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

The tangent of f(x)=((x+1)/(x+3))^2,\at x=-2 is y=-4x-7

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

tangent of

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

y = - 4 x - 7

Solution steps

Find the tangent point:

(- 2, 1)

Find the slope of

f (x) = (\frac{x + 1}{x + 3})^{2} : \frac{df ( x )}{d x} = \frac{4 ( x + 1 )}{( x + 3 ) ^{3}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 4

Find the line with slope m=

- 4

and passing through

(- 2, 1) : y = - 4 x - 7

y = - 4 x - 7

What is the tangent of f(x)=((x+1)/(x+3))^2,\at x=-2 ?
The tangent of f(x)=((x+1)/(x+3))^2,\at x=-2 is y=-4x-7