What is the tangent of y=6x^2+3x-3 ?

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

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tangent of y=6x^2+3x-3

tangent of

y = 6 x^{2} + 3 x - 3

y = (12 a_{0} + 3) x - 6 a_{0}^{2} - 3

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 6 a_{0}^{2} + 3 a_{0} - 3)

Find the slope of

y = 6 x^{2} + 3 x - 3 : \frac{d y}{d x} = 12 x + 3

Compute the slope of

y = 6 x^{2} + 3 x - 3

x = a_{0} : m = 12 a_{0} + 3

Find the line with slope m=

12 a_{0} + 3

and passing through

(a_{0}, 6 a_{0}^{2} + 3 a_{0} - 3) : y = (12 a_{0} + 3) x - 6 a_{0}^{2} - 3

y = (12 a_{0} + 3) x - 6 a_{0}^{2} - 3

(lo g_{5} (x))^{^{'}}

\int_{x}^{x} \frac{e ^{x}}{x} d x

\frac{\partial}{\partial y} (2 x cos (x^{2} + y^{2}))

y = e^{- 2 x^{3}}

\frac{x ^{2}}{3 + x}

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