assignment 29

course mth 158

Good job. See my note.

Let me know if there's anything you need me to explain further.

ؾ{]Uӂ͛ξ College Algebra 11-27-2005

......!!!!!!!!...................................

23:26:15 4.2.20 (was 4.2.10). If f(x)= (x^2-5) / x^3 a polynomial?

......!!!!!!!!...................................

RESPONSE --> no

.................................................

......!!!!!!!!...................................

23:26:37 This is not a polynomial function. It is the ratio of two polynomials, the ratio of x^2 - 5 to x^3.

......!!!!!!!!...................................

RESPONSE --> ok

.................................................

......!!!!!!!!...................................

23:49:30 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?

......!!!!!!!!...................................

RESPONSE --> f(x)=a(x-(-4))(x-0)(x-2) for a=1 f(x)=(x+4)(x)(x-2) (x^2+4x)(x-2) x^3-2x^2+4x^2-8x x^3+2x^2-8x

.................................................

......!!!!!!!!...................................

23:50:33 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).

......!!!!!!!!...................................

RESPONSE --> i solved it a little bit farther like the book shows i see

You expanded the factored form, which is very good and appropriate.

.................................................

......!!!!!!!!...................................

00:07:28 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?

......!!!!!!!!...................................

RESPONSE --> the real zeros are -3with multiplicity of1 because sqrt 3would be 1/2 and then ^2 would be 1 and then the next zero is 2 with multiplicity of 4

.................................................

......!!!!!!!!...................................

00:13:27 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.

......!!!!!!!!...................................

RESPONSE --> ok i see i went to far into the sqrt 3 . i could not find an example in the book about sqrt except for example 1 which showed g(x)=sqrt x which is showing it is raised to the 1/2 power so that is how i came up with my answer

.................................................

......!!!!!!!!...................................

00:16:01 For each zero does the graph touch or cross the x axis?

......!!!!!!!!...................................

RESPONSE --> the graph crosses the x axis -sqrt3 and touches at 2

.................................................

......!!!!!!!!...................................

00:16:11 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.

......!!!!!!!!...................................

RESPONSE -->

.................................................

......!!!!!!!!...................................

00:18:32 What power function does the graph of f resemble for large values of | x | ?

......!!!!!!!!...................................

RESPONSE --> function resembles y = x^6 for large values |x|

.................................................

......!!!!!!!!...................................

00:18:38 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. **

......!!!!!!!!...................................

RESPONSE --> ok

.................................................

......!!!!!!!!...................................

00:47:12 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.

......!!!!!!!!...................................

RESPONSE --> zeros are=1with multiplicity of 3 behavior of function=resembles that of the power function y=x^3 for large values of |x|

.................................................

......!!!!!!!!...................................

00:50:59 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.

......!!!!!!!!...................................

RESPONSE --> ok i see i do understand the zeros but the behaviour is still a little confusing

.................................................

......!!!!!!!!...................................

00:53:32 What is the maximum number of turning points on the graph of f?

......!!!!!!!!...................................

RESPONSE --> the maximum # of turning points= n-1 3-1=2

.................................................

......!!!!!!!!...................................

00:55:08 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.

......!!!!!!!!...................................

RESPONSE --> i know exactly what i have done was to take the 5x(x-1)^3 exponent of 3 not the 5x^4 exponent and minus one i do understand

.................................................

......!!!!!!!!...................................

00:59:16 Give the intervals on which the graph of f is above and below the x-axis

......!!!!!!!!...................................

RESPONSE --> (-infinity, 0)(0,1)(1,infinity)

.................................................

......!!!!!!!!...................................

01:00:53 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).

......!!!!!!!!...................................

RESPONSE --> (-infinity, 0)(0,1)(1,infinity) i believe i read the question wrong or the answer given is way into detail

You give the right intervals but don't specify whether the graph is above or below the axis on each.

.................................................