What is the tangent of 1-2x-3x^2 ?

The tangent of 1-2x-3x^2 is y=(-6a_{0}-2)x+3a_{0}^2+1

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tangent of 1-2x-3x^2

tangent of

1 - 2 x - 3 x^{2}

y = (- 6 a_{0} - 2) x + 3 a_{0}^{2} + 1

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 1 - 2 a_{0} - 3 a_{0}^{2})

Find the slope of

y = 1 - 2 x - 3 x^{2} : \frac{d y}{d x} = - 6 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 = - 6 a_{0} - 2

Find the line with slope m=

- 6 a_{0} - 2

and passing through

(a_{0}, 1 - 2 a_{0} - 3 a_{0}^{2}) : y = (- 6 a_{0} - 2) x + 3 a_{0}^{2} + 1

y = (- 6 a_{0} - 2) x + 3 a_{0}^{2} + 1

x \frac{d y}{d x} = x^{3} + 11 x^{3} y

\frac{tan ( x ) - 1}{sec ( x )}

y y^{^{'}} = 1

\int \frac{e ^{3 x}}{3 x} d x

t \to 0 lim (t \cdot ln (t))

What is the tangent of 1-2x-3x^2 ?
The tangent of 1-2x-3x^2 is y=(-6a_{0}-2)x+3a_{0}^2+1