What is the tangent of f(x)=2x^2-4x-5,\at x=2 ?

The tangent of f(x)=2x^2-4x-5,\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)=2x^2-4x-5,\at x=2

tangent of

f (x) = 2 x^{2} - 4 x - 5, at x = 2

y = 4 x - 13

Solution steps

Find the tangent point:

(2, - 5)

Find the slope of

f (x) = 2 x^{2} - 4 x - 5 : \frac{df ( x )}{d x} = 4 x - 4

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

\int sin (e^{x}) e^{x} d x

\frac{d y}{d x} = \frac{3 y ^{2} + 10 x ^{2}}{x y}

\frac{d}{d x} (7 x - 5)

\int_{0}^{4} x^{\frac{3}{2}} d x

\int_{1}^{5} \frac{5}{x} - \frac{3}{x ^{2}} d x

What is the tangent of f(x)=2x^2-4x-5,\at x=2 ?
The tangent of f(x)=2x^2-4x-5,\at x=2 is y=4x-13