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

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