Precal Blog

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.