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

The normal of y=2x^2,(-1,2) is y= 1/4 x+9/4

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

normal of

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

y = \frac{1}{4} x + \frac{9}{4}

Solution steps

Find the slope of

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

Compute the slope of the perpendicular line:

m_{p} = \frac{1}{4}

Find the line with slope m=

\frac{1}{4}

and passing through

(- 1, 2) : y = \frac{1}{4} x + \frac{9}{4}

y = \frac{1}{4} x + \frac{9}{4}

\frac{\partial}{\partial x} (- 2 e^{x})

\frac{d}{d x} (- \frac{2 x ^{2}}{5 y ^{2}})

y^{^{'}} = \frac{( 1 - x )}{y}

(u^{- 2} + u^{- 3}) (u^{5} + 5 u^{2})

\frac{d}{d x} (tan^{2} (x θ))

What is the normal of y=2x^2,(-1,2) ?
The normal of y=2x^2,(-1,2) is y= 1/4 x+9/4