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

The tangent of f(x)=-1-8x^2,\at x=-2 is y=32x+31

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=-1-8x^2,\at x=-2

tangent of

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

y = 32 x + 31

Solution steps

Find the tangent point:

(- 2, - 33)

Find the slope of

f (x) = - 1 - 8 x^{2} : \frac{df ( x )}{d x} = - 16 x

Compute the slope of

f (x) = - 1 - 8 x^{2}

x = - 2 : m = 32

Find the line with slope m=

32

and passing through

(- 2, - 33) : y = 32 x + 31

y = 32 x + 31

\int_{0}^{1} (x) - (x^{2}) d x

\int (x - 7)^{2} d x

n = 1 \sum \infty \frac{1}{2 ^{n - 1}}

x \to \infty lim (0^{0})

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

What is the tangent of f(x)=-1-8x^2,\at x=-2 ?
The tangent of f(x)=-1-8x^2,\at x=-2 is y=32x+31