What is the tangent of y=x^2-3 ?

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

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tangent of y=x^2-3

tangent of

y = x^{2} - 3

y = 2 a_{0} x - a_{0}^{2} - 3

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{2} - 3)

Find the slope of

y = x^{2} - 3 : \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} - 3) : y = 2 a_{0} x - a_{0}^{2} - 3

y = 2 a_{0} x - a_{0}^{2} - 3

y^{^{'}} e^{- x} = y^{2} + 9

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

2 e^{x^{2} - 5 x}

x \to - 3 lim (\frac{x ^{2} + 9}{x + 3})

\frac{d}{d x} (- \frac{1}{2} x^{- \frac{3}{2}})