What is the normal of y=4x^2,(-2,16) ?

The normal of y=4x^2,(-2,16) is y= 1/16 x+129/8

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normal of y=4x^2,(-2,16)

normal of

y = 4 x^{2}, (- 2, 16)

y = \frac{1}{16} x + \frac{129}{8}

Solution steps

Find the slope of

y = 4 x^{2} : \frac{d y}{d x} = 8 x

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 16

Compute the slope of the perpendicular line:

m_{p} = \frac{1}{16}

Find the line with slope m=

\frac{1}{16}

and passing through

(- 2, 16) : y = \frac{1}{16} x + \frac{129}{8}

y = \frac{1}{16} x + \frac{129}{8}

\frac{\partial}{\partial x} (8 x cos (3 x y))

\int \frac{3}{x ^{2}} - 1 d x

\int 3 + 2 x - x^{2} d x

\frac{2}{1 + 2 x}

\int \frac{2 + 3 x}{1 + x ^{2}} d x

What is the normal of y=4x^2,(-2,16) ?
The normal of y=4x^2,(-2,16) is y= 1/16 x+129/8