What is the tangent of y=3e^{-x},(0,3) ?

The tangent of y=3e^{-x},(0,3) is y=-3x+3

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

tangent of

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

y = - 3 x + 3

Solution steps

Find the slope of

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

Compute the slope of

y = 3 e^{- x}

(0, 3) : m = - 3

Find the line with slope m=

- 3

and passing through

(0, 3) : y = - 3 x + 3

y = - 3 x + 3

\frac{d y}{d x} + \frac{3 x y}{x ^{2} + 1} = \frac{6 x}{x ^{2} + 1}

\int \frac{1}{4 + 8 x} d x

\int 7 x^{\frac{2}{5}} + 3 x^{- \frac{4}{5}} d x

y^{^{'''}} - 5 y^{^{''}} + 3 y^{^{'}} + 9 y = 0

x \to - 10 lim (- 15)

What is the tangent of y=3e^{-x},(0,3) ?
The tangent of y=3e^{-x},(0,3) is y=-3x+3