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

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

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tangent of y=3x^2+4

tangent of

y = 3 x^{2} + 4

y = 6 a_{0} 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} + 4)

Find the slope of

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

Compute the slope of

y = 3 x^{2} + 4

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

Find the line with slope m=

6 a_{0}

and passing through

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

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

\int (x + 2) d x

\frac{\partial}{\partial x} (5 x^{7 y})

\int 56 x e^{7 x^{2}} d x

y^{^{'}} + 2 y^{2} cos (x + 1) = 0

\frac{d}{d x} (f (x) \cdot f (x))

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