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

The tangent of f(x)=x^3+x^2-x-3,\at x=-2 is y=7x+9

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

tangent of

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

y = 7 x + 9

Solution steps

Find the tangent point:

(- 2, - 5)

Find the slope of

f (x) = x^{3} + x^{2} - x - 3 : \frac{df ( x )}{d x} = 3 x^{2} + 2 x - 1

Compute the slope of

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

x = - 2 : m = 7

Find the line with slope m=

7

and passing through

(- 2, - 5) : y = 7 x + 9

y = 7 x + 9

3 x^{2} - 4 x + 5

f (x) = 3 sin (x) - 33 cos (x)

6 + sec (x) sin (x)

2 x^{2} + 2 x

n = 1 \sum \infty \frac{2 ^{n}}{n + 1}

What is the tangent of f(x)=x^3+x^2-x-3,\at x=-2 ?
The tangent of f(x)=x^3+x^2-x-3,\at x=-2 is y=7x+9