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

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

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tangent of 3x^2+4x

tangent of

3 x^{2} + 4 x

y = (6 a_{0} + 4) 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} + 4 a_{0})

Find the slope of

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

Compute the slope of

y = 3 x^{2} + 4 x

x = a_{0} : m = 6 a_{0} + 4

Find the line with slope m=

6 a_{0} + 4

and passing through

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

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

\int ∣ x - 5 ∣ d x

x \to 0 lim (\frac{1}{ln ∣ x ∣})

\int \frac{15 x e ^{2 x}}{( 1 + 2 x ) ^{2}} d x

\frac{d y}{d x}, e^{5 y^{2}} = x^{3} + 3

y^{^{'}} = (x - 2) e^{- 2 y}, y (2) = ln (2)

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