What is the tangent of f(x)=x^3-5x^2,\at x=2 ?

The tangent of f(x)=x^3-5x^2,\at x=2 is y=-8x+4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=x^3-5x^2,\at x=2

tangent of

f (x) = x^{3} - 5 x^{2}, at x = 2

y = - 8 x + 4

Solution steps

Find the tangent point:

(2, - 12)

Find the slope of

f (x) = x^{3} - 5 x^{2} : \frac{df ( x )}{d x} = 3 x^{2} - 10 x

Compute the slope of

f (x) = x^{3} - 5 x^{2}

x = 2 : m = - 8

Find the line with slope m=

- 8

and passing through

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

y = - 8 x + 4

\frac{d y}{d x} = \frac{x ^{2} + x y + y ^{2}}{x ^{2}}

\frac{\partial}{\partial z} (- 4 x yz)

sec^{2} (x)

y^{^{''}} + 9 y = 5 cos (2 x)

f (x) = e^{s i n (x)}

What is the tangent of f(x)=x^3-5x^2,\at x=2 ?
The tangent of f(x)=x^3-5x^2,\at x=2 is y=-8x+4