What is the tangent of f(x)=4x-x^2 ?

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

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tangent of f(x)=4x-x^2

tangent of

f (x) = 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

f (x) = 4 x - x^{2} : \frac{df ( x )}{d x} = 4 - 2 x

Compute the slope of

f (x) = 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}

\frac{d}{d x} (50 x)

\int_{25}^{50} 10 d x

6 x^{2} - 3 x + 4

\int \frac{x + 6}{x - 3} d x

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

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