What is the tangent of f(x)=12-4e^x,(0,8) ?

The tangent of f(x)=12-4e^x,(0,8) is y=-4x+8

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tangent of f(x)=12-4e^x,(0,8)

tangent of

f (x) = 12 - 4 e^{x}, (0, 8)

y = - 4 x + 8

Solution steps

Find the slope of

f (x) = 12 - 4 e^{x} : \frac{df ( x )}{d x} = - 4 e^{x}

Compute the slope of

f (x) = 12 - 4 e^{x}

(0, 8) : m = - 4

Find the line with slope m=

- 4

and passing through

(0, 8) : y = - 4 x + 8

y = - 4 x + 8

\frac{d}{d z} (\frac{x ln ( \frac{x y}{z} )}{z})

\frac{d}{d x} (3 x + 7 x)

\frac{z}{z + \frac{a}{z}}

\int_{0}^{x^{2} + 3} \frac{1}{x ^{2} + 3} d x

y = 2 x^{3} + 3 x^{2} - 12 x + 3

What is the tangent of f(x)=12-4e^x,(0,8) ?
The tangent of f(x)=12-4e^x,(0,8) is y=-4x+8