What is the tangent of f(x)=3x^3-3x^2+7,(2,19) ?

The tangent of f(x)=3x^3-3x^2+7,(2,19) is y=24x-29

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tangent of f(x)=3x^3-3x^2+7,(2,19)

tangent of

f (x) = 3 x^{3} - 3 x^{2} + 7, (2, 19)

y = 24 x - 29

Solution steps

Find the slope of

f (x) = 3 x^{3} - 3 x^{2} + 7 : \frac{df ( x )}{d x} = 9 x^{2} - 6 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 = 24

Find the line with slope m=

24

and passing through

(2, 19) : y = 24 x - 29

y = 24 x - 29

\frac{d}{d x} (3^{x} lo g_{5} (x))

\int_{0}^{x} \frac{1}{1 - x} d x

\frac{d}{d θ} (\frac{1}{θ})

\frac{\partial}{\partial x} (6 e^{x^{3} y})

\int 26 d x

What is the tangent of f(x)=3x^3-3x^2+7,(2,19) ?
The tangent of f(x)=3x^3-3x^2+7,(2,19) is y=24x-29