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

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

d er i v a t i v e x^{2} + 2 x + 4

in v erse l a pl a ce \frac{1}{s ( s + 2 ) ( s + 6 )}

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

t an g e n t f (x) = 4 x e^{x}, (0, 0)

\int x (4 + x)^{10} d x

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