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

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

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tangent of-x^2-x+3

tangent of

- x^{2} - x + 3

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

y = - x^{2} - x + 3 : \frac{d y}{d x} = - 2 x - 1

Compute the slope of

y = - x^{2} - x + 3

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

Find the line with slope m=

- 2 a_{0} - 1

and passing through

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

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

y^{^{''}} - 12 y^{^{'}} + 85 y = 0

\frac{d}{d x} (\frac{x ^{2} + 2 x + 1}{x ^{2} - 1})

f (x) = \frac{1}{x ^{6}}

\frac{\partial}{\partial y} (y^{2} + y)

\int 10 tan^{3} (x) sec (x) d x

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