What is the tangent of f(x)=e^x(-7-3x+5x^2),\at x=0 ?

The tangent of f(x)=e^x(-7-3x+5x^2),\at x=0 is y=-10x-7

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tangent of f(x)=e^x(-7-3x+5x^2),\at x=0

tangent of

f (x) = e^{x} (- 7 - 3 x + 5 x^{2}), at x = 0

y = - 10 x - 7

Solution steps

Find the tangent point:

(0, - 7)

Find the slope of

f (x) = e^{x} (- 7 - 3 x + 5 x^{2}) : \frac{df ( x )}{d x} = 5 e^{x} x^{2} + 7 e^{x} x - 10 e^{x}

Compute the slope of

f (x) = e^{x} (- 7 - 3 x + 5 x^{2})

x = 0 : m = - 10

Find the line with slope m=

- 10

and passing through

(0, - 7) : y = - 10 x - 7

y = - 10 x - 7

0.5 (x + a)^{p} - x

\int sin (3 x - 7) d x

\frac{d}{d t} (9 cos (t))

\int 8 sec^{4} (x) d x

\int 1 + sin (θ) d θ

What is the tangent of f(x)=e^x(-7-3x+5x^2),\at x=0 ?
The tangent of f(x)=e^x(-7-3x+5x^2),\at x=0 is y=-10x-7