What is the tangent of f(x)=8x-x^2 ?

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

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tangent of f(x)=8x-x^2

tangent of

f (x) = 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

f (x) = 8 x - x^{2} : \frac{df ( x )}{d x} = 8 - 2 x

Compute the slope of

f (x) = 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}

x \to 0 + lim (\frac{1}{x + 1})

\frac{\partial}{\partial x} (1 - x^{2} - y^{2})

x \to 1 lim (x - x)

\frac{\partial}{\partial x} (4 x^{- 1})

f (x) = x^{2} + 6 x, at a = 3

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