What is the tangent of y=(x^2-3x+3),(2,1) ?

The tangent of y=(x^2-3x+3),(2,1) is y=x-1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y=(x^2-3x+3),(2,1)

tangent of

y = (x^{2} - 3 x + 3), (2, 1)

y = x - 1

Solution steps

Find the slope of

y = (x^{2} - 3 x + 3) : \frac{d y}{d x} = 2 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 = 1

Find the line with slope m=

1

and passing through

(2, 1) : y = x - 1

y = x - 1

\int 45 + 4 x - x^{2} d x

y = (ln (x))^{x}

\int \frac{6 x ^{2} - 3 x}{x ^{2}} d x

y^{^{'}} = 2 t

f (x) = x + 11

What is the tangent of y=(x^2-3x+3),(2,1) ?
The tangent of y=(x^2-3x+3),(2,1) is y=x-1