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: The denominator is x^4 + 1, which could be zero only if x^4 = -1. However x^4 being an even power of x, it cannot be negative. So the denominator cannot be zero and there are no vertical asymptotes. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): There are no discrepancies ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: * 5.2.43 / 7th edition 4.3.43. Find the vertical and horizontal asymptotes, if any, of H(x)= (x^4+2x^2+1) / (x^2-x+1). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I think 5.2.43 is a different problem, but I will solve the one given here to see if I can get it right by the given solution. H(x)= (x^4+2x^2+1) / (x^2-x+1) factors into (x^2 + 1)^2 / (x – 1 )^2 The zeros of the denominator are x= 1 so the vertical asymptote is when x = 1 The degree of the numerator is larger than that of the denominator so we use long division. (Here is where I have such a hard time with this. I don’t understand how to do long division like this. I have gone back to the review chapters and I still don’t understand how to do it.) When I tried doing the long division, I came up with an answer like: x^2 – 2 + 2x^2 – 1/x^2 – x + 1 If that is the correct way, which I don’t think it is, how would you find the horizontal asymptote from that? confidence rating: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The function (x^4+2x^2+1) / (x^2-x+1) factors into (x^2 + 1)^2 / (x-1)^2. The denominator is zero if x = 1. The numerator is not zero when x = 1 so there is a vertical asymptote at x = 1. The degree of the numerator is greater than that of the denominator so the function has no horizontal asymptotes. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): So, if the degree of the numerator is greater than that of the denominator, is this saying it will always be no horizontal asymptotes? Where does the long division come in?
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Given Solution: * * The expression R(x)= (6x^2+x+12) / (3x^2-5x-2) factors as (6•x^2 + x + 12)/((x - 2)•(3•x + 1)). The denominator is zero when x = 2 and when x = -1/3. The numerator is not zero at either of these x values so there are vertical asymptotes at x = 2 and x = -1/3. The degree of the numerator is the same as that of the denominator so the leading terms yield horizontal asymptote y = 6 x^2 / (3 x^2) = 2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): There are no discrepancies ------------------------------------------------ Self-critique Rating: 3 "