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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

tangent of

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

y = - 4 a_{0} x + 2 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

f (x) = - 2 x^{2} : \frac{df ( x )}{d x} = - 4 x

Compute the slope of

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

x = a_{0} : m = - 4 a_{0}

Find the line with slope m=

- 4 a_{0}

and passing through

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

y = - 4 a_{0} x + 2 a_{0}^{2}

(y^{2} + y x) d x + (- x^{2}) d y = 0

y^{^{'}} = \frac{y ^{2} + 4 x x ^{2} + y ^{2}}{x y}

ln (4 x + 1)

3 x^{2} - 4 x + 1

\frac{d}{d x} ((8 (3 x)^{4} + 5)^{3})

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