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

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

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tangent of x^2-3x-2

tangent of

x^{2} - 3 x - 2

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

Compute the slope of

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

x = a_{0} : m = 2 a_{0} - 3

Find the line with slope m=

2 a_{0} - 3

and passing through

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

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

\frac{d}{d x} (arccos (\frac{x + 1}{3}))

y^{^{'}} = x + y, y (0) = 0

\int u^{2} d u

\int_{0}^{3} 3 x d x

\frac{d}{d x} (2 - 3 x^{2})

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