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