Assignment 30

course Mth 158

???E??-?????????assignment #030

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030. Query 30

College Algebra

08-02-2007

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14:42:39

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

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RESPONSE -->

The denominator is 0 if x^4 = -1, but x^4 is an even power x and it cannot be negative. The denominator cannot be 0 so therefore there are no asymptotes.

confidence assessment: 2

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14:43:45

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 knew that x^4 could not be negative and equation has no asymptotes.

self critique assessment: 2

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14:44:56

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 -->

This function factors into ( x^2 +1)^2 / (x-1)^2. The denominator is 0 if x= 1 so there is a vertical asymptote at x= 1.

confidence assessment: 2

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14:46:19

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 -->

I found the vertical asymptote, but I forgot to find the horizontal one. I understand that the degree of the numerator is greater than the denominator so there are no horizontal asymptotes.

self critique assessment: 2

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14:48:54

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 -->

The function factors into (6x^2 + x + 12) / ( (x-2)(3x+1)

The denominator is 0 when x = 2 and x = -1/3. So there are vertcial asymptotes at x= 2 and x = -1/3. The degree of the numerator is the same as the denominator so there is horizontal asymptote at y= 6x^2 / (3x^2) = 2.

confidence assessment: 2

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14:49:12

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 -->

I found the same asymptotes.

self critique assessment: 3

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