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

The tangent of f(x)=2x^2-3x,\at x=2 is y=5x-8

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^2-3x,\at x=2

tangent of

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

y = 5 x - 8

Solution steps

Find the tangent point:

(2, 2)

Find the slope of

f (x) = 2 x^{2} - 3 x : \frac{df ( x )}{d x} = 4 x - 3

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 = 5

Find the line with slope m=

5

and passing through

(2, 2) : y = 5 x - 8

y = 5 x - 8

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

n = 1 \sum \infty e^{- \frac{3}{n ^{2}}}

\int_{1}^{\infty} \frac{3}{x ^{2} + 1} d x

\int \frac{( 2 x + 1 )}{x} d x

n = 8 \sum \infty (\frac{7}{8})^{- n}

What is the tangent of f(x)=2x^2-3x,\at x=2 ?
The tangent of f(x)=2x^2-3x,\at x=2 is y=5x-8