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

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

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tangent of y=3x^2+3x

tangent of

y = 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

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

Compute the slope of

y = 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}

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