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

y = x4 – 6x3 + 13x2 – 10x + 5 at (0, 5)


It is given that equation of curve is 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 (0, 5) is -10.


Then, the equation of the tangent is


y – 5 = -10(x – 0)


y – 5 = 10x


10x +y =5


Then, slope of normal at (0,5)



Now, equation of the normal at (0,5)



10y -50 = x


x -10y +50 = 0


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