Precal Blog

Today we continued our discussion of exponential functions from whenever we talked about them last. We started out by solving for A and B from the general form of an exponential function (y= ab^x) when given two points. Mr. B gave us (2,10) and (5,6). You first insert these into the general equation.10=ab^ 2 and 6=ab^ 5

You then divide the first equation by the second and get 5/3=b^-3. To solve for b you must multiply both sides by -1/3, which will cancel the ^-3. You then solve for b and find it to be about .843. Insert this into either one of the regular equations and solve for a.

10=a.843^2

a should be about 14.06 Your final equation should then be y=14.06*.843^x

In addition to solving these equations we finished filling out the chart of different exponential functions. We found that making a negative results in a reflection over the x-axis. We then used our knowledge of what the variable do to sketch out quick graphs of some simple equations (a=y-intercept)

The claims for tomorrow are 7, 14-18 from chapter 7-2

Tomorrow’s scribe is Nick.

Mr. Bieniek had another meeting during six hour today; coughcoughskippingcoughcough. So Math Man (that is what Shelby calls him) taught us about the other 3 cases of exponential functions and then went over some examples of how to take 2 points and then create an exponential equation.

The first case we went over in class today was `a>0, 0<b<1`
Math Man made us convince him whether the graph was concave up or down and if it was increasing or decreasing. This created much debate, but after we went over an example we determined
`agraph

width=300;height=200;xmin=0;xmax=10;xscl=1;axes();

plot(1.5*(.5^x));

endgraph
it was decreasing, concave up. He also gave us the limit of this type of exponential function: `lim x->infty` `f(x)->0`
Don’t worry about the limit. Math Man said Mr. B didn’t really think we needed to focus on it right now.

The second case was `a<0, 0<b<1`
`agraph

width=300;height=200;xmin=-10;xmax=10;xscl=1;axes();

plot((-2)*(2^x));

endgraph
It is decreasing and concave down. The limit is: `lim x->infty` `f(x)->-infty`

The third and last case we looked at today was `a<0,0<binfty` `f(x)->0`
`agraph

width=200;height=300;xmin=0;xmax=10;xscl=1;axes();

plot(-2(.5^x));

endgraph
It is increasing and concave down. The limit is: `lim x->infty` `f(x)->0`

After we finished discussing the information the Mr. B might use for a test, Math Man gave us a set of points to make an exponential function through. Those points were `(2,10) and (5,6)`

Using `y=ab^x` , the general form of the exponential function, you substitute x and y with the appropiate coordinates. By doing that you get two equations: `10=a*b^2` and `6=a*b^5`.
To solve for b, you can divide the two equations to cancel a, and are left with: `.6=b^3`.
To get rid of the 3rd power, you can take both sides to the 1/3 power(you take the cubed root to both sides ). You are then left with `.6^(1/3)` which rounds to .84.
To solve for a you just but b in for an equation: `10=a*.84^2`
`a=14.17`
The final equation looks like: `y=14.17*.84^x`
You can check your equation by creating the graph:
`agraph

width=200;height=300;xmin=0;xmax=15;xscl=1;axes();

plot(14.17*(.84^x));

endgraph

The graph might be a little off because I rounded, but it should be close to the orginial points `(2,10) and (5,6)`.

We did one more example in class, and then we were given homework. Claims from section 7-2 are due tomorrow. 7, and then 14-18. Hopefully all of them will be claimable tomorrow.

Tomorrow’s scribe will be Julie

After our 75 minute math contest for a place on the U.S. Olympic Team to compete in Madrid, we had another day of hard core learning. We started off the class with presentations. Everyone was disappointed when there were only two problems to claim #9 and #16. Luke presented #9. Pretty much the whole point of # 9 was to understand the Ladder’s of Powers is a number line with exponents, or as Kari like to say a timeline. Today Justin drew this nifty “ladder” that showed how all the exponets fit in.

This just helps you determine that the bigger the exponent the farther it is going to be from an exponent of zero and closer you are going to be when you have a smaller negative exponent value.

Then Blair presented #16. After he plugged all the points into the tables, he graphed them out. Then he created an equation from guess and test. y=90x^.33. While making the equation, you observed how the A and B were affected. The higher the A value got the closer the line got to the points thus effecting the slope. (Hey if anyone has a good way to remember the difference between affect and effect let me know). Then for the B value the closer the value got to 1 the less nonlinear the lin became thus “affecting when the line flattens out” (Mr. B).

This was the perfect transition into our notes about exponential functions. We looked at the graph of the function f(x)=3*.7^x.

What we decided was that the if the B=1 then there would be a horizontal line at 1 and if B=0 then it would just be zero. However, the function would never hit zero in the x-axis because there is a horizontal asymptote. You can get really close to it, but the line will never touch it.

The parent function: y=B^x

The general form: y=AB^x (not to confused with the powers equation y=AX^b)

The transformed parent function: y=AB^(x-C)+D
A controls the y intercept as long as no other transformations occur. In the equation y=6*.7^x, the graph of this equation intercepted the y axis at (0,6).
Then we looked at two different equations to see how the D value affected the placement of the graphs. For the equation y=6*.7^x+3 is graphed where the line has an asymptote of 3( the line never quite hits 3). The graph in this equation also intercepted the y-axis at 9 because about 6+3=9. It looked something similar to this.

The other equation we looked at was y=2*.7^-3 where the asymptote of the line was -3 and the graph intercepted the y-axis at (0,-1) because of about 2+-3=-1. The graph would appear something like this.

Thus we discovered that D controls the horizontal asymptote.

Finally we ended the class with determining what each graph would end up looking like if when it fit between the given cases. For the case a>0, B>1 the graph was increasing concave up (the y intercept is positive and the B gets bigger). Your graph will end up looking something similar to this. It just depends upon what your equation is.
.

For homework we were to solve for the rest of the cases, and I will leave for Kari to demonstrate those for you on tomorrow nights blog.

We started off the class with claims, there were only two today, which made the rest of the class actually somewhat productive.

Doug did the first problem, which he had to put the different powers of x in an accending order. As Doug explained, it was just like putting fractions on a numberline. And with that the problem was solved

Grant did the next and final problem. he graphed a set of points and had to fit a line to it. We reasoned that the bigger b was the bigger the slope was with an increasing downward curve. Then once we found and equation that was close to it, we had solved the problem. Then we used that to find more points.

After that, we started going over exponential and power functions we went over how an exponential function will not contain the origin since x = 0 and b^0 = 1 thus always having a value for b. Though it can be metnioned that the larget b gets the smaller your result will be. the general form of the power function is y=ab^x. We decided that the a defines the y intercept. We proved this through explination and graphing.

Next we showed how this function could indeed pass through the origin at a certain point, by shifting. We made our new transformed equation y=ab^(x+c) + d. The c will provide a horizontal translation, and the d serves the purpose as the verticle shift and the location of the horizontal asymptote.

We also discussed limits a bit. We said that the limit for a general equation was the closer x gets to infinity, the closer the y will get to zero without actually crossing it. makin the limit infinity.

We then did a chart filled with statements on how these funcitons behave. To do this, we went back to the base equation without any sort of shift. For an equation containing a is greater than zero and b is greater than one, it would be a steep increasing concave up curve. making the limit for x to negative infiniy, because of the way that it is shown.

For a being greater than 0 and b being between 0 and 1, we found that it was a decresasince curve that was concave down. It never did touch the x axis, unlike the one before it, it’s limit was found to be positive infinity.

When a was less than 0, and and b was greater than 0, things got a little more tricky. It was a small concave down with asymptotes at the x AND the y axis. the x was negative, putting this curve in the thrid quadrent. its limit was as x approaches infinity, the function will approach negative infinity.

The last one we were (as dr. no would say) so rudely interrupted by the bell. So that will have to be saved until tomorrow. Also we were told that when writing a limit, we just write lim, because mathematicians are lazy. Just to let you know there is no homework for tomorrow!!!!

Tomorrow’s scribe will be Tyler.

So, today we had a sub right…wrong, Mr. B skipped because he’s a rebel. Well anyways we “learned” about the Ladder of Powers, like Birttany said it is just a way to classify your equations, yadda yadda yadda.

On to something more important, I just got done playing COD4, and I thought I had a good game….was i wrong. Adam went 72 and like 19, are you serious? sure i was like 53 and 13, but c’mon, everybody should igve it up for Adam. Luckily for Adam he needs no classification only that he ruled the game tonight, and will also be scribe tomorrow. Well thats all the time i have today, hopefully tomorrow we will have a teacher. Later Gator

Today in class we were introduced to The Ladder of Powers with a packet of activities and problems.
I think it was a little difficult at first to understand the material of the packet, because it can always be challenging to just read something and know what important points we should be taking out of it, and how to apply the new information to problems. However, I think most people at the end of the class had some sort of an understanding on what exactly the Ladder of Powers is…

Basically, the Ladder of Powers classifies power functions into two categories. Power functions
that lie above y=x, and increase more rapidly, have a higher “curvature” and being more bent, make it easier to work with mathematical modeling (See Activity 1.4)
Power functions lying under y=x, in comparison, decrease more rapidly.

So the Ladder of Powers is sort of a Spectrum, in showing trends of how rapidly functions decrease or increase.

And knowing this new information about Power functions, can help us interpret and work with data in a new way.

For tomorrow, we all should have read through the packet, and be ready for the claimable problems: 5-10, and 16.

And be ready for that Math Test tomorrow too!

GOODNIGHT,

Tomorrows scribe will be Juliana.

I don’t know about everyone else, but i plan on study hardcore for this test on Wed. How can you pass up a trip to Spain….for a math test?!?

Today in class, for 6th hour we got our 3rd quarter tests back to look at and go over. After we had spent about 20 minutes of the period going over any small questions relating to our tests, next we went over the AMC math test which we will be taking this Wednesday, and finally we got back to learning about power functions. We had determined on Friday the the general power function equation was `y=ax^b`. The example from last week involved the points (2,3) and (4,7), which when solving for looks like `3=a*2^b` and `7=a*4^b`. The question then was could you divide those two equations to solve, and the answer is yes. Being that the equations involve an equal statement, both sides are equal to each other, and therefore you can take 7 and divide it by 3, then go to the other side of each equation and divide `a*4^b` by `a*2^b`: `7=a*4^b` divided by `3=a*2^b` equals `7/3=2^b`. Since we do not yet know how to undo powers, you simply guestimate in order to solve for b, in which case we decided b is about 1.2. You then solve the equation by plugging in `3=7/3*a` to get `a=9/7`, and that leaves you with `y=9/7*x^1.2` as the equation. We went over one other power function example in class and that was it for the day.
The next scribe will be Nick

So, we spent the first half hour of class bubbling in our information for the AMC test we are taking on Wednesday. If any of us turn out to be complete math wizards we will move on to the AIME test, then a math summer camp I believe, finally we would go to Madrid, Spain for an international test. But considering most of us can’t find a corner in a square room without our calculator and this is a non calculator test, I don’t think any of us will be going to Madrid anytime soon… unless it’s for the spanish trip. So really all we did today was finish off power functions. When we left off on friday we were given the general equation `y=ax^b` and the points (2,1) and (4,7) and asked to solve for “a” and “b”. The two equations we came up with are `7=a4^b` and `1=a2^b`. On friday just before the bell rang we were discussing if it was kosher to divide `7=a4^b` by `1=a2^b` in order to cancel out the “a” and solve for “b”. Apparently we are able to do so and when you divide you get the equation `7=2^b`then because we have not learned to mathematically solve for “b” we need to use guess and test. since `2^2` is 4 and `2^3` is 8 “b” is between `^2` and `^3` so by random testing we find that b=2.8. Finally we put 2.8 in for “b” and we get the equation `1=2^2.8*a` then solving for “a” we get `1/7` and the final equation of `y=1/7*x^2.8`. For now that is the best method we have to solve for a power function equation given two points. Tomorrow’s scribe is Britney.

We started class by trying to convince Mr. B that we should watch Numbers or the Office because of the anatomy field trip. He didn’t buy it and we ended up learning about power functions. Alot of informarion was thrown at us really fast so hopefully this will help you understand what we did on friday.

The first thing we did was learn the parent function and general form of power functions.

Parent Function:y=x^b
General Form: y=ax^b

In the general form a cannot be zero because it makes the entire equation equal zero. b cannot be zero as well because it makes the graph a horizontal line.

The transformed parent function is y=a(x-c)^b +d Where a is the dilation or stretch, c is the horizontal translation, and d is the vertical translation.

The value of b greatly affects the shape of the graph. When b>1 the graph is concave up, and increasing. When b is small like 1.5 the graph has a very slow increasing curve. When b is bigger the graph has a much steeper curve.

When b is between 0 and 1 the graph is concave down and increasing. Again when b is small the curve is not as steep, and when b id bigger the curve is very steep.

The last value b can be is b<0. When b is less than zero the graph os concave down and decreasing. When b is negative inverse variation and power functions are closely related. The inverse variation function is 1/x^n which is equal to x^-n. The power function y=ax^-b is equal to y=a*1/x^b. Because of this x cannot be zero because you cannot divide by zero. When b is negative the graph will never cross the x axis. You also cannot get the equation to equal zero.

Also when b is negative there are two asymtotes. The first being x=0. Since x^-b=1/x^b and 1/0^b is undefined. The second asymtote is y=0 because 1/x^b gets smaller as x gets bigger.

After that we started a sample problem. We were given the two points (2,1) and (4,7). From these two points we created the equations 1=a*2^b and 7=a*4^b . Then we divided the two equations giving us 7=4^b/2^b. This left us with 7=2^b. This is when the bell rang.

For homework Mr. B wanted us to study our notes, find out if you can actually divide a system of equations, and figure out how to find b.

I tried to put in graphs of the functions but I couldn’t get my scanned pictures to paste into the blog. So sorry, I hope everyone has a good imagination.

The scribe for monday will be… Rachel.