What is the tangent of 4x^2+2x ?

The tangent of 4x^2+2x is y=(8a_{0}+2)x-4a_{0}^2

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tangent of 4x^2+2x

tangent of

4 x^{2} + 2 x

y = (8 a_{0} + 2) x - 4 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 4 a_{0}^{2} + 2 a_{0})

Find the slope of

y = 4 x^{2} + 2 x : \frac{d y}{d x} = 8 x + 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 = 8 a_{0} + 2

Find the line with slope m=

8 a_{0} + 2

and passing through

(a_{0}, 4 a_{0}^{2} + 2 a_{0}) : y = (8 a_{0} + 2) x - 4 a_{0}^{2}

y = (8 a_{0} + 2) x - 4 a_{0}^{2}

y = 5 x - x^{2}

\frac{d}{d x} ((x - 1) e^{x})

\frac{d}{d x} (\frac{ln ( 4 x )}{x})

\frac{d}{d x} (x e^{1})

\int \frac{2}{x ^{2}} cos (\frac{2}{x}) d x

What is the tangent of 4x^2+2x ?
The tangent of 4x^2+2x is y=(8a_{0}+2)x-4a_{0}^2