as you already know, the slope of the tangent line is simply the derivative of the function, so
![r=2cos(3\theta )\implies \cfrac{dr}{d\theta }=2\stackrel{chain~rule}{\left[ -sin(3\theta )\cdot 3 \right]} \\\\\\ \left. \cfrac{dr}{d\theta }=-6sin(3\theta ) \right|_{\theta =\frac{\pi }{6}}\implies -6sin\left( 3\cdot \frac{\pi }{6} \right)\implies -6sin\left( \frac{\pi }{2} \right)\implies -6](https://tex.z-dn.net/?f=r%3D2cos%283%5Ctheta%20%29%5Cimplies%20%5Ccfrac%7Bdr%7D%7Bd%5Ctheta%20%7D%3D2%5Cstackrel%7Bchain~rule%7D%7B%5Cleft%5B%20-sin%283%5Ctheta%20%29%5Ccdot%203%20%5Cright%5D%7D%20%5C%5C%5C%5C%5C%5C%20%5Cleft.%20%5Ccfrac%7Bdr%7D%7Bd%5Ctheta%20%7D%3D-6sin%283%5Ctheta%20%29%20%5Cright%7C_%7B%5Ctheta%20%3D%5Cfrac%7B%5Cpi%20%7D%7B6%7D%7D%5Cimplies%20-6sin%5Cleft%28%203%5Ccdot%20%5Cfrac%7B%5Cpi%20%7D%7B6%7D%20%5Cright%29%5Cimplies%20-6sin%5Cleft%28%20%5Cfrac%7B%5Cpi%20%7D%7B2%7D%20%5Cright%29%5Cimplies%20-6)
Could you word this better? Would love to help you out, but I'm not sure what the details for the question are.
The equation is y=1x-2
Step-by-step explanation:
Is this the answer u needed?
Here is my process for solving this.
First I drew arrows that indicated I was moving the whole triangle 5 units to the left.
*Look at first attachment*
Then I drew another triangle using those new points. (The new triangle is in pink)
*Look at second attachment*
Then I drew arrows that moved this new triangle 4 units up. (The new arrows are in pink)
*Look at third attachment*
Then I drew the new triangle in blue using the new points.
*Look at fourth attachment*
Then I mirrored / reflected the triangle over the x axis (the horizontal line) In green.
*Look at fifth attachment*
The fifth attachment in green is the final product! Hope that helps.