For the absolute value of |x| of x becoming large the y gets positive crossing the x at intercepts 7, -12 and a y intercept at -84. Confidence rating: mostly ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSTUDENT 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 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok, the polynomial is increasing for x > 2.5 and decreasing for x < 2.5 Self-critique rating: ********************************************* Question: `q6. 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For y = f(x) = (x-3)(x+20(x+1) the zeros are x = 3, -2, -1 If x = large positive # all products will be large positive numbers with the final product as a large positive #. If x = large negative number all products will be large negative numbers with final product as a large negative number. The graph will pass through its zeros at x = -2, -1, an 3, increasing between x = -2 and x = -1 and then decreasing between x = -1 and x = 3. It will increase more and more rapidly through x = 3 at a faster rate as x becomes larger. Confidence rating: mostly ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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 negative 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): In your answer you have a large negative x gives you three large positive xs with the final product of a negative which would be impossible and the three products can only be three negatives.
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique rating: ********************************************* Question: `q2. Describe how the two graphs of y = (x-1)(x+3)(x-4) and y = (1/12) * (x-1)(x+3)(x-4) compare. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: They will both have zeros at x = 1, -3, 4 with the second one being 12 times as close to the x axis than the first one. Confidence rating: mostly ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): y = ((1/12) * (x-1)(x+3)(x-4) having a slope of 1/12 at every point.
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): x 3 depends on how close x is to 3, whether it is less or greater than 3, Graph (x-3)^2 will increase at an increasing rate as it moves away from 3. Basic points are (3, 0) (4, 7)(2, 7) and acts very much like a parabola whose vertex is at that zero. Self-critique rating: ********************************************* Question: `qWhy 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: because with the zero being at 3, leaving x + 4 = 7 giving an appearance of a parabola. Confidence rating: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `awith the zero of 3, x+4 will equal 7, so that portion of the graph will appear as a quadratic equation or a parabola &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): it also appears as a quadratic equation. Self-critique rating: ********************************************* Question: `qDescribe the graph of 7(x-3)^2. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since we can say this is like a parabola we can use the function y = x^2 with a horizontal shift of 3 and a vertical stretch of 7 with a vertex of (3,0) and the basic poin6ts of (4,7) and (2,7). Confidence rating: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThis 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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Make sure you describe the shifts in units. Self-critique rating: ********************************************* Question: `qHow do the graphs made on your calculator or computer compare? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I still have not figured out the graphing process on my phone, dont have a clue how to do it on the computer. I do them the old fashioned way by hand and graph paper.
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Given Solution: `aThe 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Should match very closely near(3,0) to the right of the graph it will gradually move higher and to the left it will gradually move lower. Self-critique rating: ********************************************* Question: `qWhat does the graph of a polynomial look like near a second-degree zero and why? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: a second degree polynomial is quadratic with each factor very near the zero. Confidence rating: mostly ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSTUDENT 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): each factor remains nearly constantly close to the zero. Self-critique rating: ********************************************* Question: `q5. 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: In y = (x-2)^2 * (x+3)^2 *(x-1): zeros = x = 2 and -3 and passes through x axis at x = 1 *near x = 2 we have y = (x 2)^2 *(2 +3)^2 * (2 -1) = 25(x 2)^2 this has a vertex at (2,0) and is a upward parabola *near x = -3 we have y = (-3 2)^2 * (x +3)^2 *(-3 -1) = -100(x + 3)^2 this has a vertex of(-3, 0) and is a downward parabola. - if we have a large positive x value we will have a product with a positive value and the graph will increase quickly in the upward direction. - if we have a large negative x value, we will have a product with a positive value and the graph will decrease quickly in the downward direction. - the graph will rise from the large neg. x until it touches zero at x = -3, it crosses y axis at (0, -36) then goes up and crosses at x = 1, then goes and just touches x axis between x = 1 and 2, where it begins to increase more and more up . In y = -.5(x 3) (x +2)^3: zeros = x = 3 and -2 *near x = 2 we have y = -.5(-2 -3)(x +2)^3 = 2.5(x+2)^3 this has a vertex of (2,0) with other points of (3, 2.5) ( -3, -2.5) - if we have a large positive x we will have a product with a positive value and the negative graph will decrease quickly down. - If we have a large negative x we will have a product with a negative value and the graph will move up and decrease quickly as it moves in the negative direction. - the graph will rise from the large negative value until it reaches x = -2 where it will go straight and then increase again until right before it reaches zero at x = 3, where it begins to decrease more and more quickly through x = 3 towards the negative values. Confidence rating: unsure ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok, there were so much that I could have added that I actually copied this and stapled it into my notes to further study. Self-critique rating: ok "