What is the tangent of f(x)=2x^3-2x^2+8,(3,44) ?

The tangent of f(x)=2x^3-2x^2+8,(3,44) is y=42x-82

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tangent of f(x)=2x^3-2x^2+8,(3,44)

tangent of

f (x) = 2 x^{3} - 2 x^{2} + 8, (3, 44)

y = 42 x - 82

Solution steps

Find the slope of

f (x) = 2 x^{3} - 2 x^{2} + 8 : \frac{df ( x )}{d x} = 6 x^{2} - 4 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 = 42

Find the line with slope m=

42

and passing through

(3, 44) : y = 42 x - 82

y = 42 x - 82

y = (x - 1)^{\frac{3}{2}} + \frac{e ^{x - 2}}{2}

\frac{d}{d t} (t^{2} + 4)

\int \frac{ln ( 10 x )}{3 x} d x

\int - 2 x e^{- 2 x} d x

\frac{1}{2} t^{2} - 6 t + 23

What is the tangent of f(x)=2x^3-2x^2+8,(3,44) ?
The tangent of f(x)=2x^3-2x^2+8,(3,44) is y=42x-82