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}

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 = 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

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