What is the normal of y=5x^2,(-2,20) ?

The normal of y=5x^2,(-2,20) is y= 1/20 x+201/10

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normal of y=5x^2,(-2,20)

normal of

y = 5 x^{2}, (- 2, 20)

y = \frac{1}{20} x + \frac{201}{10}

Solution steps

Find the slope of

y = 5 x^{2} : \frac{d y}{d x} = 10 x

Compute the slope of

y = 5 x^{2}

(- 2, 20) : m = - 20

Compute the slope of the perpendicular line:

m_{p} = \frac{1}{20}

Find the line with slope m=

\frac{1}{20}

and passing through

(- 2, 20) : y = \frac{1}{20} x + \frac{201}{10}

y = \frac{1}{20} x + \frac{201}{10}

\frac{d}{d x} (f (x) (x^{2}))

\int \frac{5 - 50 x}{1 - 25 x ^{2}} d x

n = 0 \sum \infty 2 \cdot 6^{n}

x^{2} sin (5 x)

\frac{d y}{d x} = \frac{( x - 4 )}{y + 5}

What is the normal of y=5x^2,(-2,20) ?
The normal of y=5x^2,(-2,20) is y= 1/20 x+201/10