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

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

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

tangent of

f (x) = - 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

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

Compute the slope of

f (x) = - 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

e^{r x}

\frac{d}{d x} ((x^{2} - 3 x)^{2})

\int_{1}^{\infty} \frac{4}{1 + x ^{2}} d x

\int \frac{1}{2 cos ( x )} d x

\frac{110}{x ^{12}}

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