What is the tangent of f(x)=x^2-4 ?

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

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tangent of f(x)=x^2-4

tangent of

f (x) = 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

f (x) = x^{2} - 4 : \frac{df ( x )}{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

- 2 y^{^{''}} + 9 y = 0, y (- 2) = 2, y^{^{'}} (2) = 1

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

f (x) = 2 x^{3} - x^{2} + 1, (1, 2)

\frac{( 4 x - 2 ) ^{4}}{( - 5 x + 1 ) ^{5}}

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

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