What is the tangent of 5-x-x^2 ?

The tangent of 5-x-x^2 is y=(-2a_{0}-1)x+a_{0}^2+5

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tangent of 5-x-x^2

tangent of

5 - x - x^{2}

y = (- 2 a_{0} - 1) x + a_{0}^{2} + 5

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 5 - a_{0} - a_{0}^{2})

Find the slope of

y = 5 - x - x^{2} : \frac{d y}{d x} = - 2 x - 1

Compute the slope of

y = 5 - x - x^{2}

x = a_{0} : m = - 2 a_{0} - 1

Find the line with slope m=

- 2 a_{0} - 1

and passing through

(a_{0}, 5 - a_{0} - a_{0}^{2}) : y = (- 2 a_{0} - 1) x + a_{0}^{2} + 5

y = (- 2 a_{0} - 1) x + a_{0}^{2} + 5

\int \frac{5 x}{x ( x - 1 )} d x

\int 11 ln (x) d x

\frac{2}{( s + 2 ) ^{2}}

\frac{d}{d x} (\frac{2 x}{9 - cot ( x )})

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

What is the tangent of 5-x-x^2 ?
The tangent of 5-x-x^2 is y=(-2a_{0}-1)x+a_{0}^2+5