What is the slope of the line normal to the tangent line of #f(x) = 2x^2-3sqrt(x^2-x) # at # x= 12 #?

Ar x = 12, y= 254, nearly.

The first graph shows the trend towards P(12,254).

The equation to the normal at P.

y=-1/45 x+254 (rounded).

The suitably-scaled second graph depicts the curve and the normal,

in the neighborhood of the high-level point P.

graph{(y-2x^2+3sqrt(x^2-x))(45y+x-45266)=0 [-10, 10, -5, 5]}

graph{(y-2x^2+3sqrt(x^2-x))(y+x/45-254)=0 [0, 100, 200, 300]} P(12, 253.5)

To find the slope of the line normal to the tangent line of f(x) at x=12, we need to find the derivative of f(x) and evaluate it at x=12. The derivative of f(x) is given by f'(x) = 4x - (3/2)((2x-1)/sqrt(x^2-x)). Evaluating f'(x) at x=12, we get f'(12) = 4(12) - (3/2)((2(12)-1)/sqrt(12^2-12)). Simplifying this expression gives us f'(12) = 48 - (3/2)((23)/sqrt(132)). Therefore, the slope of the line normal to the tangent line of f(x) at x=12 is f'(12).