What is the tangent of f(x)=x^2-3 ?

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

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tangent of f(x)=x^2-3

tangent of

f (x) = 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

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

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

x \to \infty lim ((1 + \frac{1}{x ^{3}})^{2 x^{2}})

\frac{d}{d x} (e^{\frac{3}{2} x} (\frac{13}{17} cos (2 x) + b sin (2 x)) + \frac{4}{17} e^{2 x})

\int x^{2} - 6 x + 9 d x

\int x (x - 3)^{5} d x

y = 1 + 3 x, 3 \leq x \leq 9

What is the tangent of f(x)=x^2-3 ?
The tangent of f(x)=x^2-3 is y=2a_{0}x-a_{0}^2-3