What is the tangent of f(x)=14-x^2,(2,10) ?

The tangent of f(x)=14-x^2,(2,10) is y=-4x+18

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=14-x^2,(2,10)

tangent of

f (x) = 14 - x^{2}, (2, 10)

y = - 4 x + 18

Solution steps

Find the slope of

f (x) = 14 - x^{2} : \frac{df ( x )}{d x} = - 2 x

Compute the slope of

f (x) = 14 - x^{2}

(2, 10) : m = - 4

Find the line with slope m=

- 4

and passing through

(2, 10) : y = - 4 x + 18

y = - 4 x + 18

f (x) = x^{3} + 9 x - 1

n = 1 \sum \infty \frac{e ^{n}}{n ^{n}}

\int_{- 4}^{0} 16 - x^{2} d x

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

\int (x + 1) x + 6 d x

What is the tangent of f(x)=14-x^2,(2,10) ?
The tangent of f(x)=14-x^2,(2,10) is y=-4x+18