What is the tangent of 3x^2+8 ?

The tangent of 3x^2+8 is y=6a_{0}x-3a_{0}^2+8

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tangent of 3x^2+8

tangent

y = 6 a_{0} x - 3 a_{0}^{2} + 8

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 3 a_{0}^{2} + 8)

Find the slope of

y = 3 x^{2} + 8 : \frac{d y}{d x} = 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 = 6 a_{0}

Find the line with slope m=

6 a_{0}

and passing through

(a_{0}, 3 a_{0}^{2} + 8) : y = 6 a_{0} x - 3 a_{0}^{2} + 8

y = 6 a_{0} x - 3 a_{0}^{2} + 8