What is the tangent of f(x)=x^3+3x-6,(1,-2) ?

The tangent of f(x)=x^3+3x-6,(1,-2) is y=6x-8

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tangent of f(x)=x^3+3x-6,(1,-2)

tangent of

f (x) = x^{3} + 3 x - 6, (1, - 2)

y = 6 x - 8

Solution steps

Find the slope of

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

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 = 6

Find the line with slope m=

6

and passing through

(1, - 2) : y = 6 x - 8

y = 6 x - 8

\frac{1}{s - 8}

\int \frac{8}{x ^{- 1} + 4} d x

\int_{0}^{1} \frac{2}{2 - x ^{2}} d x

\int \frac{1}{( x - 4 ) ( x + 3 )} d x

(x^{2} - 9 x + 9) 25 - x^{2}

What is the tangent of f(x)=x^3+3x-6,(1,-2) ?
The tangent of f(x)=x^3+3x-6,(1,-2) is y=6x-8