What is the tangent of y=x^2-4 ?

The tangent of y=x^2-4 is y=2a_{0}x-a_{0}^2-4

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tangent of y=x^2-4

tangent of

y = x^{2} - 4

y = 2 a_{0} x - a_{0}^{2} - 4

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

y = x^{2} - 4 : \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} - 4) : y = 2 a_{0} x - a_{0}^{2} - 4

y = 2 a_{0} x - a_{0}^{2} - 4

\frac{d}{d x} (4^{- 3} (\frac{2}{x ^{7}} + 3 x)^{4})

\frac{d}{d x} ((9 x + 4) x^{2})

(40 q^{2} + q + 500)^{^{'}}

x \to - 1 + lim (\frac{3}{x + 1})

y = e^{x} + 1