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

The tangent of y=x^2-2x-3,(2,-3) is y=2x-7

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tangent of y=x^2-2x-3,(2,-3)

tangent of

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

y = 2 x - 7

Solution steps

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

Find the line with slope m=

2

and passing through

(2, - 3) : y = 2 x - 7

y = 2 x - 7

n = 0 \sum \infty c n (x - 8)^{n}

d er i v a t i v e \frac{x}{3 x ^{2} + 1} - \frac{1}{8}

\frac{\partial}{\partial t} (\frac{t}{2})

\frac{d}{d x} (2 x^{3} - 6)

\frac{d}{d x} (24 x + sin (x))

What is the tangent of y=x^2-2x-3,(2,-3) ?
The tangent of y=x^2-2x-3,(2,-3) is y=2x-7