What is the normal of y=x^4+8e^x,(0,8) ?

The normal of y=x^4+8e^x,(0,8) is y=-1/8 x+8

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normal of y=x^4+8e^x,(0,8)

normal of

y = x^{4} + 8 e^{x}, (0, 8)

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

Solution steps

Find the slope of

y = x^{4} + 8 e^{x} : \frac{d y}{d x} = 4 x^{3} + 8 e^{x}

Compute the slope of

y = x^{4} + 8 e^{x}

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

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

\int cos (2 x + 3) d x

\frac{d}{d x} (4 - x^{3})

\frac{d y}{d t} = \frac{t}{t ^{2} y + y}, y (0) = 7

x \to 7 - lim (∣ x - 7 ∣)

n = 1 \sum \infty \frac{7}{n 8 ^{n}}

What is the normal of y=x^4+8e^x,(0,8) ?
The normal of y=x^4+8e^x,(0,8) is y=-1/8 x+8