What is the tangent of y=x^2-6x+9,(3,0) ?

The tangent of y=x^2-6x+9,(3,0) is y=0

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y=x^2-6x+9,(3,0)

tangent of

y = x^{2} - 6 x + 9, (3, 0)

y = 0

Solution steps

Find the slope of

y = x^{2} - 6 x + 9 : \frac{d y}{d x} = 2 x - 6

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 = 0

Find the line with slope m=

0

and passing through

(3, 0) : y = 0

y = 0

x \to \infty lim (x^{2} - 2 x)

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

\int \frac{1}{x 3 ln ( x ) - 1} d x

\int_{- 2}^{1} (- 2 x + 3) d x

\int \frac{1}{4 y ^{3}} d y

What is the tangent of y=x^2-6x+9,(3,0) ?
The tangent of y=x^2-6x+9,(3,0) is y=0