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

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

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tangent of y=x^2+2

tangent of

y = x^{2} + 2

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

Find the line with slope m=

2 a_{0}

and passing through

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

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

\int \frac{2}{x ^{2} + 16} d x

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

\frac{5 u ^{2}}{( u ^{2} + u ) ^{3}}

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\int \frac{1}{( x + 1 ) ( x ^{2} + 2 x + 2 ) ^{2}} d x