What is the tangent of f(x)=20x-16x^2,\at x=3 ?

The tangent of f(x)=20x-16x^2,\at x=3 is y=-76x+144

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=20x-16x^2,\at x=3

tangent of

f (x) = 20 x - 16 x^{2}, at x = 3

y = - 76 x + 144

Solution steps

Find the tangent point:

(3, - 84)

Find the slope of

f (x) = 20 x - 16 x^{2} : \frac{df ( x )}{d x} = 20 - 32 x

Compute the slope of

f (x) = 20 x - 16 x^{2}

x = 3 : m = - 76

Find the line with slope m=

- 76

and passing through

(3, - 84) : y = - 76 x + 144

y = - 76 x + 144

y^{^{'}} + \frac{2}{t + 1} y = 3 t

\frac{\partial}{\partial x} (\frac{v}{2 u + v})

\frac{d}{d x} (3 e^{- 3 x})

\frac{d}{d x} (9 k lo g_{e} (\frac{( x + 9 )}{x - 9}))

\frac{5 s + 50.1}{( s + 5 ) ^{2} - 5}

What is the tangent of f(x)=20x-16x^2,\at x=3 ?
The tangent of f(x)=20x-16x^2,\at x=3 is y=-76x+144