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

The tangent of f(x)=x^2-3x is y=(2a_{0}-3)x-a_{0}^2

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^2-3x

tangent of

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

y = (2 a_{0} - 3) x - a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{2} - 3 a_{0})

Find the slope of

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

Compute the slope of

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

x = a_{0} : m = 2 a_{0} - 3

Find the line with slope m=

2 a_{0} - 3

and passing through

(a_{0}, a_{0}^{2} - 3 a_{0}) : y = (2 a_{0} - 3) x - a_{0}^{2}

y = (2 a_{0} - 3) x - a_{0}^{2}

\int 21 t e^{t} d t

\int 3 x x^{2} + 4 d x

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

\int ln (7 x + 1) d x

\frac{d}{d x} ((3 x^{5} - 1) (2 - x^{4}))

What is the tangent of f(x)=x^2-3x ?
The tangent of f(x)=x^2-3x is y=(2a_{0}-3)x-a_{0}^2