What is the tangent of f(x)=-9x^2,\at x=2 ?

The tangent of f(x)=-9x^2,\at x=2 is y=-36x+36

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=-9x^2,\at x=2

tangent of

f (x) = - 9 x^{2}, at x = 2

y = - 36 x + 36

Solution steps

Find the tangent point:

(2, - 36)

Find the slope of

f (x) = - 9 x^{2} : \frac{df ( x )}{d x} = - 18 x

Compute the slope of

f (x) = - 9 x^{2}

x = 2 : m = - 36

Find the line with slope m=

- 36

and passing through

(2, - 36) : y = - 36 x + 36

y = - 36 x + 36

e^{\frac{- x}{2}}

f (t) = 2 - \frac{4}{t}

e^{3 x}

\frac{d}{d x} (arccos (arcsin (x)))

\frac{\partial}{\partial x} (3 x ln (y) + y ln (z) - 4 x)

What is the tangent of f(x)=-9x^2,\at x=2 ?
The tangent of f(x)=-9x^2,\at x=2 is y=-36x+36