What is the tangent of-8x^2+5 ?

The tangent of-8x^2+5 is y=-16a_{0}x+8a_{0}^2+5

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tangent of-8x^2+5

tangent of

- 8 x^{2} + 5

y = - 16 a_{0} x + 8 a_{0}^{2} + 5

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - 8 a_{0}^{2} + 5)

Find the slope of

y = - 8 x^{2} + 5 : \frac{d y}{d x} = - 16 x

Compute the slope of

y = - 8 x^{2} + 5

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

Find the line with slope m=

- 16 a_{0}

and passing through

(a_{0}, - 8 a_{0}^{2} + 5) : y = - 16 a_{0} x + 8 a_{0}^{2} + 5

y = - 16 a_{0} x + 8 a_{0}^{2} + 5

\int \frac{4 t}{1 - 16 t ^{4}} d t

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\int \frac{1}{1 + u ^{2}} d u