What is the tangent of f(x)=7e^{x^2-9x+8},\at x=0 ?

The tangent of f(x)=7e^{x^2-9x+8},\at x=0 is y=-63e^8x+7e^8

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tangent of f(x)=7e^{x^2-9x+8},\at x=0

tangent of

f (x) = 7 e^{x^{2} - 9 x + 8}, at x = 0

y = - 63 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} - 9 x + 8} : \frac{df ( x )}{d x} = 7 e^{x^{2} - 9 x + 8} (2 x - 9)

Compute the slope of

f (x) = 7 e^{x^{2} - 9 x + 8}

x = 0 : m = - 63 e^{8}

Find the line with slope m=

- 63 e^{8}

and passing through

(0, 7 e^{8}) : y = - 63 e^{8} x + 7 e^{8}

y = - 63 e^{8} x + 7 e^{8}

y = 164 4 x^{4} + 4

y^{^{''}} + 6 y^{^{'}} + 5 y = 10

\int \frac{1}{h} d h

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

\int x^{2} \cdot e^{- x^{3}} d x

What is the tangent of f(x)=7e^{x^2-9x+8},\at x=0 ?
The tangent of f(x)=7e^{x^2-9x+8},\at x=0 is y=-63e^8x+7e^8