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

The tangent of f(x)=x^2-9,\at x=2 is y=4x-13

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

tangent of

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

y = 4 x - 13

Solution steps

Find the tangent point:

(2, - 5)

Find the slope of

f (x) = x^{2} - 9 : \frac{df ( x )}{d x} = 2 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 = 4

Find the line with slope m=

4

and passing through

(2, - 5) : y = 4 x - 13

y = 4 x - 13

\frac{8}{x ^{3}}

\frac{d}{d x} ((5 x^{3})^{x})

x \frac{d y}{d x} + y = 3 x y

\frac{\partial}{\partial y} (cos (3 x + 4 y))

\frac{d y}{d x} = \frac{3 y}{2 x}

What is the tangent of f(x)=x^2-9,\at x=2 ?
The tangent of f(x)=x^2-9,\at x=2 is y=4x-13