What is the tangent of f(x)=x^2-8,(-4,8) ?

The tangent of f(x)=x^2-8,(-4,8) is y=-8x-24

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^2-8,(-4,8)

tangent of

f (x) = x^{2} - 8, (- 4, 8)

y = - 8 x - 24

Solution steps

Find the slope of

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

Compute the slope of

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

(- 4, 8) : m = - 8

Find the line with slope m=

- 8

and passing through

(- 4, 8) : y = - 8 x - 24

y = - 8 x - 24

\int_{1}^{4} - 8 t ln (t) d t

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

\frac{d}{d x} (x^{7} sin (x))

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

\frac{d ^{2}}{d x ^{2}} ((x^{2} + 8)^{5})

What is the tangent of f(x)=x^2-8,(-4,8) ?
The tangent of f(x)=x^2-8,(-4,8) is y=-8x-24