What is the tangent of f(x)=7e^{x^2-5x+8},\at x=0 ?

The tangent of f(x)=7e^{x^2-5x+8},\at x=0 is y=-35e^8x+7e^8

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tangent of f(x)=7e^{x^2-5x+8},\at x=0

tangent of

f (x) = 7 e^{x^{2} - 5 x + 8}, at x = 0

y = - 35 e^{8} x + 7 e^{8}

Solution steps

Find the tangent point:

(0, 7 e^{8})

Find the slope of

f (x) = 7 e^{x^{2} - 5 x + 8} : \frac{df ( x )}{d x} = 7 e^{x^{2} - 5 x + 8} (2 x - 5)

Compute the slope of

f (x) = 7 e^{x^{2} - 5 x + 8}

x = 0 : m = - 35 e^{8}

Find the line with slope m=

- 35 e^{8}

and passing through

(0, 7 e^{8}) : y = - 35 e^{8} x + 7 e^{8}

y = - 35 e^{8} x + 7 e^{8}

(3 x^{2} - 6 x + 6) e^{- 8 x}

y \frac{d y}{d x} = (y^{2} - y) (2 x + 5)

a^{2} - y^{2}

(x^{2} + 9) y^{^{'}} + x y = 0

\frac{d}{d y} (x cos (y))

What is the tangent of f(x)=7e^{x^2-5x+8},\at x=0 ?
The tangent of f(x)=7e^{x^2-5x+8},\at x=0 is y=-35e^8x+7e^8