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

The tangent of f(x)=x-x^2,\at x=-4 is y=9x+16

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tangent of f(x)=x-x^2,\at x=-4

tangent of

f (x) = x - x^{2}, at x = - 4

y = 9 x + 16

Solution steps

Find the tangent point:

(- 4, - 20)

Find the slope of

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

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 = 9

Find the line with slope m=

9

and passing through

(- 4, - 20) : y = 9 x + 16

y = 9 x + 16

\int cos (x) cos^{2} (x) d x

3 x + 4

f (x) = \frac{( x + 2 )}{( x + 4 )}

\frac{d}{d x} (\frac{e ^{2 x} + e ^{- 2 x}}{4})

\frac{d}{d x} (3 x^{2} - 8 x)

What is the tangent of f(x)=x-x^2,\at x=-4 ?
The tangent of f(x)=x-x^2,\at x=-4 is y=9x+16