What is the tangent of y=2e^x-3x+1,\at x=0 ?

The tangent of y=2e^x-3x+1,\at x=0 is y=-x+3

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tangent of y=2e^x-3x+1,\at x=0

tangent of

y = 2 e^{x} - 3 x + 1, at x = 0

y = - x + 3

Solution steps

Find the tangent point:

(0, 3)

Find the slope of

y = 2 e^{x} - 3 x + 1 : \frac{d y}{d x} = 2 e^{x} - 3

Compute the slope of

y = 2 e^{x} - 3 x + 1

x = 0 : m = - 1

Find the line with slope m=

- 1

and passing through

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

y = - x + 3

4 x^{2} - 12 x + 5

\int - 2 x^{2} sin (π x) d x

\int \frac{y}{y} d y

f (x) = - 1 - 2 cos (x), at x = - \frac{π}{4}

y^{^{'}} = (x - 3) (y + 1)^{\frac{2}{3}}, y (0) = - 1

What is the tangent of y=2e^x-3x+1,\at x=0 ?
The tangent of y=2e^x-3x+1,\at x=0 is y=-x+3