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

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

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tangent of y=x^2-2x-3

tangent of

y = 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{x y}{8}

x^{2} + 2 x + 4

\frac{1}{s ( s + 2 ) ( s + 6 )}

\int_{- 1}^{1} \frac{1}{x ^{2} - 1} d x

n = 0 \sum \infty 1800 (\frac{1}{3})^{n}

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