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

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Answer 1 · 2017-01-05T08:01:51

#-1/45#, nearly

Ar x = 12, y= 254, nearly.

#y'=4x-3/2((2x-1)/sqrt(x^2-x))#= 45, nearly, at x = 12.

The slope of the normal #= -1/(f')=-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)