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

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

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

tangent of

f (x) = x^{2} - 2 x - 3

y = (2 a_{0} - 2) 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} - 2 a_{0} - 3)

Find the slope of

f (x) = x^{2} - 2 x - 3 : \frac{df ( x )}{d x} = 2 x - 2

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} - 2

Find the line with slope m=

2 a_{0} - 2

and passing through

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

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

\frac{\partial}{\partial x} (x^{2} e^{x + y} + 2 y cos (x y) - x y^{2} sin (x y))

x^{2}, \frac{2}{( x ^{2} + 1 )}

\frac{\partial}{\partial x} (cos (x + 7 y))

\frac{1}{x} \frac{d y}{d x} + 2 y - e^{- x^{2}} = 0

x \to 0 lim (\frac{\frac{6}{x + 6} - 1}{x})

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