What is the equation of the line normal to f(x)=- tan(2pix -2) at x=3?

1 Answer
Jul 21, 2016

y-(-tan(6pi-2)) = (x-3)/(2pi*sec^2(6pi-2))

Explanation:

f(x)=−tan(2πx−2)
f(3) =−tan(6π−2)

Point: (3, −tan(6π−2))

To find the slope, we need to take the derivative of f(x).

f'(x) = -2pi*sec^2(2pix-2)

The slope normal/perpendicular would be 1/(2pi*sec^2(2pix-2).

Therefore at x=3, the perpendicular slope would be 1/(2pi*sec^2(6pi-2).

Now, we just need to plugin everything in the point-slope form.

y-(-tan(6pi-2)) = (x-3)/(2pi*sec^2(6pi-2))