course MTH 158 v??????Y??G~assignment #029
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18:14:05 4.3.20. Find the domain of G(x)= (x-3) / (x^4+1).
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RESPONSE --> {x | x = any real number} confidence assessment: 3
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18:14:28 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.
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RESPONSE --> ok self critique assessment: 3
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18:39:02 The expression R(x)= (6x^2+x+12) / (3x^2-5x-2) factors as (6?^2 + x + 12)/((x - 2)?3? + 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.
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RESPONSE --> oks. self critique assessment: 1
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????_??????w|?H?assignment #030 030. Query 30 College Algebra 07-22-2008
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12:11:10 4.3.20. Find the domain of G(x)= (x-3) / (x^4+1).
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RESPONSE --> {x | x = any real number} confidence assessment: 3
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12:11:19 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.
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RESPONSE --> ok self critique assessment: 3
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12:14:58 4.3.43. Find the vertical and horizontal asymptotes, if any, of H(x)= (x^4+2x^2+1) / (x^2-x+1).
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RESPONSE --> There are no horizontal or oblique asymptotes because degree n of the numerator is more than the degree m of the denominator plus 1 (n > m + 1) confidence assessment: 2
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12:15:15 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.
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RESPONSE --> ok self critique assessment: 3
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12:45:47 4.3.50. Find the vertical and horizontal asymptotes, if any, of R(x)= (6x^2+x+12) / (3x^2-5x-2).
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RESPONSE --> since the degrees of both the numerator and the denominator are the same, it will be oblique. I don't see a way to factor 6x^2 + x + 12 into the (x + y)(x + y) form, so I'll use long division... the answer I get is 2 + (11x + 16) / (3x^2 - 5x - 2) the oblique asymptote is y = 2 confidence assessment: 1
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12:46:21 The expression R(x)= (6x^2+x+12) / (3x^2-5x-2) factors as (6?^2 + x + 12)/((x - 2)?3? + 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.
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RESPONSE --> So was my answer right? self critique assessment: 2
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