What is the tangent of f(x)=e^x(9-3x+2x^2),\at x=0 ?

The tangent of f(x)=e^x(9-3x+2x^2),\at x=0 is y=6x+9

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tangent of f(x)=e^x(9-3x+2x^2),\at x=0

tangent of

f (x) = e^{x} (9 - 3 x + 2 x^{2}), at x = 0

y = 6 x + 9

Solution steps

Find the tangent point:

(0, 9)

Find the slope of

f (x) = e^{x} (9 - 3 x + 2 x^{2}) : \frac{df ( x )}{d x} = 2 e^{x} x^{2} + e^{x} x + 6 e^{x}

Compute the slope of

f (x) = e^{x} (9 - 3 x + 2 x^{2})

x = 0 : m = 6

Find the line with slope m=

6

and passing through

(0, 9) : y = 6 x + 9

y = 6 x + 9

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What is the tangent of f(x)=e^x(9-3x+2x^2),\at x=0 ?
The tangent of f(x)=e^x(9-3x+2x^2),\at x=0 is y=6x+9