a curve is defined by y=2x^2 - 10x +7. point (3, -5) lies on this curve. find the equation of the normal to this curve
equation of tangent is y - y1 = m(x-x1). differentiating y gives us the value of m. so dy/dx = 4x-10. we know x is 3. therefore, dy/dx = m = 2 but we need equation of the normal, which is y-y1=(1/m)(x-x1). 1/m is 1/2. y1 = -5. x1 = 3 putting it all in gives us 2y = x - 13, and that is the equation of the normal to this curve.
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