What is the normal of y=4x^2,(-1,4) ?

The normal of y=4x^2,(-1,4) is y= 1/8 x+33/8

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normal of y=4x^2,(-1,4)

normal of

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

y = \frac{1}{8} x + \frac{33}{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 = - 8

Compute the slope of the perpendicular line:

m_{p} = \frac{1}{8}

Find the line with slope m=

\frac{1}{8}

and passing through

(- 1, 4) : y = \frac{1}{8} x + \frac{33}{8}

y = \frac{1}{8} x + \frac{33}{8}

x y^{^{'}} - 2 y = 2 x^{2} + 2 x + 1

(- 2, 0), (0, 1)

y^{^{'}} + \frac{3}{t ^{2}} y = 0

sin (6 (t - \frac{3 π}{6}))

\frac{d}{d x} (x (e^{- x}))

What is the normal of y=4x^2,(-1,4) ?
The normal of y=4x^2,(-1,4) is y= 1/8 x+33/8