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**precalculus** - Intermediate Algebra (Factoring, solving, graphing)
- Trigonometry (Measuring circles, angles, and triangles)
- Functions and Graphs (Linear, trig, exponential, log functions)
- (c) 1997-2001 Dan Bach and B & L Math Enterprises; all rights reserved. Download for personal use only.

**Factoring Polynomials**[ top of page ] (11/97)

You can use the distributive law to see that 3(4n + 5) = 12n + 15, and you can use FOIL to see that (n - 3)(n - 5) = n^2 - 8n + 15. But how can you start with the answer and find the factors? It's like Jeopardy: I say, "n^2 - 8n + 15," and you respond (in the form of a question), "What is (n - 3)(n - 5), Alex, uh, I mean Dan?.

- Here's a way to look at the first problem:

- 12n + 15
- = 3(4n) + 3(5)
- = 3(4n + 5) .
- This is called "factoring out a common factor."
- What about n^2 -
8n+ 15 ?- This can be done (factored) by finding two numbers
- whose sum is 8 and product is 15: use 3 and 5. So we get:

- n^2 -
8n+ 15- = n^2 -
3n-5n+ 15 . . . . now factor common factors in pairs:- = n
(n-3)- 5(n-3). . . watch the signs!- = (n - 5)
(n-3). . . the (n - 3) was the common factor!- Try to factor x^2 + 37x + 100.
- On cherche deux chiffres-- oh, pardon, en anglais:
- We need two numbers whose product is 100 and sum is 37.
100 = (100)(1); 100 + 1 = 101 100 = (50)(2) ; 50 + 2 = 52 100 = (25)(4) ; 25 + 4 = 29 100 = (20)(5) ; 20 + 5 = 25 100 = (10)(10); 10 + 10 = 20.

- It seems that 37 never came up as a sum, so x^2 + 37x + 100 doesn't factor.
- But do you see how you'd factor x^2 + 29x + 100 ?
- Cleverly hidden answer: (x + 25)(x + 4).
- [ intermed algebra | top of page ]

Solving Quadratic Equations[ top of page ] (5/98)Recall a linear equation is one that looks like ax + b = cx + d, and our strategy was to get all x terms on the left, all constants on the right, then divide by the coefficient on x to solve.

A

quadratic equationhas an x^2 (x-squared) term; "quadrat" is Latin for square. This web site is being written on a Macintosh Quadra; the word is for the "040" processor chip in the computer.The general quadratic equation looks like

a x^2 + b x + c = 0, . . . . where a /= 0.If we want to find the x or x's that work, we might guess and substitute and hope we get lucky, or we might try one of these three methods:

- Hey, let's solve the same equation three times, using each method!

Example (Factoring):When the left hand side factors, we can use the "zero product

- property" which says if a product is zero, one or more of the factors hasta be zero:

x^2-8x + 15 = 0. . . . . . . . .our given equation(x-3)(x-5) = 0. . . . . . . . . factor the left sidex-3 = 0. .or. .x-5 = 0. . the "zero product property"x = 3. .or. .x = 5. . . . . . . solving the two "little equations"- [ intermed algebra | top of page ]

Example (Completing the square):

- The idea is that: "stuff squared equals a number" is easy to solve, using
square roots.- I'll use -/(n) for square root of n, meaning the positive number whose square is n.
For example, -/49 = 7 because 7^2 = 49.

Also, -/2 is about 1.414, because 1.414^2 = 1.999396, which is close to 2. But you'll never hit -/2 exactly by squaring a fraction (or terminating decimal), the square root of 2 is an "irrational number", meaning its decimal equivalent goes on forever:

-/2 = 1.41421356237309504880168872420969807856967187537695 . . .

Now let's solve x^2 - 8x + 15 = 0 using the square root method. It will take four levels of trickiness to get there:

Level 1: Solve for x in the equation . . x^2 = 49.

x^2 = 49. . . . . . . . . our given equation- -
/(x^2) =-/49. . . . . . take square root of both sidesx = + 7 or-7. . . . . there are two numbers whose square is 49.

Level 2: Solve for x in the equation . . x^2 = 17.

x^2 = 17. . . . . . . . . . . . our given equation- -
/(x^2) =-/17. . . . . . . . . take square root of both sidesx = +-/17 or--/17. . . .there are two numbers whose square is 17.- . . . . . . . . . . . . . . . . . . . . we can leave the answer with square roots
- [ intermed algebra | top of page ]

Level 3: Solve for x in the equation . . (x + 6)^2 = 5.

(x + 6)^2 = 5. . . . . . . . . . . . our given equationx + 6 =-/5. or . - -/5. . . . . . take square root of both sidesx =-6 +-/5. or . -6- -/5. . . subtracting 6 from both sides

Level 4: Solve for x in the equation . . x^2 - 8x + 15 = 0.

x^2-8x + 15 = 0. . . . . . . . what perfect square starts with x^2 - 8x ?x^2-8x + 16-1 = 0. . . . . . take square root of both sides(x - 4)^2 = 1x - 4 = 1 or - 1x = 1 + 4 = 5 . or . x = - 1 + 4 = 3. .add 4 to both sides.So we ended up with the same solutions as before. Whew!

- [ intermed algebra | top of page ]

Example (The Quadratic Formula):There's a formula for solving a general quadratic

- equation like a x^2 + b x + c = 0 ; it's called the
quadratic formula.- It gives the possible values for x in terms of a, b, and c.
You can derive it by completeing the square; the result is this: . . . [ top of page ]

If . . a x^2 + b x + c = 0 . . (*) . then either

x = (- b +(b^2 - 4 a c)) / (2 a) . . or

x = (- b -(b^2 - 4 a c)) / (2 a) .

- The quantity D = (b^2 - 4 a c) is called the
discriminantof the quadratic equation (*) . .- So
x = (- b(+or-)D) / (2 a) .

If

D < 0the equation (*) hasno real solutions, because we can't do a real square root of a negative number. However, there are twocomplexsolutions.If D = 0the equation hasone real(rational)solutionx = - b / (2 a) because adding or subtracting 0 has no effect.If D > 0then the equation hastwo real solutions.If D is a perfect square(of a whole number or fraction) then there aretwo rational solutionsfor x. This is when factoring will work. Trust me.

- Now
back to our original equation : x^2 - 8x + 15 = 0. In the above notation,- We have a = 1, b = -8, and c = 15. Using the quadratic formula gives us
x = (- b+/-(b^2 - 4 a c)) / (2 a). = (-(-8)+/-((-8)^2 - 4(1)(15))) / (2(1)). = (8 +/-(64 - 60)) / 2 = (8 +/- 4) / 2. = (8 + 2) / 2or(8 -2) / 2 ; so again x = 5 or x = 3. . . . <;-}

Graphing parabolas(See also functions and graphs.) (5/99)

- How can we draw a picture of y = x^2 ? As with any graph, we put a buncha dots
- at (x, x^2), and notice tha pattern: We can make a table and go from there.

xx^2-24-1100112439

- [ intermed algebra | top of page ]

Functions and Graphs[ top of page ]

- What's a function?
- The graph of a function
- Moving graphs around
- "Four-way" functions
- Trig functions and graphs
- Exponentials and logs

What's a function?[ top of page | funcs ] (5/99)Ok, I'm going out on a limb here, but I'd say the concept of "function" is THE most important in mathematics. Any arguments from math teachers or anyone?

A

functionis a way of dealing with an"input", applying some"rule"(the function), and then getting an"output".What's the catch? There can be at most one output for every input. The inputs that make "sense" form the

domainof the function, and the answers or outputs form therange.

Function notation: We can call the inputx, the rulef, and then the output isf(x).This DOES NOT mean "f times x" , it's just a notation device to record the input and output. For example, if f(x) = x^2, then f(3) = 3^2 = 9, not f times 3 (meaningless).

Think of f(x) = x^2 as

f( ) = ( )^2; that way you can safely plug in negative numbers or even other expressions: f(-5) = (-5)^2 = 25 ; f(x+h) = (x+h)^2.

The graph of a function[ top of page | beg alg | funcs ] (11/99)

Graphs are a visual way of displaying numerical information.

The graph of a function y = f(x) is the set of all points (x, y) with y = f(x),

so this means

put a dot at (x, f(x))for every x that makes sense (the domain).- Here are a few functions and their graphs (plotted in Mathematica):

y = 2 xy = x^2y = 2^xy = 2f(x) = 2 xf(x) = x^2f(x) = 2^xf(x) = 2(linear) (quadratic; power) (exponential) (constant)

Moving graphs around[ top of page | funcs ] (11/99)

- If you have the graph of y = f(x) you can
move it to the rightc units by graphingy = f(x - c).- For example, to move the parabola
- y = x^2 to the right by 3 units, you'd
- graph y = f(x - 3) or y = (x - 3)^2.
- To
move a graph up and down, you'd change the y-coordinates, so graphingy = x^2 + kputs the "vertex" at (0, k).- In general do
y = f(x) + k.- By doing a combination of both, we can move the graph anywhere in the plane.
y = f(x - c) + kor(x - c)^2 + k.- My "danimation" shows both effects-->--/

Click the picture for a 72K animation in a new window (graphs done in Mathematica)

Four-way functions[ top of page | funcs ] (11/99)Part of the "reform" movement in precalculus and calculus is to look at functions

four ways:

(1) Verbally, (2) Numerically, (3) Symbolically, (4) Graphically.

(1) "The squaring function"woulddescribewhat's being done to the input to get the output.

(2)Make atable of valuesfor the inputs and outputs:

xx^21.01.001.11.211.21.441.31.691.41.961.52.25

(3)Describe the outputalgebraicallyin terms of the input:f(x) = x^2- where f(x) is the function notation described above. I like f( ) = ( )^2.

(4)Draw agraphof y = f(x) on a pair of coordinate axes; here'sy = x^2.

Exponential and Logarithm Functions[ top of page | funcs ] (11/99)

- If you get a job in June that pays $1 the first day, $2 the second day, $3 the third day, etc, earning one more cent per day each day, you'd have a total for the month of
- 1 + 2 + 3 + . . . + 30 = 30(30+1)/2 = $465. This daily salary grows
linearly, and the running total (465 is a "triangular number") growsquadratically. (See above red graphs)- But what if your salary
doubledevery day for a month? Think about it first.- Ok, you'd have 1 + 2 + 4 + 8 + . . . + 2^29 = 2^30 - 1 = $1,073,741,823 (a billion bucks)
- (The 30th day is 2^29 because you start with 1 = 2^0)

This is called

exponentialgrowth, and you can see its rapid rise just for three days in the graph. Some more realistic examples are compound interest, bacteria or population growth, and spread of disease or rumors in a large group.(The graph of

y = 2^xappears at the right-->)

Simple and Compound Interest[ top of page | funcs ] (8/00)

- Say you deposit some money, initially
Pdollars (or your favorite unit of currency), into a bank.- The bank pays you "interest" to keep it there; it can pay either simple or compound interest.
Simple interestis simple to figure out; hence the name!- If you make, say, 8% interest for 5 years, that makes 40%.
- The total amount of money
Athat you'll have aftertyears at interest rateris given byA = P(1 + rt)- So if you put in $2000 for 5 years at 8% interest, you get
A= 2000(1 + (.08)(5)) = 2000(1+.40) = 2000(1.40) =$2,800.- Notice the factor of 1.40 means you've made a profit of 40%.
- A graph of your money A as a function of t would look like a straight line; you always make
- 8% of just $2000, or $160, each year, forever. The amount
Ais alinear functionoft.Compound interestis more comp-licated. The difference is that you make interest on the interest.- After 1 year your $2000 would grow by 8%, so multiply by 1.08 to include the original deposit:
A(1) = 2000 (1.08) = $2,160.But the second year you'd want to make interest on the $2160, not- just on the $2000. So the second year balance should be 2160(1.08) = $2332.80. But this equals
- 2000(1.08)(1.08) = 2000 (1.08)^2. After 5 years you'd have a total of:
A(5) = 2000 (1.08)^5= 2000 (1.4693280768) = $2938.65615 ~=~$2,938.65(the bank rounds down)- Note the factor of 1.469... means your effective profit is now
46.9%(compared to 40% for simple interest.)- The general formula (if interest is "
compounded yearly" like this) for your balance at timetyrs is:A = P(1 + r)^t. . . (where the ^ means "to the power of")- Since the variable
tis in the exponent, this timeAis anexponential functionoft., and your balance- would grow at an ever-increasing rate (dollars per year), but a constant percentage rate.
- The graph would be an exponential curve like the y = 2^x example above.
- If the interest is
compounded n times per year(n=4 for quarterly, for example), then this- means you get paid that fraction (1/n) of the interest n times per year.
- With your 8% annual rate, each quarter you'd gain 8%/4 = 2% of your money; so you multiply
- your balance by 1.02, four times a year.
- In
tyears this makes4tquarters (3-month periods); after 5 years you'd have 20 quarters, and so: . .- A(5) = 2000 (1.02)^20 = 2000 (1.485947396) ~=~
$2,971.89.This represents a 48.6% profit.- The general formula (if interest is "
compounded yearly" like this) for your balance at time t yrs is:A = P(1 + r/n)^(nt). . . (remember to do the r/n first!)- It seems to be better when the interest is compounded more often! Let's make a table of how your
- final balance after 5 years depends on n, the number of compounding periods per year.

n(1 + r/n)^(nt)== growth factorA = P*(growth)1 (yearly) (1 + .08/1)^(1*5) 1.08^5 1.46933 2938.65 2 (semiann) (1 + .08/2)^(2*5) 1.04^10 1.48024 2960.48 4 (quarterly) (1 + .08/4)^(4*5) 1.02^20 1.48595 2971.89 12 (monthly) (1 + .08/12)^(12*5) 1.006667^60 1.46933 2938.65 52 (weekly) (1 + .08/52)^(52*5) 1..001538^260 1.49137 2982.73 365 (daily) (1 + .08/365)^(365*5) 1..000219^1825 1.49176 2983.51 - Apparently the total grows, but not very much, as n --> infinity. In the limit, we say the
- interest is
compounded continuously, and the amount is given by the easy 'pert formula':A = Pe^(rt)....where e = base of natural logs; e ~~ 2.71828...- Going back one last time to our example, we just put in r = .08 and t = 5:
A = 2000 e^(.08 * 5)= 2000 e^0.4 = 2000 (1.49182) =$2,983.65.- This is the best we can do in 5 years with the given interest rate; you get a 49.18% profit;
- it's not much stronger than the weekly compounding (hehe).

Logarithms[ top of page | funcs ] (11/99)

- What are
logarithms? And why are they misunderstood? Here's an easy way to deal:LOGS ARE EXPONENTS.(I have a t-shirt with this slogan.)Example: You know2^3 = 8because 2^3 means 2 * 2 * 2. (3 factors.)- What's the base? 2. What's the exponent? 3. What's the answer? 8.
- So we record the exponent, 3, that you need to put on a basem of 2 to get an answer of 8.
- Write this as "
log,base 2, of 8, equals 3" because the exponent (the log) is 3.- This is written as
log2(8) = 3.- I'm too cheap to put a separate
graph for y = log2(x), so just flip the above y = 2^x graph- across the line y = x to get the log graph; this is because they're inverse functions.

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