What is the tangent of y=-3x^3-2x^2-x+3,(-1,5) ?

The tangent of y=-3x^3-2x^2-x+3,(-1,5) 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 y=-3x^3-2x^2-x+3,(-1,5)

tangent of

y = - 3 x^{3} - 2 x^{2} - x + 3, (- 1, 5)

y = - 6 x - 1

Solution steps

Find the slope of

y = - 3 x^{3} - 2 x^{2} - x + 3 : \frac{d y}{d x} = - 9 x^{2} - 4 x - 1

Compute the slope of

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

(- 1, 5) : m = - 6

Find the line with slope m=

- 6

and passing through

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

y = - 6 x - 1

- \frac{1}{x ^{2}}

\frac{d y}{d x} = 20 x

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

\frac{\partial}{\partial x} (6 x^{4 y})

\int x^{3} 1 - x d x

What is the tangent of y=-3x^3-2x^2-x+3,(-1,5) ?
The tangent of y=-3x^3-2x^2-x+3,(-1,5) is y=-6x-1