What is the tangent of y=9x^2-x^3,(2,28) ?

The tangent of y=9x^2-x^3,(2,28) is y=24x-20

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y=9x^2-x^3,(2,28)

tangent of

y = 9 x^{2} - x^{3}, (2, 28)

y = 24 x - 20

Solution steps

Find the slope of

y = 9 x^{2} - x^{3} : \frac{d y}{d x} = 18 x - 3 x^{2}

Compute the slope of

y = 9 x^{2} - x^{3}

(2, 28) : m = 24

Find the line with slope m=

24

and passing through

(2, 28) : y = 24 x - 20

y = 24 x - 20

\int \frac{( cos ( 3 x ) )}{1 + sin ( 3 x )} d x

\int_{0}^{1} 2 π x (12 - 12 x) d x

e^{4}

y^{^{''}} - \frac{9}{t} y^{^{'}} + \frac{34}{t ^{2}} y = 0

\frac{\partial}{\partial x} (\frac{1}{2 1 + x + y})

What is the tangent of y=9x^2-x^3,(2,28) ?
The tangent of y=9x^2-x^3,(2,28) is y=24x-20