What is the tangent of f(x)=x^2-3x-4,(2,-6) ?

The tangent of f(x)=x^2-3x-4,(2,-6) is y=x-8

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^2-3x-4,(2,-6)

tangent of

f (x) = x^{2} - 3 x - 4, (2, - 6)

y = x - 8

Solution steps

Find the slope of

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

Compute the slope of

f (x) = x^{2} - 3 x - 4

(2, - 6) : m = 1

Find the line with slope m=

1

and passing through

(2, - 6) : y = x - 8

y = x - 8

\frac{d}{d x} (4 x^{3} + 4 x)

\frac{\partial}{\partial x} (\frac{x - 3 y}{x ^{2} + y ^{2}})

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

\frac{d}{d x} (arctan (\frac{x}{1 + x}))

\int \frac{( 4 x ^{3} - 3 x ^{2} - 8 x )}{6 x} d x

What is the tangent of f(x)=x^2-3x-4,(2,-6) ?
The tangent of f(x)=x^2-3x-4,(2,-6) is y=x-8