What is the tangent of y=4.2x-x^2 ?

The tangent of y=4.2x-x^2 is y=(4.2-2a_{0})x+a_{0}^2

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tangent of y=4.2x-x^2

tangent of

y = 4.2 x - x^{2}

y = (4.2 - 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}, 4.2 a_{0} - a_{0}^{2})

Find the slope of

y = 4.2 x - x^{2} : \frac{d y}{d x} = 4.2 - 2 x

Compute the slope of

y = 4.2 x - x^{2}

x = a_{0} : m = 4.2 - 2 a_{0}

Find the line with slope m=

4.2 - 2 a_{0}

and passing through

(a_{0}, 4.2 a_{0} - a_{0}^{2}) : y = (4.2 - 2 a_{0}) x + a_{0}^{2}

y = (4.2 - 2 a_{0}) x + a_{0}^{2}

\int \frac{700 x}{( x + 1 ) ^{\frac{1}{3}}} d x

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

x \to 0 lim (\frac{e ^{x} - 1}{9 x})

\frac{d y}{d x} = 6 x^{2} + 3

\int_{0}^{1} ln (6 x) d x

What is the tangent of y=4.2x-x^2 ?
The tangent of y=4.2x-x^2 is y=(4.2-2a_{0})x+a_{0}^2