What is the tangent of f(x)=3x^2+8x+4,\at x=-1 ?

The tangent of f(x)=3x^2+8x+4,\at x=-1 is y=2x+1

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^2+8x+4,\at x=-1

tangent of

f (x) = 3 x^{2} + 8 x + 4, at x = - 1

y = 2 x + 1

Solution steps

Find the tangent point:

(- 1, - 1)

Find the slope of

f (x) = 3 x^{2} + 8 x + 4 : \frac{df ( x )}{d x} = 6 x + 8

Compute the slope of

f (x) = 3 x^{2} + 8 x + 4

x = - 1 : m = 2

Find the line with slope m=

2

and passing through

(- 1, - 1) : y = 2 x + 1

y = 2 x + 1

x \to 0 lim (x^{3} + 2 x^{2} - x - 2)

x \to 0 - lim (\frac{x + ∣ x ∣}{x})

\frac{d}{d x} (\frac{1}{x} + \frac{1}{x ^{2}} + \frac{1}{x ^{3}})

x \to 0 lim (\frac{x - 1}{x ^{3} + x})

y = sin (5 x) + cos (6 x)

What is the tangent of f(x)=3x^2+8x+4,\at x=-1 ?
The tangent of f(x)=3x^2+8x+4,\at x=-1 is y=2x+1