Assignment 30

course MTH 158

v??????Y??G~assignment #029

029. Query 29

College Algebra

07-22-2008

??????????X??assignment #030

030. Query 30

College Algebra

07-22-2008

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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

y = 2 is an asymptote; it's horizontal, not oblique.

Don't forget the vertical asymptotes.

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&#This looks very good. Let me know if you have any questions. &#