Precal Blog

So yeah… the first and third graphs look wrong on the blog, because I forgot the trick to make them look right.  But they are supposed to include values in the 3rd quadrant, so if you can picture them by flipping the original curve over the x-axis and then over the y-axis, and then combine them together will give you the actual graph (it goes through the origin).  And for the third graph, do the same thing, except they don’t combine into one line, because they are repelled away from the origin.  The reason that power graphs touch the origin with positive exponents and don’t with negative exponents is because negative exponents would not give you an `f(x)` value if `x=0`, whereas you could get an `f(x)` output with a positive exponent when `x=0`.  Example: `0^-1=`undefined whereas `0^1=0`.  Also, Haley did an excellent job explaining why the graphs curve the way they do.  I also should mention that you can only have values in the 3rd quadrant when you have an odd number as your b value (exponent) or an odd number in the denominator of your b value because of the rules of roots of numbers.  NOW I’m done!

In my procrastination to do ANY reading tonight for homework I decided to install Firefox as my default web browser.  It is AMAZING!  So everyone that uses Internet Explorer should make the switch to Firefox.  It’s FREE and it only took about 10 minutes!  Also, the stuff inside the “quote” marks on this blog automatically become what they’re supposed to be with Firefox.  Do it, it’s easy!

Today in Pre-calculus we discussed the last 5 graphs of power functions. We already know about the graphs with a b value (`y=ax^b`) between 0 and 1 with an odd denominator, like this one: agraph plot(x^(1/3)); endagraph the graphs with a b value between -1 and 0 with an even denominator, like this one: agraph plot(x^(-1/2)); endagraph and the graphs with a b value between -1 and 0 with an odd denominator, like this one: agraph plot(x^(-1/3)); endagraph But today we learned about the last of the graphs of power functions, take for instance the graph with an exponent between 0 and 1 with an even denominator, like this graph: agraph plot(x^(1/2)); endagraph which is the graph for `f(x)=x^(1//2)`. Another new we learned today was the graph for an exponent which has an even value below -1, like this one: agraph plot(x^-4); endagraph which is the graph `f(x)=ax^(-4)`. The 6th graph is one which has an exponent with an odd value below -1, such as this graph: agraph plot(1/(x^3)); endagraph which is the graph `f(x)=ax^(-3)`. The graph that Mr. Bieniek put on the blog for 5th hour is an example of our next graph, one with an exponent that is an even number greater than 1. The one that he uses is the graph for `f(x)=ax^4`. And last but not least is the graph for an exponent that is an odd number greater than 1. Take for example this graph: agraph plot(x^3); endagraph which is the graph for `f(x)=ax^3`. Pheww! Now that I’m done taking up a whole page on the blog… that’s about it for what we did today in class! Have a nice rest of your day.

Today we began our unit/discussion on power function graphs and some of their characteristics. First of all, we outlined the two basic things we know about a power function: that they exhibit a multiply-multiply property and that the general equation is `f(x)=ax^b`. Next we did some exercises with square roots to help us understand the concept in a new light before we went on to the graphs.

We first looked at `sqrt9` and were asked to think about it in terms of what times what equals 9. Of course it is 3. We were also reminded that the value will always be positive unless you introduce the square root symbol.

Next we tried the `sqrtx` . This was a bit more difficult because we were obviously using a variable. We were asked to think of it in terms of exponents and their rules. When we did this, we saw that `x^(1/2)` times `x^(1/2)` made our desired value of x. This was true because when terms have the same base, their variables can be added.

Next we tried `sqrtx^3` For this we needed to think what times what gives `x^3`. For me, just seeing this was a little difficult, so I thought of it in terms of an equation, `2x=3` or what two identical terms gives 3. When I solved this equation, I found that `x^(3/2)` was the term I was looking for.

The last type of exercise we did was `3sqrt x^2`. To solve this, I again approached it with an equation of three identical terms that give 2; 3x=2. This was because I need three terms to give me `x^2`. We found the term to be `2/3`.

One trend that was noticed in class was that the answers to these problems could be found within the question. This meaning that for `sqrtx` for example, the answer was `x^(1/2)` and we could see that x was taken to the first power, and we wanted the square root, our two terms. If checked, this pattern holds true for all of the other examples as well. Mr. Bieniek told us that yes this was a vaild trend, but to aid in our understanding later, we would need to think of it in terms of the process we used above.

Then we moved on to the graphs… actually we did one more thing. We looked at two last examples, that of `sqrt(-8)` and the `3sqrt(-8)`. These two examples were uniques because prior knowledge told us that square roots of negative values were undefined. This held true for the square root of -8 because there were no two identical values that would make -8. For the cubed root of -8 though, there were three identical values, -2, that could make -8. This activity gave us two distinct rules to look at for our graphs… odd roots which make odd denominators, and even roots which make even denominators. These roots influence the b value in our general equation in the manner of what types of outputs are possible, 0, negative values, etc.

Now we moved on to graphing. We first determined that there were four catergories that b could fall into: `b<-1`, `-1<b<0`, `0<b<1`, or `b>1`. We then looked at three graphs ( the rest to follow at a later date).

We looked at this graph and first determined that it had an odd exponent because there were negative inputs giving values. Next we tried `b<-1` and `-1<b<0` with the examples of `x^(-1/3)` and `x^(-3)` neither of these trials produced values that were what the graph had. Next we ruled out `b>1` because we knew the slope would be much too big. This left us with `0<b<1` which was correct because a fractional exponent will greatly slow the slope and pace of the graph. This became one of the eight graphs we will look at and we characterized it as an odd `0<b<1` in which (-) in= (-) out and that the slow slope is due to the fraction in the exponent.

The next graph that we looked at was an even exponent graph because only positive values were in the graph. We did notice though that the graph avoided the origin. This meant that `0<b<1` would be ruled out because a fractional exponent puts the term in the denominator, thus the division of zero becomes an issue. Also, `b<-1` was ruled out due to negatives. Lastly, we found that the type was `-1<b<0` because through guess and test we saw that values were still given therefore this was correct. This type of even function we could now characterize as avoiding the origin, and (-) in=undefined values.

The last graph that we looked at was a different odd exponent graph in which negatives were present and the origin was avoided. We didn’t get to finish this graph because the bell rang, but were asked a question. We narrowed our choices down to two: `0<b<1` and `b<-1`. When we plugged in values for both options of our tester, `x^(1/3)` and `x^(-3)` the trends of increasing x=increasing term were both prevalant. This led us to trying to find a method to differentiate between the two.

The last thing we talked about before the end of class was the fact that the graphs of `b>1` and `0<b<1` contained the origin in their graphs while the others did not.

I hope this clears up any standing confusion you had, and I apologize that the graphs are not available quite yet. I will need some help from Mr. Bieniek to get those up by tomorrow!

Today in class we went over power functions and figured out how to find a and b in ax^b, but solving for x by using the fact that anything to the power of 1 is 1 and setting up a system of equations is too much work.  So if you use the system of equations idea, but use a variables instead of actual values.  The setup should look like this with x = any value from the table, z = any value from the table preceeding x, f(x) the output of the function when x is inputed, and f(z) the output of the function when z is inputed.

az^b=f(z)

ax^b=f(x)

(x=z/x)^b=f(z)/f(x)

now solve for b and find…

log(f(z)/f(x))/log(z/x) = b

with the b value, a is easy to find, just solve for a

f(x)=ax^b

f(x)/x^b = a

Now we have a way of finding a and b with no more effort than it takes to type a formula into your calculator!

To test this I will use the problem from class

If we use x=6 and x=8 and plug it into our formula we find…

log(192/108)/log(8/6) = 2 = b

108/6^2 = 3 = a

so we find f(x) = 3x^2 which is what we figured out in class

the formula for power functions is f(x)=ax^b and we learned that it has a multiply-multiply pattern.  We also learned how the graphs of these look when b is greater than 1, 0<b<1, and when b is less than o.  when b is greater than 1 and even the graph will be in quadrant 1 and 2, and u shaped going through the origin.  In fact of of these will go through the origin.  When it is odd it will look like a tangent graph.  When 0<b<1  and even it will be in quadrant one and have the slope be steeper at first and then level off. when odd it will do the same thing but also in quadrant three.  finally, when b<0 and even the graph will be in the first quadrant going closer to both axes and the same when it is odd but in the third axes.

I honestly have no idea whose turn it is to blog. but since quarter journals are due tomorrow….

today we started learning about power functions, whose formula is f(x)=ax^b.  we were given tables of values and had to find what the a and b values equaled in each function.  For one table, we were given an x-value of 1 and an f(x)-value of 6.  since x is one, and 1 raised to any number is still 1, we knew that x^b=1, which means that a must equal 6.  [to sum that up: if x=1, a=f(x).] we could then plug 6 in for a in another equation and get the value of b.

the other problem did not give us an x-value of 1, so for this we had to set up a system of equations and solve using logs and exponent rules.

We learned that the pattern for power functions for the x and f(x) values are multiply-multiply. [linear functions: add-add; exponential functions: add-multiply; log functions: multiply-add.]

overall, it was a pretty good day. and i’d also like to add that i very much enjoyed our discussion in 5th hour =]

Hey everyone I know that the proofs we got are very difficult, but I think that I have a hint that will really help some of you.  In order to see the substitution in the proofs it is very helpful to write out a few extra terms before your last ones.  Instead of just writing the dots, you should have a few terms after the dots instead of just one or to.  This was the only way I was able to do the proofs, so if I help this helps anyone who is still working on the proofs!

Oh and our quarter journal is coming up and the end of the quarter is on Thursday, so don’t forget! Enjoy the rest of your weekend everyone!

Ah yes, another quarter is coming to an end, and everyone knows what that means….QUARTER JOURNALS!!!  So, yellow sheets.  I personally am not entirely sure if I know all of them…in fact, I’m pretty sure I don’t.  This is what I do know:

A-1, A-2, A-3

B-1, B-2 (what are these exactly?)

14-2, 14-3

Correct me, please on my errors.  Also, blog posts and a letter are included again.  Consider this a friendly reminder, especially to those without the sheet that has everything we need on it.

Otherwise, enjoy mathing!

Well, I haven’t blogged in a while so I figured I better catch up. Today we had a substitute, but we had to work on proofs of the equations we received yesterday. These equations involved arithmetic and geometric sequences and series. I can’t reveal too much more because they are take home assignment problems, but I will say that the clue that was given did a good job of showing who we substituted the basic arthimetic sequence formula into the arithmetic series formula before proceeding to prove the equation by induction. We all worked pretty well with each other on this task, and there was a lot of learning going on through the explanations. I do have to say though that many of us, including myself are more than a little lost on the whole problem and will most likely need some clarification tomorrow!

Good Luck everybody!