Calculus 1-->Curve of the Witch

Groan... you do a heck of alot of algebra...
well once again dy/dx (slope) = [(1 + x^2)(0) - (2x)(1)]/(1 + x^2)^2
therefore we get -2x/(1 + x^2)^2
now slot in the value for x...
(-2*-1)/(1 + (-1)^2)^2 = (2)/(2)^2
which is 2/4 = 1/2 for the slope
now y - y' = m(x - x')
y - 1/2 = 1/2(x - (-1))
2y - 1 = x + 1

x - 2y + 2 = 0 is your tangent
need anymore help msn me..

ok
I thank you for your notes. I can use your steps to solve similar problems in my textbook.

ur welcome