aapd30809ch7

toc =Chapter 7: Powers= Chapter 7 Preview: [|aa Chapter 7 Preview.pdf]

Wiki Summary Assignments:
7-1: Dominic F. and Mitch K. 7-2/7-3: Marisa M. and Thomas S. 7-4: Gage A. and Shannon S. 7-5: Austin R. and Autumn T. 7-6: Sarah K. and Kim K. 7-7/7-8: Nick H. and Tarah L.

Section 7-1: Power Functions
Notes: [|aanotes7-1.pdf] media type="custom" key="3658811"media type="custom" key="3674241"

Dominic F. Lesson 7-1 Power Functions

X to the power of n, in this power function x is the base and n is the exponent. This whole thing is called a power. There are many different kinds of functions. There is x to the power of 1 and thats the identity function. You can raise a number to any power. The graph of every nth power function passes through the origin. the graph of every power function has symmetry.

Lesson 7-1 teaches power functions, different types of power functions, how to identify functions, and some properties of power functions. First off, say x^5. The base of that would be x, or the number being multiplied exponentially. The exponent is the small number up to the right of the base, which indicates how many times the base will be multiplied by itself. The whole thing together is a power. The general power function is f(x)=x^n, with n greater than zero. An identity function is x^1, a squaring function is x^2 a cubing function is x^3, and soon. A few properties of power functions: 1. The graph always goes through the origin. 2. The domain is all real numbers-no matter what. 3. The range is all real numbers if the exponent is odd, and zero up if the exponent is even. If the exponent is odd, there's 180 degree rotational symmetry about the origin, and if the exponent is even, there's reflection symmetry over the y axis. -- Mitch K.

Section 7-2/7-3: Properties of Powers
Notes: [|aanotes7-2.pdf] media type="custom" key="3674075"media type="custom" key="3674243" In these lessons there are many postulates in order to solve these problems. one of them is the Product of Powers postulate which is b^m * b^n = b ^ m +n an example would be 5^2 * 5^3 = 5^5 the only way this postulate works is if the base is the same number or variable. Another postulate is Power of a Power postulate which is (b^m)^n = b^mn an example would be (b^2c)^3 = (b^6c^3). There is also another postulate called the Power of a Product postulate which looks like (ab)^m = a^m b^m an example of this would be (4y^2v^3)^4 = 324y^8v^12. Theres also the Quotient of Powers postulate which says b^m/b^n in this case the problem always has to have the same base and then you just subtract the exponents. Zero exponent theorum states that b to the zero power always equals 1. In lesson 7-3 deals with negative integar exponent an example of this would be y^8/y^10 = y^ -2 = 1/y^2 Marisa M. and Thomas S.

Section 7-4: Compound Interest
Notes: [|aanotes7-4.pdf] media type="custom" key="3674071"media type="custom" key="3674247" In Lesson 7-4 we revisited the idea of interest. Previously, we had learned to use simple interest but now we are using compound interest which calculates earning or losing interest on previous interest. The amount of interest in now based on the previous amounts that interest had added to the original invested amount or principal. Our annual compound interest formula is A (which is the amount) = P (principal) times the quantity or 1+r (the rate of interest) times t (the time in this case the year). But throw out that equation and remember the general compound interest formula which allows you to calculate compound interest for any amount of time. A and P remain the same but the rate of interest will now be divided by the times interest is calculated per year. The years, or the exponent, is then also multiplied by the number of times the interest is calculated each year as well. This gives us the final equation of A=P(1+ r/n) to the power of nt. -Shannon S

Section 7-5: Geometric Sequences
Notes: [|aanotes7-5.pdf] media type="custom" key="3698831"media type="custom" key="3699447"

In this lesson we worked on geometric sequences, which are much like sequences we worked on in chapter 1. A geometric sequence is a sequence that has a first term (g sub 1) and a constant ratio r that you multiply to the first term and each term the succeeds it. The ratio is multiply each time, for each term to get the next term in the sequence (it will never equal zero). The recursive formula for the geometric sequence is  (a is the same as g) for where n is greater than or equal to 2. A good example would be: the first term is equal to 4 and the equation is g(x)=x(sub n-1)*2, the next five terms would be 8, 16, 32, 64, and 128. ~Austin R.

In the lesson 7-5 we learned how to work with different geometric sequences such as, explicit and recursive formula's. This lesson was a lot like a previous lesson earlier in the year.

The Explicit Formula: The geometric sequence with first term (g sub 1), and constant ratio of r, for integers n greater than or equal to 1.

The Recursive Formula: ( //a//1 (  //an= r////an//-1, for integers greater than or equal to 2. Where r is a nonzero constant, is the geometric, or exponential sequence with the first term (g sub 1) and the constant multiplier r. Autumn T. 

Section 7-6: nth Roots
Notes: [|aanotes7-6.pdf] media type="custom" key="3710121"media type="custom" key="3745971" 7-6 In this chapter we learned about nth roots. The definition is- b is an nth root of x if and only if b^n=x We also learned about the 1/n exponent theorem which is when x is greater or less then 0 and n is an integer greater than 1, x^(1/n) is an nth root of x. For example 81^(1/4) would be 3, because 3^4 is equal to 81. To find this we would take 1/4 of 81. There is also the Number of Real Roots Theorem where every positive real number has: 2 real nth roots, when n is even or 1 real nth root, when n is odd. Every negative real number has: 0 real nth roots, when n is even or 1 real nth root odd. Sarah K.

Section 7-7/7-8: Rational Exponents
Notes: [|aanotes7-7-8.pdf] media type="custom" key="3718369"media type="custom" key="3745973" In these lesson we learned about rational exponents. Rational exponents are number like 4^1/4, 7^1/5. When using these types of numbers in the exponent the answer to these question are going to be smaller then the starting number. When you go about to solve a question like this it you have to plug it into your calculator like this. 4^(1/4) if you put it in like 4^1/4 you will not get the right answer. Another type of question we learned about in this lesson were something like this. 4^3/5 or 7^3/4. The way you can go about solving a question like this is plugging it into your calculator like so 4^(3/5) but if the teacher is requiring you to show work you have to write it out like this.. ((4)^1/5)^3 Another type of question we got in these lessons were asking us to simplify the equation 4^-3/4 to solve this we did... (((4)^1/4)^3)^-1 Those are the basic equations we learned about in these lessons. -Nick H. user:SeedSage Rational Exponents Morning everyone, 7-7 & 7-8 were put together due to being quite similar. The main difference is whether it's a positive exponent or a negative one, both which are easy to solve for. For a start, a problem proposes to simplify 36^3\2. First, you would take the denominator, make 1 the numerator of it, & take the first numerator (that was 3) & make it the exponent of the whole problem. It's more complex in words than it really is. The problem would end up looking like this; (36^1\2)^3. Then, since 2 is the number in the denominator, you take the square (2) root of 36, & that so happens to equal 6!. Once again, the problem evolves & it turns out to be 6^3. Yes, you carry that 3 that was there from before, since you only squared what was in the parentheses. Now all you do is ...Well solve with the exponent, 6^3, which equals 216. Wow, that was easy. 7-8 has to do with negative rational exponents. Now, for example, there's a problem where you have to simplify 25^(-3\2). That negative shouldn't be too threatening, should it? Yes, yes it should be. No seriously though, you start by solving it regularly, putting the denominator over 1 & making 3 the exponent, only this time you carry the negative with it. It'll end up looking like (25^1-2)^-3. Square root of 25 is 5, which makes the parantheses go away & you get 5^-3. Okay, this is important here, so... You talk the reciprocal of 5^-3, which would be 1\5^3. Since you took the reciprocal, it got rid of that negative exponent. That's some pretty awesome magic Reciprocal. Anyways, you then solve 1\5^3, which would equal 1\125, which would turn out to be the final answer. Now go throw a party, you're officially at the end of this chapter.
 * Reciprocal!**

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