What is the normal of y=3e^x-7x,(0,3) ?

The normal of y=3e^x-7x,(0,3) is y= 1/4 x+3

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normal of y=3e^x-7x,(0,3)

normal of

y = 3 e^{x} - 7 x, (0, 3)

y = \frac{1}{4} x + 3

Solution steps

Find the slope of

y = 3 e^{x} - 7 x : \frac{d y}{d x} = 3 e^{x} - 7

Compute the slope of

y = 3 e^{x} - 7 x

(0, 3) : m = - 4

Compute the slope of the perpendicular line:

m_{p} = \frac{1}{4}

Find the line with slope m=

\frac{1}{4}

and passing through

(0, 3) : y = \frac{1}{4} x + 3

y = \frac{1}{4} x + 3

\frac{2 x - 1}{x + 2}

ln (sin^{2} (x))

\frac{d y}{d x} = x + 2 x y

\frac{d}{d x} ((3 x + 1)^{3})

n = 2 \sum \infty \frac{1}{2 n \cdot ln ( n )}

What is the normal of y=3e^x-7x,(0,3) ?
The normal of y=3e^x-7x,(0,3) is y= 1/4 x+3