What is the tangent of y=8x-x^2 ?

The tangent of y=8x-x^2 is y=(8-2a_{0})x+a_{0}^2

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tangent of y=8x-x^2

tangent of

y = 8 x - x^{2}

y = (8 - 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}, 8 a_{0} - a_{0}^{2})

Find the slope of

y = 8 x - x^{2} : \frac{d y}{d x} = 8 - 2 x

Compute the slope of

y = 8 x - x^{2}

x = a_{0} : m = 8 - 2 a_{0}

Find the line with slope m=

8 - 2 a_{0}

and passing through

(a_{0}, 8 a_{0} - a_{0}^{2}) : y = (8 - 2 a_{0}) x + a_{0}^{2}

y = (8 - 2 a_{0}) x + a_{0}^{2}

\int (x + 4)^{\frac{1}{2}} d x

\frac{d y}{d x} = (2 x)

x \to \infty lim (\frac{4 x ^{2}}{e ^{4 x}})

\frac{\partial}{\partial x} (x^{3} + x^{2} \cdot x^{3} - 2 y^{2})

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

What is the tangent of y=8x-x^2 ?
The tangent of y=8x-x^2 is y=(8-2a_{0})x+a_{0}^2