What is the normal of y=x^4+9e^x,(0,5) ?

The normal of y=x^4+9e^x,(0,5) is Normal line to y=x^4+9e^x, with a tangent of point (0,5)

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normal of y=x^4+9e^x,(0,5)

normal of

y = x^{4} + 9 e^{x}, (0, 5)

Normal line to y = x^{4} + 9 e^{x}, with a tangent point (0, 5)

Solution steps

Slope of the normal line at

(0, 5) : m = \frac{x ^{4} + 9 e ^{x} - 5}{x}

Find the slope of

y = x^{4} + 9 e^{x} : m_{y} = 4 x^{3} + 9 e^{x}

Compute the slope of the normal line:

m = - \frac{1}{4 x ^{3} + 9 e ^{x}}

What is the normal of y=x^4+9e^x,(0,5) ?
The normal of y=x^4+9e^x,(0,5) is Normal line to y=x^4+9e^x, with a tangent of point (0,5)