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

##### 1 Answer

Jan 5, 2017

#### Explanation:

Ar x = 12, y= 254, nearly.

The slope of the normal

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)