What is the tangent of y=x^3-2x+2(3.23) ?

The tangent of y=x^3-2x+2(3.23) is y=(3a_{0}^2-2)x-2a_{0}^3+6.46

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tangent of y=x^3-2x+2(3.23)

tangent of

y = x^{3} - 2 x + 2 (3.23)

y = (3 a_{0}^{2} - 2) x - 2 a_{0}^{3} + 6.46

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{3} - 2 a_{0} + 6.46)

Find the slope of

y = x^{3} - 2 x + 2 \cdot 3.23 : \frac{d y}{d x} = 3 x^{2} - 2

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 = 3 a_{0}^{2} - 2

Find the line with slope m=

3 a_{0}^{2} - 2

and passing through

(a_{0}, a_{0}^{3} - 2 a_{0} + 6.46) : y = (3 a_{0}^{2} - 2) x - 2 a_{0}^{3} + 6.46

y = (3 a_{0}^{2} - 2) x - 2 a_{0}^{3} + 6.46

What is the tangent of y=x^3-2x+2(3.23) ?
The tangent of y=x^3-2x+2(3.23) is y=(3a_{0}^2-2)x-2a_{0}^3+6.46