What is the normal of y=F4x-x^2,(1,3) ?

The normal of y=F4x-x^2,(1,3) is Normal line to y=F(y,x)4x-x^2, with a tangent of point (1,3)

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normal of y=F4x-x^2,(1,3)

normal of

y = F (y, x) 4 x - x^{2}, (1, 3)

Normal line to y = F (y, x) 4 x - x^{2}, with a tangent point (1, 3)

Solution steps

Slope of the normal line at

(1, 3) : m = \frac{F ( y , x ) \cdot 4 x - x ^{2} - 3}{x - 1}

Find the slope of

y = F (y, x) 4 x - x^{2} : m_{y} = 4 (x \frac{d}{d x} (F (y, x)) + F (y, x)) - 2 x

Compute the slope of the normal line:

m = - \frac{1}{4 ( x \frac{d}{d x} ( F ( y , x ) ) + F ( y , x ) ) - 2 x}

What is the normal of y=F4x-x^2,(1,3) ?
The normal of y=F4x-x^2,(1,3) is Normal line to y=F(y,x)4x-x^2, with a tangent of point (1,3)