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)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Compute the slope of

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

x = a_{0} : 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}

\int_{21}^{\infty} x e^{- x} d x

\frac{d}{d x} (14 ln (x))

x^{2} + x - 5

\int_{1}^{2} \frac{1}{x} d x

\frac{\partial}{\partial x} (3 x yz)

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)