What is the tangent of 8x^2+3x ?

The tangent of 8x^2+3x is y=(16a_{0}+3)x-8a_{0}^2

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tangent of 8x^2+3x

tangent of

8 x^{2} + 3 x

y = (16 a_{0} + 3) x - 8 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 8 a_{0}^{2} + 3 a_{0})

Find the slope of

y = 8 x^{2} + 3 x : \frac{d y}{d x} = 16 x + 3

Compute the slope of

y = 8 x^{2} + 3 x

x = a_{0} : m = 16 a_{0} + 3

Find the line with slope m=

16 a_{0} + 3

and passing through

(a_{0}, 8 a_{0}^{2} + 3 a_{0}) : y = (16 a_{0} + 3) x - 8 a_{0}^{2}

y = (16 a_{0} + 3) x - 8 a_{0}^{2}

\frac{d}{d x} (2 c x + c^{2})

f (x) = 2 x^{2} - x^{3}

(2 t - 1) (4 t - 6)^{- 1}

x^{2} y^{^{''}} - 2 y = 0, y (1) = 0, y^{^{'}} (1) = 3

x \to + 0 lim (x^{8} - \frac{1}{x})

What is the tangent of 8x^2+3x ?
The tangent of 8x^2+3x is y=(16a_{0}+3)x-8a_{0}^2