What is the tangent of 5x-4x^2 ?

The tangent of 5x-4x^2 is y=(5-8a_{0})x+4a_{0}^2

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tangent of 5x-4x^2

tangent of

5 x - 4 x^{2}

y = (5 - 8 a_{0}) x + 4 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 5 a_{0} - 4 a_{0}^{2})

Find the slope of

y = 5 x - 4 x^{2} : \frac{d y}{d x} = 5 - 8 x

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 5 - 8 a_{0}

Find the line with slope m=

5 - 8 a_{0}

and passing through

(a_{0}, 5 a_{0} - 4 a_{0}^{2}) : y = (5 - 8 a_{0}) x + 4 a_{0}^{2}

y = (5 - 8 a_{0}) x + 4 a_{0}^{2}

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

\int x e^{- 3 x} d x

\frac{d}{d y} (\frac{1}{3} y (y - 3))

\frac{d}{d x} (x^{l n (x^{2} + 1)})

\frac{d ^{2}}{d x ^{2}} (e^{s i n (x)})

What is the tangent of 5x-4x^2 ?
The tangent of 5x-4x^2 is y=(5-8a_{0})x+4a_{0}^2