What is the normal of y=8xe^x,(0,0) ?

The normal of y=8xe^x,(0,0) is y=-1/8 x

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normal of y=8xe^x,(0,0)

normal of

y = 8 x e^{x}, (0, 0)

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

Solution steps

Find the slope of

y = 8 x e^{x} : \frac{d y}{d x} = 8 (e^{x} + e^{x} x)

Compute the slope of

y = 8 x e^{x}

(0, 0) : 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

(0, 0) : y = - \frac{1}{8} x

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

f (x) = \frac{x ^{2} - 2 e ^{x}}{x + 3 e ^{x}}

\int \frac{1}{3} sec^{2} (8 x) d x

\int sin (2 t) cos (t) d t

y = (5 x - 4) (2 x^{3} - x^{2} + 1)

y^{^{''}} + 3 y^{^{'}} + 0 = 5