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.