What is the normal of y=x^4+9e^x,(0,9) ?

The normal of y=x^4+9e^x,(0,9) is y=-1/9 x+9

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normal of y=x^4+9e^x,(0,9)

normal of

y = x^{4} + 9 e^{x}, (0, 9)

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

Solution steps

Find the slope of

y = x^{4} + 9 e^{x} : \frac{d y}{d x} = 4 x^{3} + 9 e^{x}

Compute the slope of

y = x^{4} + 9 e^{x}

(0, 9) : m = 9

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{9}

Find the line with slope m=

- \frac{1}{9}

and passing through

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

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

9 x e^{- k x}

y = 4 - x^{2}, y = 14 - 7 x

\int 8 cos (x) + 3 x - 8 d x

y = ln (\frac{1}{( x - 2 ) ^{3}})

- 5 e^{4 t}

What is the normal of y=x^4+9e^x,(0,9) ?
The normal of y=x^4+9e^x,(0,9) is y=-1/9 x+9