What is the tangent of y=x^2-4x-3,(3,-6) ?

The tangent of y=x^2-4x-3,(3,-6) is y=2x-12

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tangent of y=x^2-4x-3,(3,-6)

tangent of

y = x^{2} - 4 x - 3, (3, - 6)

y = 2 x - 12

Solution steps

Find the slope of

y = x^{2} - 4 x - 3 : \frac{d y}{d x} = 2 x - 4

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

(3, - 6) : y = 2 x - 12

y = 2 x - 12

(\frac{1}{x}), (\frac{1}{x ^{2}}), 2

y = 4 x^{2} + 3 y^{2} + 8 x - 6 y - 45, (- 2, - 3)

\frac{4 x ^{2} + 6 x + 4}{x}

f (x) = \frac{x ^{3} e ^{x}}{1 - x}

\frac{\partial}{\partial x} (ln (y))

What is the tangent of y=x^2-4x-3,(3,-6) ?
The tangent of y=x^2-4x-3,(3,-6) is y=2x-12