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

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

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tangent of x^2+2x

tangent of

x^{2} + 2 x

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

Find the line with slope m=

2 a_{0} + 2

and passing through

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

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

\int (6 x^{2} - 8 x + 2) d x

y^{^{'''}} + y^{^{'}} = cot (x)

y^{^{''}} + 5 \frac{d y}{d x} + 6 y = 2 e^{x}

n = 1 \sum \infty \frac{9 n}{2 n + 1}

f (x) = \frac{1}{arctan ( x ^{2} )}