# What is the gradient of the normal to the curve y = x^3 - 2x^2 + 5 at the point (2, 5).

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We have the equation of the curve as y = x^3 - 2x^2 + 5

To find the slope or the gradient of the tangent drawn at any point on this curve, we need to find the differential.

y’ = 3x^2 – 4x

For the point (2, 5), y’ = 3*2^2 – 4*2 = 12 – 8 = 4.

To find the slope of the normal we use the relation that the product of the slopes of two perpendicular lines is given as -1.

Therefore the slope of the normal is -1 / 4

**The gradient of the normal to the curve y = x^3 - 2x^2 + 5 at the point (2, 5) is -1/4**

The gradient to the of the normal to any curve f(x) is -1/m where m is the gadient of the tangent to the curve .

m is given by:

m = f'(x).

Therefore m = f'(x) = (x^3-2x^2+5)'

f'(x) = (x^3)'-(2x^2)' + (5)'

f'(x) = 3x^2-4x +0

Therefore the gradient of the tangent m at (x=2) = f'(2) = (3*2^2-4*2 = 12 - 8 = 4.

Therefore the gradient of the normal = -1/m (at x=2) = -1/4.

Therefore the gradient of the normal to the given curve f(x) = x^3-2x^2+5 is -1/4.