What is the tangent of f(x)=-0.025x^2 ?

The tangent of f(x)=-0.025x^2 is y=-0.05a_{0}x+(a_{0}^2)/(40)

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tangent of f(x)=-0.025x^2

tangent of

f (x) = - 0.025 x^{2}

y = - 0.05 a_{0} x + \frac{a _{0}^{2}}{40}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

f (x) = - 0.025 x^{2} : \frac{df ( x )}{d x} = - 0.05 x

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}

Find the line with slope m=

- 0.05 a_{0}

and passing through

(a_{0}, - 0.025 a_{0}^{2}) : y = - 0.05 a_{0} x + \frac{a _{0}^{2}}{40}

y = - 0.05 a_{0} x + \frac{a _{0}^{2}}{40}

What is the tangent of f(x)=-0.025x^2 ?
The tangent of f(x)=-0.025x^2 is y=-0.05a_{0}x+(a_{0}^2)/(40)