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

The tangent of f(x)=x^2+2x+1,\at x=-3 is y=-4x-8

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

tangent of

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

y = - 4 x - 8

Solution steps

Find the tangent point:

(- 3, 4)

Find the slope of

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

Compute the slope of

f (x) = x^{2} + 2 x + 1

x = - 3 : m = - 4

Find the line with slope m=

- 4

and passing through

(- 3, 4) : y = - 4 x - 8

y = - 4 x - 8

(8 - 4) (89)

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

x^{2} + y^{2} = 169, (5, - 12)

\int \frac{x - arctan ( x )}{x} d x

f (x) = lo g_{10} (x + 1)

What is the tangent of f(x)=x^2+2x+1,\at x=-3 ?
The tangent of f(x)=x^2+2x+1,\at x=-3 is y=-4x-8