Friday's Chapter 15 Test Topics

Here’s a list of topics for Friday’s test. Can you believe it’s our last one?

Precalculus Chapter 15 Test Topics:
Identify the degree, number of real and complex zeros and the leading coefficient of a polynomial from its graph
Sketch the graph of a rational function.
Identify transformations of functions.
Classify discontinuities.
Simplify rational functions.
Determine the zeros, the sum of the zeros, the product of the zeros, the sum of the pairwise product of the zeros, and a possible equation from the graph of a cubic function.
Sketch the graph of a given polynomial.
Identify the zeros of the polynomial.
Factor the polynomial.
Prove that a quadratic has no real zeros.
Show that a value is a zero of a polynomial.
Find zeros of a polynomial.
Discuss the implications of nonreal zeros.
Determine an average rate of change.
Provide a formula for an average rate of change.
Determine an instantaneous rate of change.


13 questions on the non-calculator portion, 11 on the calculator portion. That’s it! I’ll be around after school on Thursday afternoon, and in early on Friday.

Mathematicians are like Frenchmen: whatever you say to them, they translate it into their own language, and forthwith it means something entirely different.
- Johann Wolfgang von Goethe

Is this me?

Here’s a problem to consider:
Driving problem: Hezzy Tate drives through an intersection. At time t = 2 sec she crosses the stripe at the beginning of the intersection. She slows down a bit, but does not stop, and then speeds up again. Hezzy is good at mathematics, and she figures that her displacement, , in feet, from the first stripe is given by

Use synthetic substitution to show that is a zero of d(t).
Use the results of the synthetic substitution and the quadratic formula to find the other two zeros of d(t).
How do the zeros of confirm the fact that Hezzy does not stop and go back across the stripe?
What is Hezzy’s average velocity from t = 3 to 3.01 sec?
Write the equation for the rational algebraic function equal to Hezzy’s average velocity from 3 sec to t sec.
By appropriate simplification of the fraction in Problem 16, calculate Hezzy’s instantaneous velocity at time t = 3 .

15-5: Instantaneous Rate of Change of a Function: The Derivative

Chapter 15 Section 5:

Review:

So, this is the last section of the last chapter of the year! I think that it is appropriate that I am the last person to post a blog seeing as I was the first person as well.

This section is our introduction to Calculus. Calculus is all about the rate of change and through this section we are figuring out how to predict it.

Now something to remember:

Instantaneous rate of change = Derivative = Slope of a tangent line

for the point x sub 0

Definitions:

Derivative= instantaneous rate, velocity

--derivative of the time-height function

Average rate of change of r(x) of function f(x) on an interval starting at x=c is the change in the y-value of the function divided by the corresponding change in the y-value of the function divided by the corresponding change in the x-value.

Instantaneous rate of change of f(x) at x=c is called the derivative and is denoted f '(x) "f prime of x." It is equal to the limit of the average rate as x approaches c.

The value of the derivative of f(x) at x=c equals the slope of the tangent line to the graph of f at x=c.

Example Problem:

Example 1:

Given: f(x)= x^3 + 2x + 3 what is the instantaneous rate of change of f(x) at x=1?

Remember, first you plug in 1 to f(x), so you have 1^3 + 2(1)+3 = 6

f(x)= f(x)-6/(x-1) = ((x^3 + 2x + 3)-6)/(x-1)

Now, use synthetic substitution

1 (pretend it is in the box thing) ___ 1 __ 0 __ 2__-3

______________________________________1 __ 1 __ 3

________________________________------------------------

________________________________1___1___ 3 __0

(sorry if all of the marks above confused you, but when I tried to publish the post, the computer automatically pushed all of the numbers to the left and it just looked like a jumble of numbers)

Yay! We got zero! which means that it is removable~

so:

((x-1)(x^2 + x + 3))/(x-1) the x-1 cancel out so you are left with

x^2 + x + 3. If you plug in 1 for x you get 1^2 + 1 + 3 = 5

So 5 = the Instantaneous rate of change = derivative = slope!

Now, use this information to create the equation for the line. From the information in the problem, you know the slope (= the Instantaneous rate of change = derivative) which is 5 and you also have a point, (1, 6) and as Mr. French said, it is always better to use point slope form because it is so easy!

So, y - 6 = 5 (x - 2).

Example 2:

You are at Six Flags and you are riding a roller coaster. On your favorite ride, there is a dip that has the equation y= x^4 + 5 where the x-axis is the ground and the y-axis cuts the dip in half. You are such a good Precalculus student that you want to use the knowledge that you learned in class to calculate your average velocity between 1 and 1.5 seconds after you reach the minimum of the dip.

Given this information, you know only your x values, 1 and 1.5. To find the y-values, plug the information into the equation. When you plug in 1, you get 6 and when you plug in 1.5, you get 10.0625. Using this information, you can calculate the average velocity. Your points are (1, 6) and (1.5, 10.0625). (10.0625 - 6)/ (1.5-1) = (4.0625/.5) = 8.125 = the average velocity!

For some extra help visit:

http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/tutorials/frames2_1.html
It is actually pretty cool! If you click on "previous tutorial" and "next tutorial" in the left hand corner there is more information about what this section covers.

This is a picture of me, Juan, and Alex. Juan Alderete de la Pena is the base player from my favorite band, The Mars Volta. I think Henry may have told this story on a previous blog, but I will tell you again anyway. So, ok, we were going to the Detour Music Festival in LA and we took the metro there and when were were in Union Station, we saw him and were like "Juan" and he turned around. It was amazing. We took the metro with Juan and walked down the streets of LA with him looking for the entrance. I couldn't believe that night; it was awesome! Anyway, we became buds and it was pretty amazing.

Reminder:

Oh yeah!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! No more blogs!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

15.4 - Rational Functions: Discontinuities, Limits, Partial Fractions

Hey everybody!

to update from Anna's post: two and a half weeks until summer! yay, the greatest moment ever!

down to business...when you divide a polynomial by a linear factor, the result usually resembles a quadratic function:

f(x) = (x^3 - 5x^2 + 8x - 6)/(x-3)

when you use synthetic substitution:
3} 1 -5 8 -6
3 -6 6
1 -2 2 0 is the answer,
so the net equation is x^2 - 2x + 2. there is a discontinuity at x = 3. this means that the answer is undefined because when you plug 3 in for x, the denominator is equal to 0. you can use this later for finding limits.The dot above the 3 on the x axis is the representation of the discontinuity.

LIMITS! If you want to find the y value that the discontinuity should would be, plug the equation into your calculator. then go to TBLSET, change TBLstart to a few decimal places before the desired number, and change the scale to 0.1, or some very small number. then go to TABLE and find the y values for 2.99 and 3.01. it should be obvious what the missing number is. in this case it is 5.

lim f(x) = 5

you can also find this algebraically by plugging 3 back into the net equation:
x^2 - 2x + 2
3^2 - 2(3) + 2
9-6+2 = 5

You can remove the discontinuity algebraically by simplifying the following:

(x-3)(x^2-2x +2)/(x-3) you can cancel the two (x-3)'s.

there are some equations that have more complex discontinuities that are not removable:

with the equation g(x) = (x^3 - 5x^2 + 8x - 5)/(x-3) there is an asymptote at x=3.
you can use synthetic substitution to find the factored form:

3} 1 -5 8 -5
3 -6 6
1 -2 2 1 is the answer, so it has a remainder. the net equation is therefore x^2-2x+2+1/(x-3) and the graph...


You can't factor this type of problem so it has an infinite or irremovable discontinuity.
the limit of g:
lim g(x) = does not exist or, in some cases, is infinity

PARTIAL FRACTIONS: polynomial/polynomial = partial fraction

(9x-7)/(x^2-x-6) = ....

you need to factor the denominator to (x-3)(x+2). you can represent the numerator with variables, A and B.

A/(x-3) + B/(x+2) --> give them equal denominators by multiplying the first value by (x+2) and the second by (x-3).

A(x+2)/(x-3)(x+2) + B(x-3)/(x-3)(x+2) --> distribute the A and B values.

(Ax +2A +Bx+ 2B)/(x-3)(x+2) = (9x-7)/(x^2-x-6)

you know there are 9x's and that you have A + B x's, so A+B=9
you also know that... 2A-3B = -7,
so you can use matrices or substitution to solve for A and B. they equal 4 and 5, respectively.

a shorter way to do this is by making the factors in the denominator equal zero...like this:
(9x-7)/(x-3)(x+2)
plug in 3 for x... (27-7)/(3+2) --> 20/5 --> 4
plug in -2 for x... (-18-7)/(-2-3) --> -25/-5 --> 5
plug those values back in to get the partial fraction:

4/(x-3) + 5/(x+2) and that is the answer.

Examples!

What is the partial fraction for (4x-14)/(x^2 + 2x - 8)

first, factor the denominator: (x-2)(x+4).

then, set up two fractions: A/(x-2) + B/(x+4).
make equal denominators: A(x+4)/(x-2)(x+4) + B(x-2)/(x-2)(x+4)
distribute: (Ax + 4A +Bx - 2B) /(x-2)(x+4) = 4x-14
A + B = 4 and 4A - 2B = -14
solve with matrices or substitution.
A = 1, B = 5

or...

this way is better for speed.
(4x-14)/(x-2)(x+4)
plug in 2 for x to find A value, then plug in -4 to find B.
A = -6/6 --> -1
B = -30/-6 --> 5

the partial fraction is 1/(x-2) + 5/(x+4)

LINKS:

here is some extra help for...
partial fractions: http://www.sosmath.com/algebra/pfrac/pfrac.html
discontinuities: http://www.tpub.com/math2/39.htm
limits: http://en.wikipedia.org/wiki/Limit_(mathematics)

Reminder: Mariclare is next

this is good fun: it's german AND stop motion! two of my fave things!
http://youtube.com/watch?v=4RST7CYjfjo
i don't know what he's saying though. i could be exposing you to crude language without knowing it... :(

i hope all of this helped!

Thursday's Quiz Topics

Now that we’re through the worst of the AP season, we can finally get back on track (after the Junior retreat on Friday, anyway)! Here’s a list of topics for Thursday’s quiz:

As a reminder, your Weekly Challenge will be due on Monday, and the test for Chapter 15 will be next Friday.

Precalculus Quiz 15.1-3 Topics
Given a graph of a function:
Determine the degree
Determine the leading coefficient
Determine the number of real and complex (nonreal) zeros
Identify an extreme point and a point of inflection.
Given a cubic function:
Determine the product of the zeros
Determine the sum of the pairwise product of the zeros
Determine the sum of the zeros
Perform synthetic substitution.
Interpret the results of your synthetic substitution.
Given the factored form of a cubic, determine the zeros of the function.
Given data for a cubic function:
Prove the constant third difference property
Determine the equation for the function algebraically (matrices)
Verify the equation with regression techniques
Analyze the equation
Given partial information about the zeros of a cubic function:
Determine the remaining zeros
Determine the product of the zeros
Determine the sum of the pairwise product of the zeros
Determine the sum of the zeros
Determine an equation for the function
Analyze a function (word problem!)

That’s it! The format’s the same as always – ½ Non-calculator, ½ Calculator. There are 20 questions on the Non-calculator portion and 20 questions on the Calculator section (all worth 1 point each). If you know your stuff, you should be able to finish quickly. I’ll be around after school on Wednesday (today), and on campus by 7:00 AM on Thursday.

"I never did very well in math - I could never seem to persuade the teacher that I hadn't meant my answers literally."
- Calvin Trillin

Trashing math and country music - ouch!

Friday's Quest Topics

Here’s a list of topics for Friday’s test:

Precalculus Chapter 14 Test Topics:
Find terms – arithmetic sequence (determine pattern)
Find terms – geometric sequence (determine pattern)
Definitions of sequences and series
Find terms – arithmetic series (use information)
Determine type of sequence and justify
Determine terms of a sequence – unknown type
Determine terms of a series – unknown type
Explain your method of determination/pattern of series
Determine terms of a sequence (determine pattern)
Determine terms of a series (given pattern)
Determine value of n for given term in a series/sequence
Determine term(s) of a binomial expansion
Word problem – analysis of arithmetic and geometric sequences
Arithmetic series – determine formula, apply formula
Geometric series – determine formula, apply formula
Partial sums of arithmetic and geometric series
Infinite sums of geometric series
Sigma notation

11 questions on the non-calculator portion, 14 on the calculator portion. Since this chapter was so short, each question will be worth 3 points instead of 5, making this a “quest” instead of a test. That’s it! I’ll be around after school on Thursday afternoon, and in early on Friday.

"Knowledge is of no value unless you put it into practice."
- Anton Chekhov

This strategy won't work on me:

14.2 Arithmetic, Geometric, and other Sequences

Hey Guys! I hope everyone is coping sorta okay with AP's coming up and everything. In 5 weeks it will be summer!

Ok so lets define some terms first:

• Sequence is a function where the domain is a set of positive integers. The independent variable is n and the dependent is tn.
• A Recursion formula specifies tn as a function of the preceding term (tn-1)
• A Explicit formula of a sequence specifies tn as a function of n.

Arithemetic Sequence is where you add a constant to the previous term to get the next term. The constant added is called the common difference. (This can be adding a negative number)

Geometric Squence is where you multiply the previous term by a constant (the common ratio) to get to the next term. (you can multiply by a fraction which is basically dividing)

Formulas:

It is good to use recursive just to find the next few terms but for finding terms later in the sequence use the explicit formula. Also, a sequence can have a finite or an infinite number of terms depending on whether its domain is finite or infinite.

Arithmetic Linear Function:

• Explicit: Tn = To + D(n-1)
• Recursive: Tn = Tn-1 + D

so To is the initial value in the sequence, R is the common difference, Tn is a value in the sequence, and Tn-1 is the previous term.

Geometric Exponential Fution:

• Explicit: Tn = To * R^(n-1)
• Recursive: Tn = Tn-1*R

so To is the initial value in the sequence, R is the common ratio, Tn is a value in the sequence, and Tn-1 is the previous term.

Calculator!

• Make sure you are in sequential mode.
• go to y=
• nMin=1 (you will usually want to leave this =1)
• u(n)= this is where you put your equation (u(n) is the same as t(n))
• to get u go to 2nd 7 and for the n press the regular x button.
• for u(nMin)= put the first term of the sequence
• once you plug in your equation you can go to table set to find values for any term. Go to table set to start at a specific value.
• you can also graph additional sequences under (v) and (w) (the same way) in case you want to compare the sequences.

Examples!

Given the following problem how would you plug it into the calculator (i had a hard time with this so i am giving it as a example problem.)

You put $1000 into the bank with 6% interest a year. What are the steps for putting this scenario into the calculator.

1. seq mode
2. y=
3. then under nMin=1 (leave this alone)
4. u(n)= u(n-1)(1.06)
5. u(nMin)= 1000 (which is the starting value)

Now if the problem asked for a different term (for example the 15th). You can either go to the table or since you have the equation entered you can put in u(the term) on the home screen and press enter and get the answer.

Example Problem 2

For the following problem:

• Tell what kind of sequence it is
• Write the next 2 terms
• Find t100
• Find the term number of the last given term
  

2. 2, 4, 8...32
Ok so this is an geometric sequence where the common ratio is 2. So to get to the next term you multiply the previous term by 2. The next two terms are 16 and 32. Now to find t100 you can figure out the explicit formula and either plug in 100 for the n value or you can plug in the formula into y= and go to the table at 100. The explicit formula formula for this sequence is Tn=2(2) ^(n-1) . T100 = 1.267 *10^30. Now to find the term number of the last given term you can either look for the term in the table or you can plug it in for Tn in the formula.
32= 2(2) ^(n-1)

16=(2) ^(n-1)

log 16 =( n-1) log 2

n= the 5th term

Additional Help:

http://home.alltel.net/okrebs/page131.html or http://www.purplemath.com/modules/series3.htm

this site is also kinda cool :http://www.explorelearning.com/index.cfm?method=cResource.dspDetail&ResourceID=340

Reminder: Natalie you are next!

Ok so...our dance teacher told us to watch this clip. It's CRAZY. This woman is balancing and dancing in pointe shoes on this guy's head. Absolutely amazing (and crazy!) You have to watch it for a minute before she dances on his head...

http://www.youtube.com/watch?v=8uaVXmWEN-c

14-3: Series and Partial Sums

Okay, I'm sorry if this is really confusing, but I didn't really get it today in class. (ESPECIALLY not the binomial part)

The difference between a series and a sequence:
A sequence is just a list of numbers like 2,4,6,8
A series is different because it's all being added together in a string like 2+4+6+8.

Mr. Frost taught us that to find the sum of a series, you can pair up the different numbers in the series. So if you had the series 3,7,11,15,19,23,27,31,35,39 you could pair it up like:
3 + 39 = 42
7 + 35 = 42
11 + 31 = 42
15 + 27 = 42
19 + 23 = 42

42 x 5 = 210 (the complete sum for the series)
The equation for the partial sum is

S(n) = (t(1) + t(n))(n/2) -- This is for an ARITHMATIC series

The geometric partial sum equation is:

S(n) = t(1) x (1-r^n)/(1-r)

If the absolute value of r is less than 1, however, the n value will only get bigger and bigger, and this will only make the r^n value in the formula smaller and smaller until it eventually doesn't even matter anymore. So, if the absolute value of r is less than 1, the formula for the geometric partial sum is:

S(n) = t(1)/(1-r)

A geometric series will converge to a limit if the common ratio r satisfies the inequatlity "the absolute value of r is less than 1". If the absolute value of r is greater or equal to 1, the series diverges.

Binomial Series
-coefficients of a binomial series can be calculated recursively using the term before it.

(coefficient of a)(exponent of a)/(term number) = coefficient of next term

-coefficients can also be calculuated algebraically. They are equal to the numbers of combinations of n objects taken r at a time. Like in the probability section:

5C2 = 5!/3!2! = 5x4x3x2x1/(3x2x1)(2x1) = 10, so the term containing b^2 is
5!/3!2!a^3b^2

When expanding a binomial (a + b)^n the term with b^r is
(n!/(n-r)!r!)a^(n-r)b^r = (n/r)a^(n-r)b^r = nCra^(n-r)b^r

Practice Problem:
Using the series 7+12+17+22+27+32, find the 6th partial sum.
7 + 32 = 39
12 + 27 = 39
17 + 22 = 39

39 x 3 = 117 you can do it this way OR you can do it the equation way

S(n) = (t(1) + t(n))(n/2)
S(6) = (7 + 32)(3)
S(6) = (39)(3) - which is the same thing from before!
S(6) = 117

Extra Help: go to http://home.alltel.net/okrebs/page133.html

Reminder: Landry! You're the next BLOGMASTER


umm... i hope this posted. it's all just html in this format. This is a picture of one of Picasso's most famous blue period paintings. Also my favorite of his blue period.