course mth 158 If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: * * This is not a polynomial function. It is the ratio of two polynomials, the ratio of x^2 - 5 to x^3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): There are no discrepancies ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: * 5.1.20 / 7th edition 4.2.40 (was 4.2.30). If a polynomial has zeros at x = -4, 0, 2 the what is its minimum possible degree, and what form will the polynomial have? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Zeros are x = -4, 0, 2 This would mean that 0 and 2 are factors of x = -4 To set this up you would choose a leading coefficient. I will use a = 1 (x – (-4)) = a(x+4)(x-0)(x-2) = x^2+x^2-2x+4x-0+4x-8 =2x^2 + 6x -8 (which is a quadratic equation) the degree is 2 confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * The factors of the polynomial must include (x - (-4) ) = x + 4, x - 0 = x, and x - 2. So the polynomial must be a multiply of (x+4)(x)(x-2). The general form of the polynomial is therefore f(x)=a(x+4)(x-0)(x-2). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I don’t know if the given solution is the same as my solution. Did I need to work it like that or could I have just said f(x)=a(x+4)(x-0)(x-2)?
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Given Solution: * * f(x) will be zero if x + sqrt(3) = 0 or if x - 2 = 0. The solutions to these equations are x = - sqrt(3) and x = 2. The zero at x = -sqrt(3) comes from (x + sqrt(3))^2 so has degree 2. The zero at x = 2 comes from (x-2)^4 so has degree 4. For each zero does the graph touch or cross the x axis? In each case the zero is of even degree, so it just touches the x axis. Near x = -sqrt(3) the graph is nearly a constant multiple of (x+sqrt(3))^2. Near x = 2 the graph is nearly a constant multiple of (x - 2)^4. What power function does the graph of f resemble for large values of | x | ? If you multiply out all the terms you will be a polynomial with x^6 as the highest-power term, i.e., the 'leading term'. For large | x | the polynomial resembles this 'leading term'--i.e., it resembles the power function x^6. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I didn’t see the question of the power function until now. I am still a little unsure how to solve this. I have been reading ch5 again and the way I understand the power function at large values of |x| is this: There are only 4 types. 1. It opens up like that of a parabola when the power function is an even number (2, 4, 6….) and the leading number is greater than 0. 2. It open down when the power function is even and the leading number is less than 0. 3. It looks like an asymmetric graph with respect to the origin with the points (1,1) and (-1,-1) when the power function is odd and the leading number is greater than 0. 4. It looks like number 3 only it has been flipped with points (-1, -1) and (1, -1) when the power function is odd and the leading number is less than 0. If my interpretation is correct, then how do you multiply out the terms to get the highest power term? It seems like I have more trouble with the basics like (multiplying out the terms, and long division of polynomials) than grasping this.
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Given Solution: * * The zeros occur when x = 0 and when x - 1 = 0, so the zeros are at x = 0 (multiplicity 1) and x = 1 (multiplicity 3). Each zero is of odd degree so the graph crosses the x axis at each. If you multiply out all the terms you will be a polynomial with 5 x^4 as the highest-power term, i.e., the 'leading term'. For large | x | the polynomial resembles this 'leading term'--i.e., it resembles the power function 5 x^4. What is the maximum number of turning points on the graph of f? This is a polynomial of degree 4. A polynomial of degree n can have as many as n - 1 turning points. So this polynomial could possibly have as many as 4 - 1 = 3 turning points. Give the intervals on which the graph of f is above and below the x-axis this polynomial has zeros at x = 0 and x = 1. So on each of the intervals (-infinity, 0), (0, 1) and (1, infinity) the polynomial will lie either wholly above or wholly below the x axis. If x is a very large negative or positive number this fourth-degree polynomial will be positive, so on (-infinity, 0) and (1, infinity) the graph lies above the x axis. On (0, 1) we can test any point in this interval. Testing x = .5 we find that 5x ( x-1)^3 = 5 * .5 ( .5 - 1)^3 = -.00625, which is negative. So the graph lies below the x axis on the interval (0, 1). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I just saw where it asks the question…what are the maximum numbers of turning points of the graph? I don’t see how the degree is 4 in this particular problem. I know that if it is 4, then the turning points would be n-1 making it 3 turning points.