What is the tangent of f(x)=5x-4x^2,(-3,-51) ?

The tangent of f(x)=5x-4x^2,(-3,-51) is y=29x+36

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=5x-4x^2,(-3,-51)

tangent of

f (x) = 5 x - 4 x^{2}, (- 3, - 51)

y = 29 x + 36

Solution steps

Find the slope of

f (x) = 5 x - 4 x^{2} : \frac{df ( x )}{d x} = 5 - 8 x

Compute the slope of

f (x) = 5 x - 4 x^{2}

(- 3, - 51) : m = 29

Find the line with slope m=

29

and passing through

(- 3, - 51) : y = 29 x + 36

y = 29 x + 36

\int_{0}^{1} x^{n} d x

\frac{4}{e ^{x} + e ^{- x}}

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 2, y^{^{'}} (0) = 0

y^{^{''}} - 5 y^{^{'}} + 6 y = 0

\frac{x}{( x ^{2} + 1 ) ( x ^{2} + 9 )}

What is the tangent of f(x)=5x-4x^2,(-3,-51) ?
The tangent of f(x)=5x-4x^2,(-3,-51) is y=29x+36