Algebra CECC

sábado, 22 de marzo de 2014

Polynomial Factoring

Simple Polynomial Factoring

Factoring polynomial expressions is not quite the same as factoring numbers, but the concept is very similar. When factoring numbers or factoring polynomials, you are finding numbers or polynomials that divide out evenly from the original numbers or polynomials. But in the case of polynomials, you are dividing numbers and variables out of expressions, not just dividing numbers out of numbers.
Previously, you have simplified expressions by distributing through parentheses, such as:

2(x + 3) = 2(x) + 2(3) = 2x + 6

Simple factoring in the context of polynomial expressions is backwards from distributing. That is, instead of multiplying something through a parentheses, you will be seeing what you can take back out and put in front of a parentheses, such as:

2x + 6 = 2(x) + 2(3) = 2(x + 3)

The trick is to see what can be factored out of every term in the expression. Warning: Don't make the mistake of thinking that "factoring" means "dividing something off and making it magically disappear". Remember that "factoring" means "dividing out and putting in front of the parentheses". Nothing "disappears" when you factor; things merely get rearranged.

Factor 3x – 12.

The only thing common between the two terms (that is, the only thing that can be divided out of each term and then moved up front) is a "

". So I'll factor this number out to the front:

3x – 12 = 3( )

When I divided the "

" out of the "

", I was left with only the "

" remaining. I'll put that "

" as my first term inside the parentheses:

3x – 12 = 3(x )

When I divided the "

" out of the "–

", I left a "–

" behind, so I'll put that in the parentheses, too:

3x – 12 = 3(x – 4)

This is my final answer:

3(x – 4)

Warning: Be careful not to drop "minus" signs when you factor.
Some books teach this topic by using the concept of the Greatest Common Factor, or GCF. In that case, you would methodically find the GCF of all the terms in the expression, put this in front of the parentheses, and then divide each term by the GCF and put the resulting expression inside the parentheses. The result will be the same. But this seems like an awful lot of work to me, so I just go straight to the factoring.

Factor 7x – 7.

A "

" can come out of each term, so I'll factor this out front:

7x – 7 = 7( )

Dividing the

out of "

" leaves just an "

7x – 7 = 7(x )

What am I left with when I divide the

out of the second term? Well, if "nothing" is left, then "

" is left. (Remember:

7 ÷ 7 = 1

.) So I get:

7x – 7 =

7(x – 1)

Take careful note: When "nothing" is left after factoring, a "1" is left behind in the parentheses.

Factor 12y² – 5y.

In this case, no number is a common factor between the two terms (specifically, the

and the

share no common numerical factor), but I can still divide out a common

variable

factor of "

" from each of the two terms.

12y² – 5y = y( )

In the first term, I have the "

" and the other "

" factor left over:

12y² – 5y = y(12y )

(This is because

12y²

means

12×

, so taking the

and one of the

's out front leaves the second

behind.) In the second term, I have the "

" left over:

12y² – 5y =

y(12y – 5)

Don't forget the "minus" sign in the middle!

Factor x²y³ + xy

I can factor an "

" and a "

" out of each term:

x²y³=

xy(x

) = xy(xy

)

xy = xy(1).

x²y³ + xy = xy( )

= xy(xy² )

xy(xy² + 1)

Remember: When "nothing" is left after factoring, a "1" is left behind in the parentheses.

Factor 3x³ + 6x² – 15x.

I can factor a "

" and an "

" out of each term:

= 3x(x

), 6x

= 3x(2x)

, and

–15x = 3x(–5

)

. Being careful of my signs, I get:

3x³ + 6x² – 15x = 3x( )

= 3x(x² )

= 3x(x² + 2x )

3x(x² + 2x – 5)

Factor 2(x – y) – b(x – y). Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved

This may look different from what I've done above, but really it's not. The two terms,

2(x – y)

and

– b(x – y)

, do indeed have a common factor; namely, the parenthetical factor

x – y

. This may be different from what you're used to seeing referred to as being a "factor", but the factorization process works just the same as before.

First, I'll take the common factor out front:

2(x – y) – b(x – y) = (x – y)( )

From the first term, I have a "

" left over:

2(x – y) – b(x – y) = (x – y)(2 )

From the second term, I have a "

–b

" left over:

2(x – y) – b(x – y) =

(x – y)(2 – b)

Factor x(x – 2) + 3(2 – x).

This is almost the same as the previous case, but not quite, because "

x – 2

" is not quite the same as "

2 – x

". If I'd had "

x + 2

" and "

2 + x

", the factors would have been the same, because order doesn't matter for addition. But order

does

matter for subtraction, so I don't actually have a common factor here.

But I

would

have a common factor if I could just flip (or "reverse the order of") that subtraction. What would happen if I did that? Take a look at the following numerical subtraction:

5 – 3 = 2

3 – 5 = –2

When I flipped the subtraction in the second line, I got the same answer except that the sign had changed. This is always true: When you flip a subtraction, you also change the sign out front. In our case, this means:

x(x – 2) + 3(2 – x) = x(x – 2)

–

x – 2

)

By reversing the subtraction in the second parenthetical, I have created a common factor, so I can now proceed as I had in the previous example:

x(x – 2) + 3(2 – x) = x(x – 2) – 3(x – 2)

(x – 2)(x – 3)

These examples lead us to the next topic: factoring "in pairs"....

There is one special case for factoring that you may or may not need, depending upon how your book is structured and how your instructor intends to teach factoring quadratics. I call it "factoring in pairs".

Factor xy – 5y – 2x + 10.

Is there anything that factors out of all four terms? No. When you have four terms, and nothing factors out of all of them, think of factoring "in pairs". To factor "in pairs", I split the expression into two pairs of terms, and then factor the pairs separately.

xy – 5y

– 2x + 10

What can I factor out of the first pair? I can take out a "

xy – 5y

– 2x + 10

y(x – 5)

– 2x + 10

What can I factor out of the second pair? I can take out a "

–2

xy – 5y

– 2x + 10

y(x – 5)

– 2x + 10

y(x – 5)

– 2(x – 5)

(I took out a

–2

, rather than a

, because the leading sign on the pair was a "minus". And I got a "

–5

" in the result because, when I divided the positive

by the negative

, the result was a negative

. Be careful with your signs!)

Now that I do have a common factor, I can proceed as usual:

xy – 5y – 2x + 10

= y(x – 5) – 2(x – 5)

= (x – 5)(y – 2)

Factoring "in pairs" is most commonly used to introduce factoring quadratics. So you may see exercises that look like this:

Factor x² + 4x – x – 4.

This polynomial has four terms with no factor common to all four, so I'll try to factor "in pairs":

x² + 4x – x – 4

= x(x + 4) – 1(x + 4)

(x + 4)(x – 1)

In the second line above, I factored a "1" out. Why? If "nothing" factors out, a "1" factors out.

Factor x² – 4x + 6x – 24.

I'll try to factor "in pairs":

x² – 4x + 6x – 24

= x(x – 4) + 6(x – 4)

(x – 4)(x + 6)

If you will be using "factoring in pairs" for factoring quadratics (which is not the method I use), your book will refer to it as something like "factoring by grouping", and it will work like this:

Factor x² – 5x – 6.

First, I have to find factors of the last term,

–6,

that add up to the numerical coefficient of the middle term,

–5.

I'll use the number

–6

and

, because

(–6)(+1) = –6

, and

(–6) + (+1) = –5

. Using these numbers, I'll split the middle "

–5x

" term into the two terms "

–6x

" and "

+1x

". This will then allow me to factor in pairs:

x²

– 5x

– 6

= x²

– 6x + 1x

– 6

= x(x – 6) + 1(x – 6)

(x – 6)(x + 1)

Factor 6x² – 13x + 6.

This is a bit more complicated, because the leading coefficient (the number on the

x²

) is not a simple "

". But I can still factor the polynomial.

First, I need to find factors of

(6)(6) = 36

that add up to

–13

. I'll use the numbers

–9

and

–4

, because

(–9)(–4) = 36

, and

(–9) + (–4) = –13

. Then I can split the middle "

–13x

" term into the two terms "

–9x

" and "

–4x

", and then factor in pairs:

6x²

– 13x

+ 6

= 6x²

– 9x – 4x

+ 6

= 3x(2x – 3) – 2(2x – 3)

(2x – 3)(3x – 2)

For a complete explanation of these last two examples (whose method may look somewhat "magical" at the moment), please study my lesson on factoring quadratics. The "simple" and then the "hard" case pages should completely clarify the topic.

miércoles, 5 de marzo de 2014

Laws of Exponents

Exponents are also called Powers or Indices

The exponent of a number says how many times to use the number in a multiplication.

In this example: 8² = 8 × 8 = 64

In words: 8² could be called "8 to the second power", "8 to the power 2" or simply "8 squared"

Try it yourself:

So an Exponent just saves you writing out lots of multiplies!

Example: a⁷

a⁷ = a × a × a × a × a × a × a = aaaaaaa

Notice how I just wrote the letters together to mean multiply? We will do that a lot here.

Example: x⁶ = xxxxxx

The Key to the Laws

Writing all the letters down is the key to understanding the Laws

Example: x²x³ = (xx)(xxx) = xxxxx = x⁵

Which shows that x²x³ = x⁵, but more on that later!

So, when in doubt, just remember to write down all the letters (as many as the exponent tells you to) and see if you can make sense of it.

All you need to know ...

The "Laws of Exponents" (also called "Rules of Exponents") come from three ideas:

The exponent says how many times to use the number in a multiplication.

A negative exponent means divide, because the opposite of multiplying is dividing

A fractional exponent like 1/n means to take the nth root:

If you understand those, then you understand exponents!

And all the laws below are based on those ideas.

Laws of Exponents

Here are the Laws (explanations follow):

Law	Example
x¹ = x	6¹ = 6
x⁰ = 1	7⁰ = 1
x^-1 = 1/x	4^-1 = 1/4

x^mxⁿ = x^m+n	x²x³ = x²⁺³ = x⁵
x^m/xⁿ = x^m-n	x⁶/x² = x^6-2 = x⁴
(x^m)ⁿ = x^mn	(x²)³ = x^2×3 = x⁶
(xy)ⁿ = xⁿyⁿ	(xy)³ = x³y³
(x/y)ⁿ = xⁿ/yⁿ	(x/y)² = x² / y²
x^-n = 1/xⁿ	x^-3 = 1/x³
And the law about Fractional Exponents:

Laws Explained

The first three laws above (x¹ = x, x⁰ = 1 and x^-1 = 1/x) are just part of the natural sequence of exponents. Have a look at this:

Example: Powers of 5
	.. etc..
5²	1 × 5 × 5	25
5¹	1 × 5	5
5⁰	1	1
5^-1	1 ÷ 5	0.2
5^-2	1 ÷ 5 ÷ 5	0.04
	.. etc..

Look at that table for a while ... notice that positive, zero or negative exponents are really part of the same pattern, i.e. 5 times larger (or 5 times smaller) depending on whether the exponent gets larger (or smaller).

The law that x^mxⁿ = x^m+n

With x^mxⁿ, how many times will you end up multiplying "x"? Answer: first "m" times, then by another "n" times, for a total of "m+n" times.

Example: x²x³ = (xx)(xxx) = xxxxx = x⁵

So, x²x³ = x⁽²⁺³⁾ = x⁵

The law that x^m/xⁿ = x^m-n

Like the previous example, how many times will you end up multiplying "x"? Answer: "m" times, then reduce that by "n" times (because you are dividing), for a total of "m-n" times.

Example: x⁴/x² = (xxxx) / (xx) = xx = x²

So, x⁴/x² = x^(4-2) = x²

(Remember that x/x = 1, so every time you see an x "above the line" and one "below the line" you can cancel them out.)

This law can also show you why x⁰=1 :

Example: x²/x² = x^2-2 = x⁰ =1

The law that (x^m)ⁿ = x^mn

First you multiply "m" times. Then you have to do that "n" times, for a total of m×n times.

Example: (x³)⁴ = (xxx)⁴ = (xxx)(xxx)(xxx)(xxx) = xxxxxxxxxxxx = x¹²

So (x³)⁴ = x^3×4 = x¹²

The law that (xy)ⁿ = xⁿyⁿ

To show how this one works, just think of re-arranging all the "x"s and "y" as in this example:

Example: (xy)³ = (xy)(xy)(xy) = xyxyxy = xxxyyy = (xxx)(yyy) = x³y³

The law that (x/y)ⁿ = xⁿ/yⁿ

Similar to the previous example, just re-arrange the "x"s and "y"s

Example: (x/y)³ = (x/y)(x/y)(x/y) = (xxx)/(yyy) = x³/y³

The law that $fractional exponents$

OK, this one is a little more complicated!

I suggest you read Fractional Exponents first, or this may not make sense.

Anyway, the important idea is that:

x^1/n = The n-th Root of x

And so a fractional exponent like 4^3/2 is really saying to do a cube (3) and a square root (1/2), in any order.

Just remember from fractions that m/n = m × (1/n):

Example: $fractional exponents$

The order does not matter, so it also works for m/n = (1/n) × m:

Example: $fractional exponents$

And That Is It!

If you find it hard to remember all these rules, then remember this:

you can work them out when you understand the
three ideas at the top of this page

Oh, One More Thing ... What if x= 0?

Positive Exponent (n>0)		0ⁿ = 0
Negative Exponent (n<0)		Undefined! (Because dividing by 0 is undefined)
Exponent = 0		Ummm ... see below!

The Strange Case of 0⁰

There are two different arguments for the correct value of 0⁰.
0⁰ could be 1, or possibly 0, so some people say it is really "indeterminate":

	x⁰ = 1, so ...	0⁰ = 1
	0ⁿ = 0, so ...	0⁰ = 0
	When in doubt ...	0⁰ = "indeterminate"

viernes, 10 de enero de 2014

Mis mejores deseos en este nuevo año y siempre.

jueves, 5 de diciembre de 2013

The Midpoint Formula

The Midpoint Formula

Sometimes you need to find the point that is exactly between two other points. For instance, you might need to find a line that bisects (divides into equal halves) a given line segment. This middle point is called the "midpoint". The concept doesn't come up often, but the Formula is quite simple and obvious, so you should easily be able to remember it for later. Think about it this way: If you are given two numbers, you can find the number exactly between them by averaging them, by adding them together and dividing by two. For example, the number exactly halfway between 5 and 10 is ^{[5 + 10]}/₂ = ¹⁵/₂ = 7.5.
The Midpoint Formula works exactly the same way. If you need to find the point that is exactly halfway between two given points, just average the x-values and the y-values.

Find the midpoint between (–1, 2) and (3, –6).

Apply the Midpoint Formula:

So the answer is

P = (1, –2).

Technically, the Midpoint Formula is the following:

But as long as you remember that you're averaging the two points' x- and y-values, you'll do fine. It won't matter which point you pick to be the "first" point you plug in.

Find the midpoint between (6.4, 3) and (–10.7, 4).

Apply the Midpoint Formula:

So the answer is

P = (–2.15, 3.5)

Find the value of p so that (–2, 2.5) is the midpoint between (p, 2) and (–1, 3).

This reduces to needing to figure out what p is, in order to make the x-values work:

So the answer is p = –3.

Is y = 2x – 4.9 a bisector of the line segment with endpoints at (–1.8, 3.9) and (8.2, –1.1)?

If I just graph this, it's going to look like the answer is "yes". But I have to remember that a picture can suggest an answer, it can give me an idea of what is going on, but only the algebra can give me an exact answer. So I'll need to find the midpoint, and then see if the midpoint is actually a point on the given line. First, I'll apply the Midpoint Formula:

Now I'll check to see if this point is on the line:

y = 2x – 4.9

y = 2(3.2) – 4.9 = 6.4 – 4.9 = 1.5

But I needed

to equal

1.4

, so this line is

close

to being a bisector (as a picture would indicate), but it is not

exactly

a bisector (as the algebra proves). So the answer is "

No, this is not a bisector.

Find the perpendicular bisector of the line segment with endpoints at (–1.8, 3.9) and (8.2, –1.1). Copyright © Elizabeth Stapel 2000-2011 All Rights Reserved

This is a multi-part problem, and is actually typical of problems you will probably encounter at some point when you're learning about straight lines. This is an example of a question where you'll be expected to remember the Midpoint Formula from however long ago you last saw it in class. Don't be surprised if you see this kind of question on a test! Here's how to answer it:

First, I need to find the midpoint, since any bisector, perpendicular or otherwise, must pass through the midpoint. I'll apply the Midpoint Formula:

Now I need to find the

slope

of the line segment. I need this slope in order to find the perpendicular slope. I'll apply the Slope Formula:

The perpendicular slope (for my perpendicular bisector) is the negative reciprocal of the slope of the line segment. Remember that "negative reciprocal" means "flip it, and change the sign". Then the slope of the perpendicular bisector will be

+²/₁ = 2

. With the slope and a point (the midpoint, in this case), I can find the

equation of the line

–

1.4 = 2(x

–

3.2)

–

1.4 = 2x

–

6.4

y = 2x

–

6.4 + 1.4

y = 2x – 5

Find the center of the circle with a diameter having endpoints at (–4, 3) and (0, 2).

Since the center is at the midpoint of any diameter, I need to find the midpoint of the two given endpoints:

( ^{[–4
+ 0]}/₂ , ^{[3 + 2]}/₂ ) = ( ^–4/₂ , ⁵/₂ ) =

(–2, 2.5)

These examples really are fairly typical. You will have some simple "plug-n-chug" problems when the concept is first introduced, and then later, out of the blue, they'll hit you with the concept again, except it will be buried in some other type of problem. I'm telling you this now, so you'll know to remember the Formula for later.

The Distance Formula

The Distance Formula

The Distance Formula is a variant of the Pythagorean Theorem that you used back in geometry. Here's how we get from the one to the other:

Suppose you're given the two points (–2, 1) and (1, 5), and they want you to find out how far apart they are. The points look like this:
You can draw in the lines that form a right-angled triangle, using these points as two of the corners:
It's easy to find the lengths of the horizontal and vertical sides of the right triangle: just subtract the x-values and the y-values:

Then use the Pythagorean Theorem to find the length of the third side (which is the hypotenuse of the right triangle):

c² = a² + b²

This format always holds true. Given two points, you can always plot them, draw the right triangle, and then find the length of the hypotenuse. The length of the hypotenuse is the distance between the two points. Since this format always works, it can be turned into a formula:

Distance Formula:

Given the two points

(x₁, y₁)

and

(x₂, y₂)

, the distance between these points is given by the formula:

Don't let the subscripts scare you. They only indicate that there is a "first" point and a "second" point; that is, that you have two points. Whichever one you call "first" or "second" is up to you. The distance will be the same, regardless.

Find the distance between the points (–2, –3) and (–4, 4).

I just plug the coordinates into the Distance Formula:

Then the distance is sqrt(53), or about 7.28, rounded to two decimal places.
The most common mistake made when using the Formula is to accidentally mismatch the x-values and y-values. Be careful you don't subtract an x from a y, or vice versa; make sure you've paired the numbers properly.

Also, don't get careless with the square-root symbol. If you get in the habit of omitting the square root and then "remembering" to put it back in when you check your answers in the back of the book, then you'll forget the square root on the test, and you'll miss easy points.
You also don't want to be careless with the squaring inside the Formula. Remember that you simplify inside the parentheses before you square, not after, and remember that the square is on everything inside the parentheses, including the minus sign, so the square of a negative is a positive.
By the way, it is almost always better to leave the answer in "exact" form (the square root "

" above). Rounding is usually reserved for the last step of word problems. If you're not sure which format is preferred, do both, like this:

Now YOU try it!

Very often you will encounter the Distance Formula in veiled forms. That is, the exercise will not explicitly state that you need to use the Distance Formula; instead, you have to notice that you need to find the distance, and then remember (and apply) the Formula. For instance:

Find the radius of a circle, given that the center is at (2, –3) and the point (–1, –2) lies on the circle.

The radius is the distance between the center and any point on the circle, so I need to find the distance:

Then the radius is

sqrt(10)

, or about 3.16, rounded to two decimal places.

Find all points (4, y) that are 10 units from the point (–2, –1).

I'll plug the two points and the distance into the Distance Formula:

10 = sqrt[(-2 - 4)^2 + (-1 - y)^2] = [y^2 + 2y + 37]

Now I'll square both sides, so I can get to the variable:

100 = y^2 + 2y + 37, so 0 = y^2 + 2y - 63

This means

–

= 7

, so

the two points are

(4, –9)

and

(4, 7)

If you're not sure why there are two points that solve this exercise, try drawing the (–2, –1) and then drawing a circle with radius 10 around this. Then draw the vertical line through x = 4. You'll see that the vertical line crosses the circle in two spots: (4, –9) and (4, 7).

lunes, 2 de diciembre de 2013

Slope of a line

Slope of a Straight Line

One of the most important properties of a straight line is in how it angles away from the horizontal. This concept is reflected in something called the "slope" of the line. Let's take a look at the straight line y = ( ²/₃ ) x – 4. Its graph looks like this:

To find the slope, we will need two points from the line.
Pick two x's and solve for each corresponding y: If, say, x = 3, then y = ( ²/₃ )(3) – 4 = 2 – 4 = –2. If, say, x = 9, then y = ( ²/₃ )(9) – 4 = 6 – 4 = 2. (By the way, I picked the x-values to be multiples of three because of the fraction. It's not a rule that you have to do that, but it's a helpful technique.) So the two points (3, –2) and (9, 2) are on the line y = ( ²/₃ )x – 4.
To find the slope, you use the following formula:

(Why "m" for "slope", rather than, say, "s"? The official answer is: Nobody knows.)
The subscripts merely indicate that you have a "first" point (whose coordinates are subscripted with a "1") and a "second" point (whose coordinates are subscripted with a "2"); that is, the subscripts indicate nothing more than the fact that you have two points to work with. It is entirely up to you which point you label as "first" and which you label as "second". For computing slopes with the slope formula, the important thing is that you subtract the x's and y's in the same order. For our two points, if we choose (3, –2) to be the "first" point, then we get the following:

The first y-value above, the –2, was taken from the point (3, –2) ; the second y-value, the 2, came from the point (9, 2); the x-values 3 and 9 were taken from the two points in the same order. If we had taken the coordinates from the points in the opposite order, the result would have been exactly the same value:

As you can see, the order in which you list the points really doesn't matter, as long as you subtract the x-values in the same order as you subtracted the y-values. Because of this, the slope formula can be written as it is above, or alternatively it can be written as:

Let me emphasize: it does not matter which of the two formulas you use or which point you pick to be "first" and which you pick to be "second". The only thing that matters is that you subtract your x-values in the same order as you had subtracted your y-values.

Technically, the equivalence of the two slope formulas above can be proved by noting that:

y₁

–

y₂ =

–

y₂ + y₁ =

–

(y₂

–

y₁)

x₁

–

x₂ =

–

x₂ + x₁ =

–

(x₂

–

x₁)

Doing the subtraction in the so-called "wrong" order serves only to create two "minus" signs which cancel out. The upshot: Don't worry too much about which point is the "first" point, because it really doesn't matter. (And please don't send me an e-mail claiming that the order does somehow matter, or that one of the above two formulas is somehow "wrong". If you think I'm wrong, plug pairs of points into both formulas, and try to prove me wrong! And keep on plugging until you "see" that the mathematics is in fact correct.)

Let's find the slope of another line equation:

Find the slope of y = –2x + 3.

Graphing the line, it looks like this:

I'll pick a couple of values for

, and find I'll find the corresponding values for

. Picking

x = –1

, I get

y = –2(–1) + 3 = 2 + 3 = 5

. Picking

x = 2

, I get

y = –2(2) + 3 = –4 + 3 = –1

. Then the points

(–1, 5)

and

(2, –1)

are on the line

y = –2x + 3

. The slope of the line is then calculated as:

Now YOU try it!

Scroll back up this page and look at those equations and their graphs. For the first equation, y
= ( ²/₃ )x – 4, the slope was m = ²/₃. And the line, as you moved from left to right along the x-axis, was heading up toward the top of the drawing; technically, the line was "increasing". For the second line, y = –2x + 3, the slope was m = –2. And the line, as you moved from left to right along the x-axis, was heading down toward the bottom of the drawing; technically, the line was "decreasing". This relationship is always true: Increasing lines have positive slopes, and decreasing lines have negative slopes. Always!
This fact can help you check your calculations: if you calculate a slope as being negative, but you can see from the graph that the line is increasing (so the slope must be positive), you know you need to re-do your calculations. Being aware of this connection can save you points on a test because it will enable you to check your work before you hand it in.

Increasing lines have positive slopes; decreasing lines have negative slopes. With this in mind, consider the following horizontal line:

y = 4

Its graph is shown to the right.

Is the horizontal line going up; that is, is it an increasing line? No, so its slope won't be positive. Is the horizontal line going down; that is, is it a decreasing line? No, so its slope won't be negative. What number is neither positive nor negative? Zero! So the slope of this horizontal line is zero. Let's do the calculations to confirm this value. Using the points (–3, 4) and (5, 4), the slope is:

This relationship is true for every horizontal line: a slope of zero means the line is horizontal, and a horizontal line means you'll get a slope of zero. (By the way, all horizontal lines are of the form "y = some number", and the equation "y = some number" always graphs as a horizontal line.)

Now consider the vertical line x = 4:

Is the vertical line going up on one end? Well, kind of. Is the vertical line going down on the other end? Well, kind of. Is there any number that is both positive and negative? Nope.

Verdict: vertical lines have NO SLOPE. In particular, the concept of slope simply does not work for vertical lines. The slope doesn't exist! Let's do the calculations. I'll use the points (4, 5) and (4, –3); the slope is:

(We can't divide by zero, which is of course why this slope value is "undefined".)
This relationship is always true: a vertical line will have no slope, and "the slope is undefined" means that the line is vertical. (By the way, all vertical lines are of the form "x = some number", and "x = some number" means the line is vertical. Any time your line involves an undefined slope, the line is vertical, and any time the line is vertical, you'll end up dividing by zero if you try to compute the slope.)
Warning: It is very common to confuse these two lines and their slopes, but they are very different. Just as "horizontal" is not at all the same as "vertical", so also "zero slope" is not at all the same as "no slope". The number "zero" exists, so horizontal lines do indeed have a slope. But vertical lines don't have any slope; "slope" just doesn't have any meaning for vertical lines. It is very common for tests to contain questions regarding horizontals and verticals. Don't mix them up!
Slope: Parallel and Perpendicular Lines

Parallel lines and their slopes are easy. Since slope is a measure of the angle of a line from the horizontal, and since parallel lines must have the same angle, then parallel lines have the same slope — and lines with the same slope are parallel.

Perpendicular lines are a bit more complicated. If you visualize a line with positive slope (so it's an increasing line), then the perpendicular line must have negative slope (because it will be a decreasing line). So perpendicular slopes have opposite signs. The other "opposite" thing with perpendicular slopes is that their values are reciprocals; that is, you take the one slope value, and flip it upside down. Put this together with the sign change, and you get that the slope of the perpendicular line is the "negative reciprocal" of the slope of the original line — and two lines with slopes that are negative reciprocals of each other are perpendicular to each other. In numbers, if the one line's slope is m = ⁴/₅, then the perpendicular line's slope will be m = ^–5/₄. If the one line's slope is m = –2, then the perpendicular line's slope will be m = ¹/₂.
In your homework, you will probably be given some pairs of points, and be asked to state whether the lines through the pairs of points are "parallel, perpendicular, or neither". To answer the question, you'll have to calculate the slopes and compare them. Here's how that works:

One line passes through the points (–1, –2) and (1, 2); another line passes through the points (–2, 0) and (0, 4). Are these lines parallel, perpendicular, or neither?

To answer this question, I'll find the slopes.

Since these two lines have identical slopes, then these lines are parallel.

One line passes through the points (0, –4) and (–1, –7); another line passes through the points (3, 0) and (–3, 2). Are these lines parallel, perpendicular, or neither?

I'll find the values of the slopes.

If I were to flip the "

" and then change its sign, I would get

^–1

/₃"

. In other words, these slopes are negative reciprocals, so the lines through the points are perpendicular.

One line passes through the points (–4, 2) and (0, 3); another line passes through the points (–3, –2) and (3, 2). Are these lines parallel, perpendicular, or neither?

I'll find the slopes.

These slope values are not the same, so the lines are not parallel. The slope values are not negative reciprocals either, so the lines are not perpendicular. Then the answer is "neither".

Warning: When asked a question of this type ("are they parallel or perpendicular?"), do not start drawing pictures. If the lines are close to being parallel or close to being perpendicular (or if you draw the lines messily), you can very-easily get the wrong answer from your picture. Besides, they're not asking if the lines look parallel or perpendicular; they're asking if the lines actually are parallel or perpendicular. To be sure of your answer, do the algebra.

sábado, 22 de marzo de 2014

Polynomial Factoring

miércoles, 5 de marzo de 2014

Laws of Exponents

Example: a7

Example: x6 = xxxxxx

The Key to the Laws

Example: x2x3 = (xx)(xxx) = xxxxx = x5

All you need to know ...

Laws of Exponents

Laws Explained

The law that xmxn = xm+n

Example: x2x3 = (xx)(xxx) = xxxxx = x5

The law that xm/xn = xm-n

Example: x4/x2 = (xxxx) / (xx) = xx = x2

Example: x2/x2 = x2-2 = x0 =1

The law that (xm)n = xmn

Example: (x3)4 = (xxx)4 = (xxx)(xxx)(xxx)(xxx) = xxxxxxxxxxxx = x12

The law that (xy)n = xnyn

Example: (xy)3 = (xy)(xy)(xy) = xyxyxy = xxxyyy = (xxx)(yyy) = x3y3

The law that (x/y)n = xn/yn

Example: (x/y)3 = (x/y)(x/y)(x/y) = (xxx)/(yyy) = x3/y3

The law that

Example:

Example:

And That Is It!

Oh, One More Thing ... What if x= 0?

The Strange Case of 00

viernes, 10 de enero de 2014

Mis mejores deseos en este nuevo año y siempre.

jueves, 5 de diciembre de 2013

The Midpoint Formula

The Distance Formula

lunes, 2 de diciembre de 2013

Slope of a line

Example: a⁷

Example: x⁶ = xxxxxx

Example: x²x³ = (xx)(xxx) = xxxxx = x⁵

The law that x^mxⁿ = x^m+n

Example: x²x³ = (xx)(xxx) = xxxxx = x⁵

The law that x^m/xⁿ = x^m-n

Example: x⁴/x² = (xxxx) / (xx) = xx = x²

Example: x²/x² = x^2-2 = x⁰ =1

The law that (x^m)ⁿ = x^mn

Example: (x³)⁴ = (xxx)⁴ = (xxx)(xxx)(xxx)(xxx) = xxxxxxxxxxxx = x¹²

The law that (xy)ⁿ = xⁿyⁿ

Example: (xy)³ = (xy)(xy)(xy) = xyxyxy = xxxyyy = (xxx)(yyy) = x³y³

The law that (x/y)ⁿ = xⁿ/yⁿ

Example: (x/y)³ = (x/y)(x/y)(x/y) = (xxx)/(yyy) = x³/y³

The law that $fractional exponents$

Example: $fractional exponents$

Example: $fractional exponents$

The Strange Case of 0⁰