What is the tangent of y=4x-3x^2,(2,-4) ?

The tangent of y=4x-3x^2,(2,-4) is y=-8x+12

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y=4x-3x^2,(2,-4)

tangent of

y = 4 x - 3 x^{2}, (2, - 4)

y = - 8 x + 12

Solution steps

Find the slope of

y = 4 x - 3 x^{2} : \frac{d y}{d x} = 4 - 6 x

Compute the slope of

y = 4 x - 3 x^{2}

(2, - 4) : m = - 8

Find the line with slope m=

- 8

and passing through

(2, - 4) : y = - 8 x + 12

y = - 8 x + 12

(x + 2)^{2}

\frac{d y}{d x} - \frac{1}{x} y = - 9

\int \frac{5 x ^{3} + 2 x}{x - 1} d x

\frac{d}{d x} (2 + 4 cos (x^{3}) + 3 sin (x^{3}))

What is the tangent of y=4x-3x^2,(2,-4) ?
The tangent of y=4x-3x^2,(2,-4) is y=-8x+12