What is the tangent of f(x)=x^3+4x^2+5x,\at x=-1 ?

The tangent of f(x)=x^3+4x^2+5x,\at x=-1 is y=-2

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tangent of f(x)=x^3+4x^2+5x,\at x=-1

tangent of

f (x) = x^{3} + 4 x^{2} + 5 x, at x = - 1

y = - 2

Solution steps

Find the tangent point:

(- 1, - 2)

Find the slope of

f (x) = x^{3} + 4 x^{2} + 5 x : \frac{df ( x )}{d x} = 3 x^{2} + 8 x + 5

Compute the slope of

f (x) = x^{3} + 4 x^{2} + 5 x

x = - 1 : m = 0

Find the line with slope m=

0

and passing through

(- 1, - 2) : y = - 2

y = - 2

\int \frac{1}{4 x ^{2} + 12 x + 16} d x

(7, - 5), (- 7, - 5)

\frac{d}{d x} (y (e^{x} - 1))

\int (\frac{x}{5})^{n} d x

\int_{- 1}^{1} 2^{x} d x

What is the tangent of f(x)=x^3+4x^2+5x,\at x=-1 ?
The tangent of f(x)=x^3+4x^2+5x,\at x=-1 is y=-2