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

The tangent of 4x^2+3x is y=(8a_{0}+3)x-4a_{0}^2

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tangent of 4x^2+3x

tangent of

4 x^{2} + 3 x

y = (8 a_{0} + 3) x - 4 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 4 a_{0}^{2} + 3 a_{0})

Find the slope of

y = 4 x^{2} + 3 x : \frac{d y}{d x} = 8 x + 3

Compute the slope of

y = 4 x^{2} + 3 x

x = a_{0} : m = 8 a_{0} + 3

Find the line with slope m=

8 a_{0} + 3

and passing through

(a_{0}, 4 a_{0}^{2} + 3 a_{0}) : y = (8 a_{0} + 3) x - 4 a_{0}^{2}

y = (8 a_{0} + 3) x - 4 a_{0}^{2}

\frac{\partial}{\partial x} (\frac{108}{x})

\int \frac{1}{e ^{3 t} + e ^{t}} d t

\frac{d y}{d x} = 4 x - 6

\int \frac{2}{x} - \frac{3}{x ^{2}} d x

\int \frac{sin ( \frac{5}{x} )}{x ^{2}} d x

What is the tangent of 4x^2+3x ?
The tangent of 4x^2+3x is y=(8a_{0}+3)x-4a_{0}^2