What is the tangent of y=8x-x^2(1.7) ?

The tangent of y=8x-x^2(1.7) is y=(8-3.4a_{0})x+1.7a_{0}^2

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tangent of y=8x-x^2(1.7)

tangent of

y = 8 x - x^{2} (1.7)

y = (8 - 3.4 a_{0}) x + 1.7 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} \cdot 1.7)

Find the slope of

y = 8 x - x^{2} 1.7 : \frac{d y}{d x} = 8 - 3.4 x

Compute the slope of

y = 8 x - x^{2} 1.7

x = a_{0} : m = 8 - 3.4 a_{0}

Find the line with slope m=

8 - 3.4 a_{0}

and passing through

(a_{0}, 8 a_{0} - a_{0}^{2} 1.7) : y = (8 - 3.4 a_{0}) x + 1.7 a_{0}^{2}

y = (8 - 3.4 a_{0}) x + 1.7 a_{0}^{2}

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What is the tangent of y=8x-x^2(1.7) ?
The tangent of y=8x-x^2(1.7) is y=(8-3.4a_{0})x+1.7a_{0}^2