What is the tangent of y=7x-x^2,(1,6) ?

The tangent of y=7x-x^2,(1,6) is y=5x+1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y=7x-x^2,(1,6)

tangent of

y = 7 x - x^{2}, (1, 6)

y = 5 x + 1

Solution steps

Find the slope of

y = 7 x - x^{2} : \frac{d y}{d x} = 7 - 2 x

Compute the slope of

y = 7 x - x^{2}

(1, 6) : m = 5

Find the line with slope m=

5

and passing through

(1, 6) : y = 5 x + 1

y = 5 x + 1

\int_{1}^{2} x^{4} + 2 x^{2} d x

\int_{0}^{1} 5 e^{x} - 5 x e^{x^{2}} d x

\int \frac{16}{( x ^{2} + 4 ) ^{2}} d x

y = - 3 x - 9, (- 9, 18)

g (x) = 2 x^{0.8} + 4 x^{- 0.8}