How do you find the equation of the tangent and normal line to the curve #y^3-3x-2y+6=0# at (9,3)?
Equation of the tangent:
Equation of the normal:
Start by finding the derivative.
The slope of the tangent is given by substituting
Hence, the equation of the tangent, by point-slope form, is:
The equation of the normal line is simple: it is perpendicular to the tangent. A line perpendicular to another has a slope that is the negative reciprocal of the original line.
Thus, the slope of the normal is
Hopefully this helps!