What is the tangent of f(x)=-6x^2-9,\at x=0 ?

The tangent of f(x)=-6x^2-9,\at x=0 is y=-9

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tangent of f(x)=-6x^2-9,\at x=0

tangent of

f (x) = - 6 x^{2} - 9, at x = 0

y = - 9

Solution steps

Find the tangent point:

(0, - 9)

Find the slope of

f (x) = - 6 x^{2} - 9 : \frac{df ( x )}{d x} = - 12 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 = 0

Find the line with slope m=

0

and passing through

(0, - 9) : y = - 9

y = - 9

x \to - 4 - lim (f (x))

\frac{d}{d x} (e^{x} \cdot 2^{x})

x \to 0 lim (\frac{e ^{x} - 1}{sin ( 4 x )})

e^{- t} sin (t)

\frac{\partial}{\partial x} (6 y - 2 x)

What is the tangent of f(x)=-6x^2-9,\at x=0 ?
The tangent of f(x)=-6x^2-9,\at x=0 is y=-9