Find the equation of the tangent and the normal to the following curves at the indicated points:

y = x4 – 6x3 + 13x2 – 10x + 5 at x = 1 y = 3


finding slope of the tangent by differentiating the curve



m(tangent) at (x = 1) = 2


normal is perpendicular to tangent so, m1m2 = – 1



equation of tangent is given by y – y1 = m(tangent)(x – x1)


y – 3 = 2(x – 1)


y = 2x + 1


equation of normal is given by y – y1 = m(normal)(x – x1)




3
1