What is the tangent of (2x+1)^2(0.1) ?

The tangent of (2x+1)^2(0.1) is y=0.4(2a_{0}+1)x+(-4a_{0}^2+1)/(10)

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tangent of (2x+1)^2(0.1)

tangent of

(2 x + 1)^{2} (0.1)

y = 0.4 (2 a_{0} + 1) x + \frac{- 4 a _{0}^{2} + 1}{10}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, (2 a_{0} + 1)^{2} \cdot 0.1)

Find the slope of

y = (2 x + 1)^{2} 0.1 : \frac{d y}{d x} = 0.4 (2 x + 1)

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 = 0.4 (2 a_{0} + 1)

Find the line with slope m=

0.4 (2 a_{0} + 1)

and passing through

(a_{0}, (2 a_{0} + 1)^{2} 0.1) : y = 0.4 (2 a_{0} + 1) x + \frac{- 4 a _{0}^{2} + 1}{10}

y = 0.4 (2 a_{0} + 1) x + \frac{- 4 a _{0}^{2} + 1}{10}

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What is the tangent of (2x+1)^2(0.1) ?
The tangent of (2x+1)^2(0.1) is y=0.4(2a_{0}+1)x+(-4a_{0}^2+1)/(10)