What is the normal of y=8x^2+8,(2,40) ?

The normal of y=8x^2+8,(2,40) is y=-1/32 x+641/16

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

normal of y=8x^2+8,(2,40)

normal of

y = 8 x^{2} + 8, (2, 40)

y = - \frac{1}{32} x + \frac{641}{16}

Solution steps

Find the slope of

y = 8 x^{2} + 8 : \frac{d y}{d x} = 16 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 = 32

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{32}

Find the line with slope m=

- \frac{1}{32}

and passing through

(2, 40) : y = - \frac{1}{32} x + \frac{641}{16}

y = - \frac{1}{32} x + \frac{641}{16}

\int (1 + x) e^{x} d x

d er i v a t i v e 2^{x} ln (2)

d er i v a t i v e - sin (x) - 2 sec^{2} (x) tan (x)

\int \frac{1}{7 x ^{5}} d x

y^{^{'}} + 2 x y = x e^{- x^{2}}, y (0) = 1

What is the normal of y=8x^2+8,(2,40) ?
The normal of y=8x^2+8,(2,40) is y=-1/32 x+641/16