Hi guys! How do we find the normal to the curve in this problem?

**Find the equations of the tangent to the curve y = x - x^{3}at (1,0), and the normal to the same curve at (1/2, 3/8). Find where the tangent and normal intersect.**

First I found the derivative and then calculated the gradient of the tangent. After it, I determined the equation of the tangent:

y = x- x^{3}

dy/dx = 1 - 3x^{2}>>>>>(1,0)

m = 1 - 3 = -2

y - y_{0}= m.(x - x_{0})

y - 0 = -2.(x - 1)

y = -2x + 2

Normally, I would do that:

The normal to the curve = -1/m

= -1/-2

= 1/2

But I know it's false and "the normal to the same curve at (1/2, 3/8)" makes me confused. Do I need to find the slope of the tangent at that point to find the normal? Like :

m = 1 - 3x^{2}>>>>>>(1/2, 3/8)

m = 1 - 3.(1/2)^{2}

m = 1/4

the normal = -1/m

= -1/1/4

= -4

y - y_{0}= m.(x - x_{0}) >>>>>>>(1/2, 3/8)

y - 3/8 = -4.(x - 1/2)

y = -4x + 19/8

??

Any help would be appreciated.

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