Find the line that is tangent to the curve with the function y^2=x+3 at point (6,-3)
I know m=dy/dx so implicit derivative is: 2y dy/dx=1, so dy/dx = 1/2y
then equation of a line y-y_1=m(x-x_1). thus y+3=1/2y(x-6)
I can't believe that this is suppose to be my answer as this is clearly not a straight line. What am I doing wrong?