What is the tangent of y=x^3-3x+2,(3,20) ?

The tangent of y=x^3-3x+2,(3,20) is y=24x-52

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tangent of y=x^3-3x+2,(3,20)

tangent of

y = x^{3} - 3 x + 2, (3, 20)

y = 24 x - 52

Solution steps

Find the slope of

y = x^{3} - 3 x + 2 : \frac{d y}{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 = 24

Find the line with slope m=

24

and passing through

(3, 20) : y = 24 x - 52

y = 24 x - 52

y = \frac{1}{( 4 x ^{2} + 3 ) ^{5}}

\int (2 x^{3} + 5 x^{2} - 2 x + 7) d x

\frac{\partial}{\partial y} (x y + ln (\frac{x}{y}))

\int x^{4} d x

\int - x^{- 2} d x

What is the tangent of y=x^3-3x+2,(3,20) ?
The tangent of y=x^3-3x+2,(3,20) is y=24x-52