What is the tangent of y=4x-3x^2(2-4) ?

The tangent of y=4x-3x^2(2-4) is y=(4+12a_{0})x-6a_{0}^2

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tangent of y=4x-3x^2(2-4)

tangent of

y = 4 x - 3 x^{2} (2 - 4)

y = (4 + 12 a_{0}) x - 6 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 4 a_{0} + 6 a_{0}^{2})

Find the slope of

y = 4 x - 3 x^{2} (2 - 4) : \frac{d y}{d x} = 4 + 12 x

Compute the slope of

y = 4 x - 3 x^{2} (2 - 4)

x = a_{0} : m = 4 + 12 a_{0}

Find the line with slope m=

4 + 12 a_{0}

and passing through

(a_{0}, 4 a_{0} + 6 a_{0}^{2}) : y = (4 + 12 a_{0}) x - 6 a_{0}^{2}

y = (4 + 12 a_{0}) x - 6 a_{0}^{2}

3 x y^{3} d x - (x + 2) d y = 0

\frac{d}{d x} (2 cos^{2} (2 x))

x \to 3 - lim (\frac{3 x + 2}{x - 3})

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

\frac{\partial}{\partial x} (\frac{3}{3 x + y})

What is the tangent of y=4x-3x^2(2-4) ?
The tangent of y=4x-3x^2(2-4) is y=(4+12a_{0})x-6a_{0}^2