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

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

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tangent of x^2-2x-3

tangent of

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

y = x^{2} - 2 x - 3 : \frac{d y}{d x} = 2 x - 2

Compute the slope of

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

x = a_{0} : 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{d y}{d x} = \frac{2 x}{x ^{2} - 9}

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

x \to \infty lim (x n (x))

\int_{0}^{3.5} 130 x d x

\int_{0}^{2} (- x^{2} + 2 x)^{2} d x

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