Assignment 29

course Mth 158

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029. Query 29

College Algebra

08-08-2008

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06:59:24

4.2.20 (was 4.2.10). If f(x)= (x^2-5) / x^3 a polynomial?

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RESPONSE -->

No it is not a polynomial it is a ration of the two polynomials

confidence assessment: 2

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06:59:50

This is not a polynomial function. It is the ratio of two polynomials, the ratio of x^2 - 5 to x^3.

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RESPONSE -->

ok, the ratio of x^2-5 to x^3

self critique assessment: 3

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07:03:09

4.2.40 (was 4.2.30). If a polynomial has zeros at x = -4, 0, 2 the what is its minimum possible degree, and what form will the polynomial have?

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RESPONSE -->

f(x)=(x+4)(x-0)(x-2)

would be in general form

confidence assessment: 2

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07:03:20

The factors of the polynomial must include (x - (-4) ) = x + 4, x - 0 = x, and x - 2. So the polynomial must be a multiply of (x+4)(x)(x-2).

The general form of the polynomial is therefore

f(x)=a(x+4)(x-0)(x-2).

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RESPONSE -->

ok

self critique assessment: 3

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07:13:09

4/2/52 (was 4.2.40). What are the zeros of the function f(x)=(x+sqrt(3))^2 (x-2)^4 and what is the multiplicity of each?

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RESPONSE -->

The zeros of multiplicity would be -sqrt3 and 2

-sqrt3 degree of 2

2 has degree of 4

confidence assessment: 3

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07:13:20

f(x) will be zero if x + sqrt(3) = 0 or if x - 2 = 0.

The solutions to these equations are x = - sqrt(3) and x = 2.

The zero at x = -sqrt(3) comes from (x + sqrt(3))^2 so has degree 2.

The zero at x = 2 comes from (x-2)^4 so has degree 4.

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RESPONSE -->

ok

self critique assessment: 3

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07:13:58

For each zero does the graph touch or cross the x axis?

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RESPONSE -->

they would both touch the x axis

confidence assessment: 2

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07:14:21

In each case the zero is of even degree, so it just touches the x axis. Near x = -sqrt(3) the graph is nearly a constant multiple of (x+sqrt(3))^2. Near x = 2 the graph is nearly a constant multiple of (x - 2)^4.

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RESPONSE -->

ok it will touch because they are of even degree.

self critique assessment: 3

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07:19:17

What power function does the graph of f resemble for large values of | x | ?

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RESPONSE -->

x^8

confidence assessment: 1

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07:20:37

If you multiply out all the terms you will be a polynomial with x^6 as the highest-power term, i.e., the 'leading term'.

For large | x | the polynomial resembles this 'leading term'--i.e., it resembles the power function x^6. **

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RESPONSE -->

if the powers are x^2 and x^4 if we multiply the powers would it not be x^8. Im not sure where x^6 is coming from.

self critique assessment: 1

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07:23:23

4.2.62 (was 4.2.50). f(x)= 5x(x-1)^3. Give the zeros, the multiplicity of each, the behavior of the function near each zero and the large-|x| behavior of the function.

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RESPONSE -->

zeros -5 and 1

-5 degree of 1 and 1to the degree of 3

both degrees are odd so it will cross the x axis at the zeros

large lxl=5x^4

confidence assessment: 3

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07:24:00

The zeros occur when x = 0 and when x - 1 = 0, so the zeros are at x = 0 (multiplicity 1) and x = 1 (multiplicity 3).

Each zero is of odd degree so the graph crosses the x axis at each.

If you multiply out all the terms you will be a polynomial with 5 x^4 as the highest-power term, i.e., the 'leading term'.

For large | x | the polynomial resembles this 'leading term'--i.e., it resembles the power function 5 x^4.

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RESPONSE -->

I think I understand so I am multiplying the terms and adding the degrees to get the large lxl

self critique assessment: 3

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07:24:42

What is the maximum number of turning points on the graph of f?

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RESPONSE -->

n-1 would be 4-1=3 so it could have 3 turning points.

confidence assessment: 2

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07:24:56

This is a polynomial of degree 4. A polynomial of degree n can have as many as n - 1 turning points. So this polynomial could possibly have as many as 4 - 1 = 3 turning points.

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RESPONSE -->

ok

self critique assessment: 3

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07:26:50

Give the intervals on which the graph of f is above and below the x-axis

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RESPONSE -->

interval of (1,-1) to lie both above and below the x-axis

confidence assessment: 2

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07:27:30

this polynomial has zeros at x = 0 and x = 1.

So on each of the intervals (-infinity, 0), (0, 1) and (1, infinity) the polynomial will lie either wholly above or wholly below the x axis.

If x is a very large negative or positive number this fourth-degree polynomial will be positive, so on (-infinity, 0) and (1, infinity) the graph lies above the x axis.

On (0, 1) we can test any point in this interval. Testing x = .5 we find that 5x ( x-1)^3 = 5 * .5 ( .5 - 1)^3 = -.00625, which is negative.

So the graph lies below the x axis on the interval (0, 1).

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RESPONSE -->

ok, I understand

self critique assessment: 3

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