What is the tangent of 8x-x^2,\at x=1 ?

The tangent of 8x-x^2,\at x=1 is y=6x+1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of 8x-x^2,\at x=1

tangent of

8 x - x^{2}, at x = 1

y = 6 x + 1

Solution steps

Find the tangent point:

(1, 7)

Find the slope of

y = 8 x - x^{2} : \frac{d y}{d x} = 8 - 2 x

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 = 6

Find the line with slope m=

6

and passing through

(1, 7) : y = 6 x + 1

y = 6 x + 1

2 y^{^{'}} + y = 0

\int_{- 1}^{1} t^{2} - 2 d t

\int \frac{1}{2} (2 cos (θ))^{2} d θ

\int 200 - x d x

\int_{0}^{6} 2 x - (x^{2} - 4 x) d x