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

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

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tangent of f(x)=3x^2-3x

tangent of

f (x) = 3 x^{2} - 3 x

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 3 a_{0}^{2} - 3 a_{0})

Find the slope of

f (x) = 3 x^{2} - 3 x : \frac{df ( x )}{d x} = 6 x - 3

Compute the slope of

f (x) = 3 x^{2} - 3 x

x = a_{0} : m = 6 a_{0} - 3

Find the line with slope m=

6 a_{0} - 3

and passing through

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

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

y = \frac{x}{1 + x}

\int \frac{1}{x ^{2} + 10 x + 41} d x

(0, - 3), (- 3, 0)

\frac{d y}{d x} = \frac{( x + y ^{2} )}{2 y}

\frac{\partial}{\partial x} (x^{a} y^{b})

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