Haha, hmmm.... well to make a long story short, I looked at the blog briefly last night because I thought there was homework that needed to be posted (there wasn't) and while I was there, I completely forgot to check who the next scribe was. (evidently it was me). I learned from Aichelle a couple minutes after the end of class that I was, fortunately the class wasn't very complicated and no new concpets were really introduced. So, here we go with probably the easiest scribe post I've had to do in a while.

Well, this class was a "workshop" class, so we got into our groups and took a look at the first question which seemed very familiar. It has to do with the graph of the height of a thrown ball over time (Grade 11 Pre-Cal question). However he added the velocity of the ball which is the rate of change (a.k.a. derivative) of the function. Then we were asked a series of questions about the scenario. For question and answer, click the "first question link".

Question Two basically asked us to find the slopes of the tangent at each point and list them in increasing order.

The third question was a little interesting because it gave us a told us that an original function was transformed by with a vertical shift. Then it asked us about how the derivative of the transformed function related to that of the original function. Well, when a function is shifted vertically, the vertical and horizontal scale remains the same (same size), as does the shape and it's position along the x-axis (x coordinates). The only thing that differs between it and it's original function is the y coordinates. Now, we know that a derivative is a graph of the slope of the tangents of the parent function vs. the points along the x-axis. So, since the shape, size, and x coordinates of the transformed function do not change, the slope of tangent at each point remain the same meaning the derivatives are the same!!!.

When Mr. K. clicked the next slide button on the SmartBoard Question Four came on the screen. This dealt with the derivatives of even and odd functions. Basically, the derivative of a point [ie. (4,2)] on a even function will result in a derivative that is equal, but opposite at the point that is opposite about the y-axis [(-4,2)]. When dealing with an odd function, the derivative at one point [ie. (1,3)] will be the exact same as the derivative of the point opposite about the origin [(-4,-2)].

Finally, we finished with two questions that were connected. Question Five was the last one we worked with, but has the same rule as Question Six which is to be completed and posted in the comments box this evening. For an explanation of how to do these two, click here.

Well, that concludes my scribe post for the night. I shall post my answer to the Homework question in the comment box now that at least some people have tried to answer it.

Don't Forget:

-PRETEST TOMORROW!!!

-TEST ON FRIDAY!!!

-B.O.B. BEFORE THE TEST (excluding Aichelle as she has already done so)

LAST BUT NOT LEAST, THE NEXT SCRIBE WILL BEEEEEEE:

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once again I pick GREY M!!!

good night =D

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## 1 comment:

Yay pretest scribe x-)!

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