Assignment 30

course Mth 158

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030. Query 30

College Algebra

08-08-2008

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07:32:05

4.3.20. Find the domain of G(x)= (x-3) / (x^4+1).

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RESPONSE -->

x not=to 3 or 0

confidence assessment: 2

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07:34:18

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 -->

I was thinking the numerator could not be 0 as well, but it should have been x l x not=0

self critique assessment: 3

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07:45:01

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 -->

vertical asymptote at x=1 since it is a zero of the denominator and there are no horizontal asymptotes.

confidence assessment: 2

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07:45:16

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 -->

self critique assessment: 3

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09:07:21

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 -->

3x^2-5x-2=(3x+1) (x-2)

3x+1=0

3x=-1

x=-1/3

x-2=0

x=2

There are hertical asymptotes at -1/3 and 2

6x^2/3x^2=2

There is a horizontal asymptote at 2.

confidence assessment: 2

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09:07:37

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.

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RESPONSE -->

self critique assessment: 3

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Assignment 30

&#Very good work. Let me know if you have questions. &#