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

The tangent of f(x)=-7-x^2,\at x=9 is y=-18x+74

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tangent of f(x)=-7-x^2,\at x=9

tangent of

f (x) = - 7 - x^{2}, at x = 9

y = - 18 x + 74

Solution steps

Find the tangent point:

(9, - 88)

Find the slope of

f (x) = - 7 - x^{2} : \frac{df ( x )}{d x} = - 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 = - 18

Find the line with slope m=

- 18

and passing through

(9, - 88) : y = - 18 x + 74

y = - 18 x + 74

\frac{d}{d x} (\frac{x cos ( x ) - 2 π}{x - 2 π})

\frac{d}{d x} (\frac{1 - cos ( x )}{1 - sin ( x )})

\frac{d}{d x} ((3 x^{2} + 1)^{4})

k = 1 \sum \infty \frac{k ^{3}}{3 ^{k}}

y^{^{'}} + 3 y = 2 e^{- 3 x}

What is the tangent of f(x)=-7-x^2,\at x=9 ?
The tangent of f(x)=-7-x^2,\at x=9 is y=-18x+74