What is the tangent of y=4x-x^2(1.3) ?

The tangent of y=4x-x^2(1.3) is y=(4-2.6a_{0})x+1.3a_{0}^2

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tangent of y=4x-x^2(1.3)

tangent of

y = 4 x - x^{2} (1.3)

y = (4 - 2.6 a_{0}) x + 1.3 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} \cdot 1.3)

Find the slope of

y = 4 x - x^{2} 1.3 : \frac{d y}{d x} = 4 - 2.6 x

Compute the slope of

y = 4 x - x^{2} 1.3

x = a_{0} : m = 4 - 2.6 a_{0}

Find the line with slope m=

4 - 2.6 a_{0}

and passing through

(a_{0}, 4 a_{0} - a_{0}^{2} 1.3) : y = (4 - 2.6 a_{0}) x + 1.3 a_{0}^{2}

y = (4 - 2.6 a_{0}) x + 1.3 a_{0}^{2}

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y^{^{'}} = x^{2} + x, y (1) = 3

\int sin^{3} (6 x) cos^{2} (6 x) d x

\int_{0}^{3} π (\frac{1}{x + 5})^{2} d x

What is the tangent of y=4x-x^2(1.3) ?
The tangent of y=4x-x^2(1.3) is y=(4-2.6a_{0})x+1.3a_{0}^2