What is the tangent of f(x)=1260-16x^2,\at x=3 ?

The tangent of f(x)=1260-16x^2,\at x=3 is y=-96x+1404

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=1260-16x^2,\at x=3

tangent of

f (x) = 1260 - 16 x^{2}, at x = 3

y = - 96 x + 1404

Solution steps

Find the tangent point:

(3, 1116)

Find the slope of

f (x) = 1260 - 16 x^{2} : \frac{df ( x )}{d x} = - 32 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 = - 96

Find the line with slope m=

- 96

and passing through

(3, 1116) : y = - 96 x + 1404

y = - 96 x + 1404

f (x) = x^{\frac{19}{9}} + x^{\frac{10}{9}}

\int x z d y

\frac{\partial}{\partial x} (\frac{x - y}{x + y})

\frac{d}{d x} (15 x^{4} - 15 x^{2})

(u^{2} + u)^{3}

What is the tangent of f(x)=1260-16x^2,\at x=3 ?
The tangent of f(x)=1260-16x^2,\at x=3 is y=-96x+1404