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

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

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tangent of 1-x-x^2

tangent of

1 - x - x^{2}

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

y = 1 - x - x^{2} : \frac{d y}{d x} = - 2 x - 1

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} - 1

Find the line with slope m=

- 2 a_{0} - 1

and passing through

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

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

\int tan^{3} (x) sec^{9} (x) d x

\int x^{2} (x + 1)^{3} d x

(3 y^{2} + 2 x y) - (2 x y + x^{2}) \frac{d y}{d x} = 0

3 (s^{3} - s^{2})

f (r) = \frac{6}{r ^{3}}

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