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

The tangent of f(x)=-5x^2-3x,\at x=-3 is y=27x+45

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tangent of f(x)=-5x^2-3x,\at x=-3

tangent of

f (x) = - 5 x^{2} - 3 x, at x = - 3

y = 27 x + 45

Solution steps

Find the tangent point:

(- 3, - 36)

Find the slope of

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

Compute the slope of

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

x = - 3 : m = 27

Find the line with slope m=

27

and passing through

(- 3, - 36) : y = 27 x + 45

y = 27 x + 45

\frac{d}{d x} (\frac{3 x + 1}{x ^{\frac{2}{3}}})

\int e^{\frac{- 1}{x}} d x

10 \cdot 2^{- 0.274 x}

\frac{2 x}{7 - tan ( x )}

f (x) = 4 - x, x = - 2, x = 3, y = 0

What is the tangent of f(x)=-5x^2-3x,\at x=-3 ?
The tangent of f(x)=-5x^2-3x,\at x=-3 is y=27x+45