What is the tangent of-0.025x^2+8x ?

The tangent of-0.025x^2+8x is y=(-0.05a_{0}+8)x+0.025a_{0}^2

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tangent of-0.025x^2+8x

tangent of

- 0.025 x^{2} + 8 x

y = (- 0.05 a_{0} + 8) 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} + 8 a_{0})

Find the slope of

y = - 0.025 x^{2} + 8 x : \frac{d y}{d x} = - 0.05 x + 8

Compute the slope of

y = - 0.025 x^{2} + 8 x

x = a_{0} : m = - 0.05 a_{0} + 8

Find the line with slope m=

- 0.05 a_{0} + 8

and passing through

(a_{0}, - 0.025 a_{0}^{2} + 8 a_{0}) : y = (- 0.05 a_{0} + 8) x + 0.025 a_{0}^{2}

y = (- 0.05 a_{0} + 8) x + 0.025 a_{0}^{2}

y^{^{'}} = x y^{2} + 2 y^{2}, y (0) = 1

f (x) = \frac{2 x ^{2} - 6 x + 5}{2 ( 2 x - 3 )}

\frac{d y}{d x} = 4 x e^{3 y}

x y^{^{'}} + 2 y = 4 cos (x)

\frac{d y}{d x} = sec^{2} (x) (2 + y)^{2}

What is the tangent of-0.025x^2+8x ?
The tangent of-0.025x^2+8x is y=(-0.05a_{0}+8)x+0.025a_{0}^2