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Good.

If anything is not clear let me know.

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21:41:54 4.2.20 (was 4.2.10). If f(x)= (x^2-5) / x^3 a polynomial?

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RESPONSE --> no, I believe this is not a polynomial, but a ratio.

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21:41:59 This is not a polynomial function. It is the ratio of two polynomials, the ratio of x^2 - 5 to x^3.

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RESPONSE --> ok

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21:44:59 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?

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RESPONSE --> If the polynomial's zeroes are at -4,0,2, then it would have factors= x+4, x-0, and x-2, just multiplied together. I believe its minimum possible degree is 1.

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21:45:24 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).

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RESPONSE --> Yes, but what about its degree? The rest I understand.

If you multiply out (x+4)(x)(x-2) you get a polynomial with x^3, x^2, x and constant terms. Since (x+4)(x)(x-2) is part of the polynomial, the polynomial has degree at least 3. There might be other factors as well, which could increase the degree beyond 3, but the degree is at least 3.

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21:46:38 4/2/52 (was 4.2.40). What are the zeros of the function f(x)=(x+sqrt(3))^2 (x-2)^4 and what is the multiplicity of each?

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RESPONSE --> The zeroes are at -sqrt3 and 2. The multiplicities are 2 and 4.

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21:46:48 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.

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RESPONSE --> ok

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21:47:16 For each zero does the graph touch or cross the x axis?

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RESPONSE --> The degrees are both even, so each zero touches the x axis.

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21:47:25 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.

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RESPONSE --> ok

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21:50:04 What power function does the graph of f resemble for large values of | x | ?

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RESPONSE --> I know according to the definition that this graph behaves like y=a sub n *x^n. I am not sure where else to head with this.

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21:50:27 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. **

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RESPONSE --> Still a little hazy on this, but I will read and reread until I get it.

Your answer was correct. When x gets very large, only the highest-power term is significant, and the graph will resemble the graph of that term.

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21:53:03 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).

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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 Again a keystrike problem. Zeroes occur where x=1 or 0. x=0 multiplicity is 1, x=1, mult.= 3. Each zero is odd so it crosses the x axis at those zeroes. I hate the enter key, it has messed me up a lot lately.

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No problem here. As you see, the program nags you until you provide the information.