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}

\frac{\frac{s}{2} + \frac{5}{3}}{s ^{2} + 4 s + 6}

\int_{0}^{9} π (9 x - x^{2}) d x

\frac{x}{3 x ^{2} + 1} - \frac{1}{8}

x \to 3 - lim (\frac{( x ^{2} )}{x - 3})

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