What is the tangent of y=e^{-4t},\at t=0 ?

The tangent of y=e^{-4t},\at t=0 is y=-4t+1

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tangent of y=e^{-4t},\at t=0

tangent of

y = e^{- 4 t}, at t = 0

y = - 4 t + 1

Solution steps

Find the tangent point:

(0, 1)

Find the slope of

y = e^{- 4 t} : \frac{d y}{d t} = - 4 e^{- 4 t}

Compute the slope of

y = e^{- 4 t}

t = 0 : m = - 4

Find the line with slope m=

- 4

and passing through

(0, 1) : y = - 4 t + 1

y = - 4 t + 1

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What is the tangent of y=e^{-4t},\at t=0 ?
The tangent of y=e^{-4t},\at t=0 is y=-4t+1