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

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

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tangent of y=4x-3x^2

tangent of

y = 4 x - 3 x^{2}

y = (4 - 6 a_{0}) x + 3 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} - 3 a_{0}^{2})

Find the slope of

y = 4 x - 3 x^{2} : \frac{d y}{d x} = 4 - 6 x

Compute the slope of

y = 4 x - 3 x^{2}

x = a_{0} : m = 4 - 6 a_{0}

Find the line with slope m=

4 - 6 a_{0}

and passing through

(a_{0}, 4 a_{0} - 3 a_{0}^{2}) : y = (4 - 6 a_{0}) x + 3 a_{0}^{2}

y = (4 - 6 a_{0}) x + 3 a_{0}^{2}

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = 0

\int \frac{8}{4 + x ^{2}} d x

\frac{\partial}{\partial x} (x^{3} e^{x^{2} + y} - 12 y^{2})

y = - 2 x sin (x^{2})

\frac{s - 3}{s ^{2} - 3 s + 2}

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