- #1
koolkris623
- 19
- 0
Find the equation of the line tangent to the curve at (5, -3)
(x-2)^2 + (y+3)^2 = 9
I solved the derivative to be dy/dx = ((-2x+4)/ (2y+6))
when i plugged in the points (5, -3) I got the slope as -6/0...How is this possible??
How can i find the equation of this curve if the slope is undefined??
(x-2)^2 + (y+3)^2 = 9
I solved the derivative to be dy/dx = ((-2x+4)/ (2y+6))
when i plugged in the points (5, -3) I got the slope as -6/0...How is this possible??
How can i find the equation of this curve if the slope is undefined??