Verified answer

If the tangent is parallel to the x-axis then the gradient at that point is 0

f(x) = x^3 - 9x^2 + 24x - 3

f'(x) = 3x^2 - 18x + 24 (differentiation)

If gradient =0, f'(x)=0 so

3x^2 - 18x + 24 = 0

x^2 - 6x + 8 = 0

(x-4)(x-2)=0

x = 4 or 2

If x=4, y = f(4) = 4^3 - 9*4^2 + 24*4 - 3 = 13

If x=2, y = f(2) = 2^3 - 9*2^2 + 24*2 - 3 = 17

So the two points are (2, 17) and (4, 13)

The tangents which are parallel to the x-axis have a slope of zero.

he equation of the tangent to the circle isobtained by calculateing dy/dx, (ie the slope0 and equating to zero, therefore

dy/dx= 3x^2 - 18x + 24 =0, therefore after diving thru by 3 , we get

x^2 - 6x + 8=0 which factorises to

(x-4)(x-2)=0, so x=2 and x=4

Now sunstitute these values into the original equation to get the y-coordinates.