Unit 3 logarithms, Scribe post, Collin Downs

Today we started class with the super correction follow up test for the first 25 minutes of class. We then jumped to (1+1/x)^x which is also called e and what happens to it when it gets very large and how neither the base or the power wins and makes the function go towards 1 or a large number. We used this to show the transformation from A=P(1+r/n)^nt to A=Pe^rt because you can substitute n/r for x and 1/x= r/n so you can substitute e for 1+r/n and get A=Pe^rt. We then went on to homework where we talked about #21 with reference to the purple math visual of turning a log into an exponential. The last thing we learned today was the big three log properties with reference to the mystery function worksheet.
1. log(a*b)=log(a) * log(b)
2. log a^n= n*log a
3. log a/b= log a - log b
These will be used in our homework tonight which is p. 243/11-45 odd, 61-67 odd.

 HW:

• p. 243/11-45 odd, 61-67 odd

Today we did a warm-up with a liftoff function. example: g5(125)= g5(5^3)=3. Then we had to sketch y=4^x and use it to sketch a graph of a) y= 4^(-x) - 3, b) y=16*4^x and c) y=4^(0.5x). In b) 16=4^2 which makes y=4^2 + 4^x = 4^(x+2). In c) 4=2^2 which makes y= (2^2)^(0.5x)= 2^x. Then we discussed exercises 55 and 59 from hw. We used a formula A=P(1 + (r/n))^nt and GDC. Then we also used A=Pe^(rt) formula. Then we did the last page of Mystery function. Then we realized that mystery function and liftoff function is log. Definition: x=b^y means the same as y=log (b)^x. We went to http://www.purplemath.com/modules/logs.htm.
example: log(27)^9, in words: "The power you raise 27 to to get 9"
Then we found the inverse of exponential: f(x)=b^x then: y=b^x, then inverse: x=b^y, then solve for y: y= log(b)^x. So f^(-1)(x)=log (b)^x

Monday, November 23rd, 2009

Lift-off function

Check out the log definition graphic over at Purple Math.


HW:

• p. 236/1-29 odd, 33, 39-44 all, 49-69 odd
• Be sure you are ready for the Supercorrection Follow-up Test (blank copy of the test available here)

Today we did a warm-up with rational functions. We reviewed graphing and simplifying, and molly wrote up a google-doc on 'steps to graph rational functions.' We also discussed the equation 0/0=0, and pondered whether or not the answer would be 1, or whether it would result in an error on your calculators. We calculated our asymptotes for our warm-up using division to find the end-behavior. We explored rational functions using grapher, and added extra x's on the numerator's using our previous equation's from our warm-ups. We also fooled around with adding 7x, 5x, 12x, etc. and 2x, etc. on the numerator and denominator (respectively) to change the horizontal asymptotes. What we ascertained is that when there is a larger power on the denominator than on the numerator, the end behavior reached closer and closer to 0, regardless of the numerator's behavior. We can find the end behavior of our graphs by using division if the numerator's power is higher than the denominator's power, if the powers are the same then the lead coefficient of the numerator divided by the lead coefficient of the denominator is the horizontal asymptote for the end behavior, and if the power in the denominator is higher than the highest power in the numerator, then the end behavior is y=0. We finished by graphing the functions with all of the above information. We then calculated f(x) is less than or equal to 0. We used molly's steps to solve and graph our second equation. We revised our steps by adding that you can cancel portions of equations if it is possible. We changed our equation from g(x)=(2-x)(2+x)/(x-2) =-x-2, x does not equal 2. Mr. O'Brien went over what to expect for the test, and gave us the revision problems for the test, and encouraged us to do extra problems if we wanted to. Mr. O'Brien corrected the notion that Juniors will be in class on the 17th when the Supercorrections are due, and we spent the rest of the class reviewing any homework problems we wanted to.

We started class with a radical function warm up. After we did this problem, but before we went over it, we went over the quiz for the first half of class, and learned some ways to use our calculators to solve for i or find a complex conjugate. Next we went over the homework problems from two nights ago, on page 179. We went over problems 43 and 47, finding the roots over rational numbers, real numbers, and complex numbers for #43. For #47, we solved by factoring by grouping, then solving for zero. We learned how to factor a number that is the sum of two squares. So (x^2 +25) becomes (x+ 5i) (x-5i), which is the difference between two squares. When a complex number is a root, another root will be the complex conjugate. So since 5i is a root, - 5i is too. After going over those two problems we went back to the warm up. To help solve this problem, we went to http://calc101.com/webMathematica/long-divide.jsp This website can do long division, and it will show you the steps as well. We then went over how to find a slant asymptote. To find the slant asymptote, you can use the form of the radical function where there is and x + c outside of the fraction. To find the y-intercept it is easiest to use the original, unfactored form. All the x values will cancel out, and you will be left with the c values, which have no x value. This is the y intercept. We then talked about a removable asymptote, which is when a factor cancels out of a radical function, but should still be in the domain. For example: (x^2 + 1)(x + 1) ( x-3)/(x-2) (x-1) (x-3) The (x-3) will cancel out, but x cannot equal 3, so you must draw an open circle at this point on the graph.
There is no quiz on Friday, but the Unit 2 test is on Tuesday.