What is the tangent of-0.025x^2+9x ?

The tangent of-0.025x^2+9x is y=(-0.05a_{0}+9)x+0.025a_{0}^2

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tangent of-0.025x^2+9x

tangent of

- 0.025 x^{2} + 9 x

y = (- 0.05 a_{0} + 9) x + 0.025 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - 0.025 a_{0}^{2} + 9 a_{0})

Find the slope of

y = - 0.025 x^{2} + 9 x : \frac{d y}{d x} = - 0.05 x + 9

Compute the slope of

y = - 0.025 x^{2} + 9 x

x = a_{0} : m = - 0.05 a_{0} + 9

Find the line with slope m=

- 0.05 a_{0} + 9

and passing through

(a_{0}, - 0.025 a_{0}^{2} + 9 a_{0}) : y = (- 0.05 a_{0} + 9) x + 0.025 a_{0}^{2}

y = (- 0.05 a_{0} + 9) x + 0.025 a_{0}^{2}

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What is the tangent of-0.025x^2+9x ?
The tangent of-0.025x^2+9x is y=(-0.05a_{0}+9)x+0.025a_{0}^2