Monday, October 1, 2007


Hello everybody this is Alanna with the scribe post for today.

A.M Class:
This morning Mr. K gave us some time to work on a few questions to start off the period.

For part A the last two questions are the ones the I'll be concentrating on because those are the ones that the class somewhat had trouble on.

The coordinates of point, A, on the graph y= f(x) are (-2, -3). What are the coordinates of its image on each of the following graphs?
1. y= 3f(x)
First off we were given the points for A in the beginning of the question: (-2, -3).
Since the 3 is outside of the brackets this will cause Only the output to be tripled. So the x values with remain the same.
(-3)(3)= -9

Leaving us with the coordinates of (-2, -9)

2. y= f(1/2x)
We take the x value of coordinate A, which is (-2) and multiply it by the reciprocal in the equation which is (2/1) or just the number 2.

(-2)(2)= 4
Leaving us with the coordinates of (-4, -3)

*note- Mr.K told us that since we are in a higher class of math we don't divide anymore, we multiply everything.

For part B of the questions we were given the same thing but except this time we have to work backwards.
e.x We are given coordinates (5, -4) and equation y= f(x - 4)-5, now we have to find the original coordinates of B.
Take the x coordinate 5 and add -4, which will give you 1. For the y coordinate take -4 and subtract -5 and it will give you 1. This will give you the coordinates (1, 1).

After doing these questions he gave one more long question and I'm pretty sure everyone will agree that the last part of the question was complicated.

Not until one of the students pointed out an easier way to solve the question and pretty much everybody liked his way better. Not sure of the guys name but good job!
The questions are on slide 2 of todays lesson and "that guys" method is on slide 3. What he did was just fill in the green equation on slide 2 then just worked on it from there by isolating the x variable. That gives you the original coordinates of B that was asked for.

P.M Class

To begin the afternoon class Mr. K
rambled on about the calculator and how they can have technical limitations.

After that we applied our knowledge of transformations onto graph. Thank fully we had that excerise because it made things even more clearer.

Later on we learned about odd and even functions.

Even Functions:
To tell if its an even function what you get is kinda like a mirror image on the Y axis.

Even Only IF f(-x) = (x)
f(x)= x^2
f(-x)= (-x)^2

f(-x) = x^2

This is equal because the end result is the same thing as the beginning equation.

g(x)= x^2 + 2x
g(-x)= (-x)^2 + 2(-x)

g(-x)= x^2 - 2x

Not equal because in the end you end up with x^2 - 2x which is not the same as the original.

Odd Functions:
To tell if its an odd function by a glance is if you are able to get the same image if you flipped it over.


A function is odd only IF f(-x) = -f(x)
f(x) = x^3 - x

f(-x) = (-x)^3 - (-x)

f(x) = -x^3 + x

not even

-f(x) = -(x^3 - x)

-f(x) = -x^3 + x

since f(-x) = -f(x) this is an odd function.

The bell rang for class change so that was the end of the class. Mr. K assigned exercise 9 today.

Tomorrows scribe will be Mary Ann.

1 comment:

Mary Ann said...

Good job on the scribe!
I hope i'll deliver a good scribe for tomorrow..