So this was supposed to be done like Tuesday? night I believe, but I forgot, and then I left my notebook at school, so finally here is my post on power functions.

The general equation for a power function is f(x)=ab^x.  There are a few rules to follow for power functions: a ?0 because then f(x)=0, b?0 because then f(x)=a, and when b=1 then f(x)=ax (which is linear).  However, b can be a positive, negative or fraction.

Power functions can be various shapes.  However, the two main shapes are u-shaped and s-shaped.  Even functions, such as x^2 or x^4, are general u-shapes.  Odd functions, like x^3 or x^7, tend to form s-shapes.

A variety of things can occur when b is a fraction.  When b is a fraction you take the denominater-root of x^numerator.  As a general rule, n?x^m (the n-root)=x^(m/n).  So for example, if b=4/3, it would become ³?(x^4).  The four options in the senario are even/even, odd/even, odd/odd, and even/odd.  However, it also depends whether the fraction is greater than or less than 1. 

 For each situation the domain, range, and end behavior are affected.  As an example, say you had x^(11/2), which is the same as ²?x^11.  Its an odd/even fraction greater than 1.  For the domain, x?0 because if you take a negative to an odd power you get a negative, and you cant take the even root of a negative number.  Also, the range is y?0 because if you cant plug in negatives, you will never get negatives out in this situation.  The end behavior is: x??, f(x)??.

Some generalizing rules are:

for odd roots: negative inputs?negative outputs, positives?positives

for even roots: NO negative inputs, positives?positives

for odd powers: negative inputs?negative outputs, positives?positives

for even powers: negative inputs?positive outputs, positives?positives   

And finally, the numeric pattern for power functions is multiply-multiply.  So say as you double x, f(x) is multiplied by a constant.

That pretty much covers everything for power functions.  Maybe this will help even though its kinda a few days late.  I probably couldve done it earlier if I didnt have an english paper to write. Hope everybody manages to survive this rough night. Peace

Alrighty so today we had a exploration worksheet to help us understand a new funtion that we are leanring. These new funtions are trinomials because they have 3 x values. Such as
y= x^3 + 2x^2 -x + 18

The worksheet the required us to find the 3 point where y=0. Usually one of the values is a whole number while the other two are slightly more difficult to find. The way Mr. B taught us to find out what these other two values where was to divide the function by the expression you know. This may sound confusing at first but hopefully me elaborating will make it a little bit clearer.

So say the y-intercept you know is 2. You can use this information to create the function y=x-2 because you know that x-2 is a factor of your overall function you can divide the overall function by this one. This way you will get the number that you multiply x-2 by to get your starting function. This function that you find will be a quadratic formula. You can then just plug that into your calculator and it will give you the 2 remaining x values. Hopefully that made sense. Tomorrow we will be having claims so be ready! And good luck to all the forensics people! The next scribe will be Adam

-Pat out

A power function has the general equation of f(x)=ax^b. The restrictions on this equation are b cannot =0, because otherwise you get a straight line, and a cannot =0, because then f(x)=0. B can =1. This results in a straight line. B can also be a negative number. In the equation f(x)=ax^b, a produces a vertical stretch.

We learned that power functions can be all kinds of graphs, such as parabolas, straight lines, a w-shape, or a v-shape, to name a few. These graphs result from the different b-values in the equation.

We learned that the b-values can equal a fraction. When they equal a fraction, you simply take the square root of x^numerator. For example, if the power is 3/5, you would then take the fifth root of x cubed. When the exponents are fractions, they follow a pattern. This pattern is the type of number in the numerator and denominator. For example, if the power is 4/10, then that would be an even-even type of graph. All fractions that have an even numerator and denominator result in a graph that is a bent v-shape. We found the domain, range, and end behavior of each type of graph. For example, the domain of an even-even power is all real numbers. The range is y must be greater than or equal to 0. The end behavior is as x approaches positive or negative infinity, f(x) approaches infinity.

We also learned what different types of inputs result in different types of outputs. For example, even roots cannot have negative inputs, and positive inputs come out positive. We learned the general pattern for even roots, odd roots, odd powers, and even powers.

The numerical pattern for power functions is multiply-multiply. So, when you double f(x), you multiply x my a constant.

That’s it.

Today in class, Mr. B gave us a set of problems from the book to work on for claims. These problems are 1-16 and 20-27 on page 287 of our books. He suggested that we only spend as much time on 1-16 as we feel is needed, and focus mainly on problems 20-27, which involve more of the functions themselves instead of just finding a pattern. Personally, I will need a bit more work on finding a pattern and creating/solving a power function equation for that pattern. If anyone else is still struggeling on this, I’ll post a problem from our book to hopefully shed some light on the issue.

Sec. 7-3, Problem #2 (bear with me because I’m not sure how to make tables on here…)

x: 2,4,6,8,10

f(x):1500,750,500,375,300

So just try to image this crazy sideways setup I have going on as one of our vertical tables. Using what we know from the last unit, we see that this table involves a multiply-multiply pattern. This is because when 2 is multiplied by 2, its corresponding value 1500 is multiplied by 1/2. When 4 is multiplied by 2, its corresponding value 750 is multiplied by 1/2, and matches up with 8′s corresponding value 375. I know this is probably a little hard to picture without helpful arrows and such, but lets keep going. So now that we know every time an x value is multiplied by 2, its corresponding value is multiplied by 1/2, we know this pattern will result in an exponential power function who’s equation will follow the basic ya*x^b format. Now we can take some of the values from the table and form two equations, since we will be dealing with two variables. First, we can plug 1500 in for y and 2 in for x, giving us 1500=a*2^b. We can use the other pair of values that we used to determine our pattern, 750 and 4. This will give us the equation 750=a*4^b. Since these two equations share similar forms, we can set them equal to each other by getting a by itself. We do this by divide the left side by everything that is on the right side with the exception of a. This gives us two new equations, a=1500/2^b and a=750/4^b. We can then set them equal to each other, and do some simple cross multiplication to get us 750*2^b=1500*4^b. Next, we have to break this equation down further so that we can apply what we know about natural logs in order to solve for the exponent. We do that by dividing both sides by 750, and dividing both sides by 4^b. This simplifies everything down to .5^b=2. Next we can use natural logs to solve for b, by writing the equation as b(ln.5)=(ln2). Finally we just divide ln of 2 by ln of .5, which gives us that b=-1. It may seem like a lengthy process, but it’s simply built on other concepts we’ve learned this year. So that’s pretty much it as far as the identifying functions part of this unit goes. Just remember that claims are due Friday!