What is the tangent of-x^3-5x^2+2,\at x=-1 ?

The tangent of-x^3-5x^2+2,\at x=-1 is y=7x+5

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tangent of-x^3-5x^2+2,\at x=-1

tangent of

- x^{3} - 5 x^{2} + 2, at x = - 1

y = 7 x + 5

Solution steps

Find the tangent point:

(- 1, - 2)

Find the slope of

y = - x^{3} - 5 x^{2} + 2 : \frac{d y}{d x} = - 3 x^{2} - 10 x

Compute the slope of

y = - x^{3} - 5 x^{2} + 2

x = - 1 : m = 7

Find the line with slope m=

7

and passing through

(- 1, - 2) : y = 7 x + 5

y = 7 x + 5

\int cos (x) + 2 d x

\frac{d}{d x} (6 arcsin (x^{2}))

f (x) = x - \frac{1}{x ^{2}}

\int 9 + 9 tan^{2} (x) d x

\frac{d}{d x} (x (7 x + 9)^{\frac{1}{3}})

What is the tangent of-x^3-5x^2+2,\at x=-1 ?
The tangent of-x^3-5x^2+2,\at x=-1 is y=7x+5