Find the equations of the tangent and normal to the given curves at the indicated points:

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


It is given that equation of curve is y = y = x4 – 6x3 + 13x2– 10x + 5


On differentiating with respect to x, we get


= 4x3 - 18x2 +26x - 10



Therefore, the slope of the tangent at (1, 3) is 2.


Then, the equation of the tangent is


y – 3 = 2(x – 1)


y – 3 = 2x - 2


y = 2x +1


Then, slope of normal at (1,3)


=


Now, equation of the normal at (1,3)


y – 3 = (x – 1)


2y -6 =- x + 1


x + 2y - 7 = 0


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