What is the tangent of f(x)=5-2x^2,(1,3) ?

The tangent of f(x)=5-2x^2,(1,3) is y=-4x+7

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=5-2x^2,(1,3)

tangent of

f (x) = 5 - 2 x^{2}, (1, 3)

y = - 4 x + 7

Solution steps

Find the slope of

f (x) = 5 - 2 x^{2} : \frac{df ( x )}{d x} = - 4 x

Compute the slope of

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

(1, 3) : m = - 4

Find the line with slope m=

- 4

and passing through

(1, 3) : y = - 4 x + 7

y = - 4 x + 7

\frac{d}{d x} (7 e^{- x} - 8 e^{- 7 x})

f (x) = \frac{( x + 3 )}{2 x + 1}, at x = - 1

x \to 2 lim (\frac{1}{∣ x - 2 ∣})

\int \frac{7 x ^{3} - 1}{x ^{2}} d x

\frac{\partial}{\partial x} (e^{x} sec^{2} (y))

What is the tangent of f(x)=5-2x^2,(1,3) ?
The tangent of f(x)=5-2x^2,(1,3) is y=-4x+7