Today in class we went over the Exploration 3-3b sheet, which we did in class on friday.  We went over how to properly determine the horizontal dilation which is covered in Haley’s previous post.  We then got another sheet that practiced more transformations.

The first was a transformation of a Tangent graph: y=2+5tan3(`theta`-5°)

As Lauryn and I worked through this transformation we thought we were right on track.  After we compared with Zak and Matt, we found that we had made a huge mistake, instead of having the middle point start at x=5 (point 5,2) we had started the period at x=5.  This threw our graph off completely.  The lesson was learned that the horizontal dilation factor effects where the middle point will be located, not the asymptotes.

The second transformation was for a cotangent graph: y=-1+3cot2(`theta`-30°)

To make this graph we used our knowledge of tangent graphs and essentially flipped it, so that the graph started high with an asymptote and ended with a low one.  We knew we could do this because cotangent is the inverse of tangent which is sin?/cos?, making cotangent cos?/sin?.

The third transformation was for a secant graph:y=4+6sec.5(`theta`+50°)

I think that most of the people in our class find it easier to draw a cosine graph first, then draw a secant graph based off of that.

The fourth transformation was for a cosecant graph: y=3+2csc4(`theta`+10°)

Like the third transformation, I found it easier to draw this graph by first drawing a sine graph.

Hopefully everyone feels a little more confident about transformations of tangent, cotangent, secant, and co secant graphs.

Tomorrows Scribe is Lauryn.

We spent the last half of class yesterday discussing an equation for the following function:

We started by recognizing that one cycle completes in 30 degrees. This means we are fitting 6 cycles into the normal 180 degrees that a cotangent cycles in. So our “B” value is 6. Next we made a mistake. We “forgot” that the parent cotangent has asymptotes at 0 plus or minus 180n degrees so we thought the horizontal translation was 6 degrees to the right.

We mistakenly used the parent tangent function and thought of our asymptotes as centered around the origin.

Had we correctly remembered that parent cotangent asymptotes start at 0 we would have seen that the horizontal shift is actually 9 degrees to the left. This makes our D value -9.

Next we noted that the “mid-line” (although it isn’t really a mid-line right?) is at -1. So our C value is -1. Lastly, since one quarter of the way through the cycle (where sine and cosine are equal) we find the y value two units away from the “mid-line”. In the parent cotangent (and tangent) we are one unit away from the “mid-line” so we must have stretched by a factor of 2, making our A value 2. Note that three quarters of the way through the cycle (where sine and cosine are opposites) we are also 2 units from the “mid-line”.

Putting this all together makes our equation y = -1 + 2cot 6(x+9). Remember that typing this equation into a computer or grapher requires the non-factored form:
y = -1 + 2cot (6x+54). We like a the factored form because the 9 is more meaningful than the 54 – yeah?

Your final challenge is to write an equation for this function in terms of tangent.