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

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

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tangent of y=3x^2-3x+4

tangent of

y = 3 x^{2} - 3 x + 4

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

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} + 4)

Find the slope of

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

Compute the slope of

y = 3 x^{2} - 3 x + 4

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

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

\frac{d y}{d x} = 15 + 5 y + 3 x + y x

f^{^{'}} (x) = x^{2}

a \to 0 lim (\frac{1}{a})

\frac{\partial}{\partial x} (\frac{x}{1 - ln ( x - 1 )})

y = \frac{1}{20 x ^{2}}, y = x, y = \frac{x}{6}

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