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