What is the tangent of-0.025x^2+4x ?

The tangent of-0.025x^2+4x is y=(-0.05a_{0}+4)x+0.025a_{0}^2

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tangent of-0.025x^2+4x

tangent of

- 0.025 x^{2} + 4 x

y = (- 0.05 a_{0} + 4) 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} + 4 a_{0})

Find the slope of

y = - 0.025 x^{2} + 4 x : \frac{d y}{d x} = - 0.05 x + 4

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 0.05 a_{0} + 4

Find the line with slope m=

- 0.05 a_{0} + 4

and passing through

(a_{0}, - 0.025 a_{0}^{2} + 4 a_{0}) : y = (- 0.05 a_{0} + 4) x + 0.025 a_{0}^{2}

y = (- 0.05 a_{0} + 4) x + 0.025 a_{0}^{2}

What is the tangent of-0.025x^2+4x ?
The tangent of-0.025x^2+4x is y=(-0.05a_{0}+4)x+0.025a_{0}^2