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

Jan 5, 2017

$- \frac{1}{45}$, nearly

Explanation:

Ar x = 12, y= 254, nearly.

$y ' = 4 x - \frac{3}{2} \left(\frac{2 x - 1}{\sqrt{{x}^{2} - x}}\right)$= 45, nearly, at x = 12.

The slope of the normal $= - \frac{1}{f '} = - \frac{1}{45}$

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)