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Curve point differentation

The curve has the equation y= 3x-x^2. Find the equation of the normal at the point (1,2)
So I have differentiated the equation to get dy/dx= 3-2x
now what? I need to get it to be perpendicular somehow?

Reply 1

whatacrydonnie

The normal is perpendicular to the tangent at that point. This means the gradient of the normal is the negative reciprocal of the gradient of the tangent.

Reply 2

username1560589

Sub in x = 1 to find the gradient of the tangent. The gradient of the normal is the negative reciprocal of the gradient of the tangent. m_n = -1 / m_t

Reply 3

jadechapell

Original post by whatacrydonnie

The normal is perpendicular to the tangent at that point. This means the gradient of the normal is the negative reciprocal of the gradient of the tangent.