What is the tangent of y=4x-x^2 ?

The tangent of y=4x-x^2 is y=(4-2a_{0})x+a_{0}^2

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tangent of y=4x-x^2

tangent of

y = 4 x - x^{2}

y = (4 - 2 a_{0}) x + a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 4 a_{0} - a_{0}^{2})

Find the slope of

y = 4 x - x^{2} : \frac{d y}{d x} = 4 - 2 x

Compute the slope of

y = 4 x - x^{2}

x = a_{0} : m = 4 - 2 a_{0}

Find the line with slope m=

4 - 2 a_{0}

and passing through

(a_{0}, 4 a_{0} - a_{0}^{2}) : y = (4 - 2 a_{0}) x + a_{0}^{2}

y = (4 - 2 a_{0}) x + a_{0}^{2}

x \cdot sin (x) \cdot d x + cos (y) \cdot d y = 0

\int y^{2} - 2 y + 1 d y

\frac{\partial}{\partial v} (v cos (u))

\int cot^{7} (2 x) d x

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

What is the tangent of y=4x-x^2 ?
The tangent of y=4x-x^2 is y=(4-2a_{0})x+a_{0}^2