Rational Zero theorem

If P(x)= the anxn+an-1 xn-1 dah dah dah a1x a0 has integer coefficients (an cannot be zero)

and p/q is a rational zero in lowest terms of P then


p is a factor of the constant term and

q is a factor of the leading coefficient

Guidelines for FInding the zeros of a polynomial function with integer coefficients

1. Gather general information. Determine the degree N of the polynomial function. The number of the distinct zeros of the polynomial function is at most N
Apply Descartes' Rule of Signs to find the possible number of positive zeros and also the possible number of negative zeros

2. Check suspects. Apply the Rational Zero Theorem to list rational numbers that are possible zeros. Use syn div to test the numbers in your list. If you find an upper or lower bound then eliminate from your list any number that is greater than the upper bound or less than the lower bound

3 Work with reduced polynomials. each time a s zero is found, you obtain a reduced polynomial

if a reduced polynomial is of degree 2, find its zeros either by factoring or by applying the quadratic formula.

if the degree of a reduced polynomial is 3 or greater repeat the above steps for this polynomial.

Descartes' Rule of Signs

Let P be be a polynomial function with real coefficients and with the terms arranged in order of decreasing powers of x

1. The number of positive real zeros of P is equal to the number of variations in the sign of P(x), or to that number decreased b an even integer.

2. The number of negative real zeros of P is equal to t he n umber of variations in sign of P(-x), or to that number decreased by an even integer.

Properties of Exponential and Logarithimic functions

Quiz 4

Exponential Functions

f(x)=b to the x power

where b>0

b cannot = 1

b is the base


x is any real number

Inverse Functions

To find the inverse of a function you interchange x for y or in other words (6,0) = (0,6) in inverse

essentially what you are doing is taking the domain y and switching it with the range x therefore an inverse function has some of these properties

The domain of f(x) = f(x)^-1 range and vice versa

this causes the x intercepts and y intercepts to be switched along with Horizontal and Vertical asymptotes to be switched.

Functions

A function is a set of ordered pairs in which no two ordered pairs have the same first values

this can be determined by a vertical line test.

Domain is the "first value"

The range is the plug in of the first value being evaluated by the function

Notation

y=f(x)

f(x)=2x+1

f(1)=2+1= 3

ordered pair is then (1,3)


1 to 1 function No two ordered pairs have the same 2nd value or range?

this can be determined by a horizontal line test.

if always increasing or decreasing a function satisfies the 1 to 1 definition of a function

Properties of Real Numbers

Let a,b, and c be real numbers



Closure a+b is a unique real number ab is a unique real number

Commutative a+b=b+a ab=ba

Associative (a+b)+c=a+(b+c) (ab)c=a(bc)

Identity There exists a unique real There exists a unique real
number 0 such that number 1 such that
a+0=0+a=a a*1==1*a=a

Inverse For each real number a, For each nonzero real
there is a unique real number a, there is a number -a such that unique real number 1/a
a+(-a)=(-a)+a=0 such that
a*1/a=1/a*a=1

Distributive a(b+c)=ab=ac

Holes

Holes are in a graph if the restricted values in the domain of the function are not restricted in a function equal to that function.

i.e.

so the first function has a restricted value of 2 but the function equal to it does not have a restriction at 2 there would be a hole there not an asymptote

Asymptotes

First you need to find the domain of the function. You do this by looking at the denominator of the function and then factoring it out. You will then look for real zeros. Non real zeros do not affect the domain since they are not on the real number line

After you find the domain of the function you should find the x value of the problem. you find the x value by factoring out the numerator and then you will find your x values by comparing them, or zeroing them.

I.E. (x-5)(x+1) To zero these you would set one of the x's= to (5-5=0) (-1+1=0)



find your y value by setting x to zero.

Then you need to find your Vertical Asymptote. You find this by using the the theorem of vertical asymptotes which states

If the real number @ is a zero of the denominator then the graph of the function
p/q where p and q have no common factors has the vertical asymptote of

x=@

Then find the Horizontal Asymptotes by using this theorem

a function with unlimited degree on numerator and denominator both degrading to a 0 degree(a real number usually) ie

be a rational function with the numerator of degree m and denominator of degree n

1.if the degree of of the numerator is less than the degree of the denominator then the x axis is the horizontal asymptote of graph F

2. If numerator is equal to the denominator then the line given by divide the numerator coefficient by the denominators coefficient

3. if the numerator is greater than the denominator there is no horizontal asymptote but

there is the possibility of having a slant asymptote if degree numerator is = degree denominator +1

remember they cannot have any common factors for a slant asymptote

Vertex of a parabola

-2b/a = h

4ac-b squared
________________
4a

=k

h,k is the vertex

Sequences

a_n=2n=2,4,6,8,10,12,14,16,18...

A series

(4i - 5) = - 1 + 3 + 7 + 11 + 15 + 19 = 54.

5(k + 2) = 5(k + 2) = 5[6 + 7 + 8 + 9] = 150.

functions

F(x) where x is the variable of one of the numbers within the domain. A domain is a number set. Domain is the input and range is the output (result) of the function.

For example you would have a function that goes like this.

f(x)=2x+20

your domain list would be (this would represent x)

X=1,3,5,7,9,11

Your range (the output) for this functions domain would be

f(1)=22
f(3)=26
f(5)=30
f(7)=34
f(9)=38
f(11)=42

Precalculus assessement

Today I was taking an assessment test and realized something... I did not remember functions! F(X) had no meaning to me and thus was my predicament. I decided that because of this I was going to start this blog and put down information that I would find personally useful later on.