What is the tangent of f(x)=3x^3-x^2+8,(2,28) ?

The tangent of f(x)=3x^3-x^2+8,(2,28) is y=32x-36

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=3x^3-x^2+8,(2,28)

tangent of

f (x) = 3 x^{3} - x^{2} + 8, (2, 28)

y = 32 x - 36

Solution steps

Find the slope of

f (x) = 3 x^{3} - x^{2} + 8 : \frac{df ( x )}{d x} = 9 x^{2} - 2 x

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 32

Find the line with slope m=

32

and passing through

(2, 28) : y = 32 x - 36

y = 32 x - 36

\frac{d}{d x} (8 - x^{2})

y = (t)^{t}

x \to \frac{1}{3} lim (\frac{x + 3}{x + 4})

\frac{d}{d x} ((x^{2}) d x)

x \to - 3 lim (\frac{x ^{2} + x}{x + 3})

What is the tangent of f(x)=3x^3-x^2+8,(2,28) ?
The tangent of f(x)=3x^3-x^2+8,(2,28) is y=32x-36