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

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

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tangent of 3-8x^2

tangent of

3 - 8 x^{2}

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

Compute the slope of

y = 3 - 8 x^{2}

x = a_{0} : m = - 16 a_{0}

Find the line with slope m=

- 16 a_{0}

and passing through

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

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

x (\frac{d y}{d x}) + \frac{y}{( x + 1 )} = x (x + 1)

\int x (3 x + 1)^{9} d x

\int_{- \infty}^{6} x e^{\frac{x}{3}} d x

y^{^{'}} = cos^{2} (t) cos^{2} (2 y)

x \to 0 lim (1 - cot (x))