What is the tangent of f(x)=x^2-2,(5,23) ?

The tangent of f(x)=x^2-2,(5,23) is y=10x-27

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^2-2,(5,23)

tangent of

f (x) = x^{2} - 2, (5, 23)

y = 10 x - 27

Solution steps

Find the slope of

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

Compute the slope of

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

(5, 23) : m = 10

Find the line with slope m=

10

and passing through

(5, 23) : y = 10 x - 27

y = 10 x - 27

\frac{10 \cdot e ^{- s}}{5 s + 1}

\frac{\partial}{\partial x} (x^{2 y} + 4 y)

\int (cot^{2} (x) + cot^{4} (x)) d x

\frac{d}{d x} (ln (\frac{x + 2}{x - 2}))

y = 6 x^{4} ln (x)

What is the tangent of f(x)=x^2-2,(5,23) ?
The tangent of f(x)=x^2-2,(5,23) is y=10x-27