What is the normal of y=2x^2,(-2,8) ?

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

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normal of y=2x^2,(-2,8)

normal of

y = 2 x^{2}, (- 2, 8)

y = \frac{1}{8} x + \frac{33}{4}

Solution steps

Find the slope of

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

Compute the slope of

y = 2 x^{2}

(- 2, 8) : 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

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

y = \frac{1}{8} x + \frac{33}{4}

\int_{0}^{5} (t^{2} + 2 t - 15) d t

t \to 0 lim (\frac{5}{t} - \frac{5}{e ^{t} - 1})

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

\frac{d}{d x} (3 (4 - 9 x)^{4})

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