Wednesday, October 27, 2010

4.2: Trigonometry



Okay, so when theres like a right triangle, its got 6 different trigonometric things, these are called:



Sin (oposite leg / hypotenuse)



Cosine (adjacent leg / hypotenuse)



Tangent (opposite leg / adjacent leg)
(also simply sin / cos)



Each of these has a reciprocal:

cosecant = 1 / sin

Secant = 1 / cosine

Cotangent = 1 / tangent


These are reciprocals NOT inverses


On the unit circle, the sin = y value and the cos = x value


a trick to remeber: every 15 degrees = pie / 12
examples:
  • 30 degrees = 2 pi / 12 = pi / 6
  • 45 degrees = 3 pi / 12 = pi / 4
  • 90 degrees = 6 pi / 12 = pi / 2

Tuesday, October 19, 2010

2.6: Rational Functions

Rational Functions


A rational function is a function that can be written as


Where N(x) and D(x) are polynomials.
  • Example of a rational function:

A function likeis considered a rational function, because it can be rewritten as

*NOTE: When you divide by a fraction, the answer gets larger.
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Vertical Asymptotes
Consider the equation .

The denominator cannot be 0, so x ≠ 3.


 Notice that the graph avoids x = 3, and

There can be more than one vertical asymptote, as is demonstrated in the equation, because x = ± 1.

 
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Horizontal Asymptotes
Unlike vertical asymptotes, there can only be one horizontal asymptote.

In order to determine what the horizontal asymptote is
  • Pretend that x is a large number like 1,000,000,000
  • Consider the highest degree in the numerator and denominator
Because x is so large, everything but the highest degree polynomial is irrelevant. There are three different horizontal asymptote scenarios.

1. Higher degree polynomial in the denominator
 
 Look only at the highest degree polynomials, 3x and 2x2

, so the H.A. is 0.

2. Highest degree polynomial same in both numerator and denominator

Look at only the highest degree polynomials, 3x2 and 2x2. The x2s cancel each other out, and , so the H.A. is 3/2.

3. Higher degree polynomial in the numerator

It does not level off, so and there is NO H.A.


 Still confused? CLICK HERE to watch a video explaining rational function asymptotes.

Wednesday, October 13, 2010

Section 2.4 Complex Numbers




A complex number is the combination of a real number
and an imaginary number.

Adding Complex Numbers



Subtracting Complex Numbers




Multiplying Complex Numbers



Dividing Complex Numbers

eq=\frac{2+3i}{4-2i}= \frac{2+3i}{4-2i}(\frac{4+2i}{4+2i}
=\frac{8+4i+12i+6i^2 }{16-4i^2 }=
\frac{8+16i+(6)(-1)}{16-4(-1)}=
\frac{2+16i}{20}=\frac{1}{10}+\frac{4i}{5
}

Conjugate
use when there is an imaginary number in the denominator.
2+3i    the conjugate is 2-3i




Monday, October 11, 2010

Solutions to review

Answer key

Don't forget to check out Assignment 4 on your sheet for a nice overview of the kinds of problems you should be able to do.

Sunday, October 10, 2010

Well, here ya go.

2.3: Real Zeros of Polynomial Functions

To find the real zeros of polynomial functions, one can use both long division as well as synthetic division.

Here is an example of how to use long division:

Divide f(x) = 6x^3 - 19x^2 + 16x - 4 by x - 2.

 |  | \nx | - | 2 |  | 6 x^2 | - | 7 x | + | 2\n6 x^3 | - | 19 x^2 | + | 16 x | - | 4\n6 x^3 | - | 12 x^2 |  |  |  | \n |  | -7 x^2 | + | 16 x |  | \n |  | -7 x^2 | + | 14 x |  | \n |  |  |  | 2 x | - | 4\n |  |  |  | 2 x | - | 4\n |  |  |  |  |  | 0

Now you know that both x - 2, 2x - 1, and 3x - 2 are the factors of f(x) (2x - 1 and 3x - 2 being the factors of 6x^2 - 7x + 2). You then set these values equal to zero and solve for x, giving you the zeros of the f(x) equation.

Here is a video that will better explain long division of polynomials:

Here is a video that will go through the process of synthetic division of polynomials.

Note 1: The end of the video about synthetic division also goes over the remainder theorem at the end. The remainder theorem reads as follows:

If a polynomial f(x) is divided by x - k, the remainder is r = f(k).

Note 2: When using synthetic division with polynomials that have 0 as some of their coefficients, you must include the 0 in the problem in order to ensure you get the right answer.

The Rational Zero Test

If the polynomial f(x) = ax^n + ax^n-1 + ... + ax^2 + ax + a has integer coefficients, every rational zero of f has the form:

rational zero = p/q

where p and q have no common factors other than 1, p is a factor of the constant term, and q is a factor of the leading coefficient a.


For your enjoyment:

Wednesday, October 6, 2010

PhD a.k.a. A Higher Degree

The graph of a polynomial has no breaks, holes, or gaps.

Pretending like i didnt use paint and my lines were actually acurate, we can see what i mean.  The first graph is good cause it has no breaks, the second is not a polynomial because it has breaks.  The third is "theoretically" good cause it has no sharp turns unlike the 4th, which has many sharp turns.

THE LEADING COEFFICIENT








Now, we examine End Behavior.  End behavior is where the right and left ends of a polynomial graph points too.  In the first of the two graphs above, the two ends point in different directions, this lets us know that the leading coefficient is negative.  In the second graph, the two ends are pointed in the same direction, thus, the leading coefficient is positive.

The notation we use to show the first graph's end behavior is:
for the left end and

               Those last 6 lines took me just about 45 minutes
Now we can apply the things we have learned to some equations.
Because the degree is odd and the leading coefficient is negative, the graph rises to the left and falls to the right. Like this:
It can be shown that for a polynomial function f of degree n, the following statements are true.

1.  The graph of f has at most n real zeros.
2.  The function f has at most n-1 relative extrema (relative minimums or maximums).