Question

Find the equation of the tangent line and the normal line at a the given x-value.
y=cotx x=3pi/4

Answer 1

\[y=\cot x \]
\[x=\frac{ 3\pi }{ 4 } \]

Answer 2

y=-1

Answer 3

1/ find y'(3pi/4) =m. That is the slope of the tangent line
2/ (3pi/4, -1) is the point. Plug m, and that point into the equation of the line y -y_0 =m (x-x_0. That is the equation of the tangent line
3/ the normal line has the slope reciprocal with the slope of the tangent line. the leftover is the same step 2