What is the tangent of 5-9+8x^2 ?

The tangent of 5-9+8x^2 is y=16a_{0}x-8a_{0}^2-4

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tangent of 5-9+8x^2

tangent of

5 - 9 + 8 x^{2}

y = 16 a_{0} x - 8 a_{0}^{2} - 4

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 8 a_{0}^{2} - 4)

Find the slope of

y = 5 - 9 + 8 x^{2} : \frac{d y}{d x} = 16 x

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 16 a_{0}

Find the line with slope m=

16 a_{0}

and passing through

(a_{0}, 8 a_{0}^{2} - 4) : y = 16 a_{0} x - 8 a_{0}^{2} - 4

y = 16 a_{0} x - 8 a_{0}^{2} - 4

\int sec^{2} (2 x) tan^{4} (2 x) d x

\int \frac{1}{5 x ^{2} + 1} d x

x \to + 100 lim (7)

\int \frac{4 + x - 3 x}{x} d x

f (x) = - 2 x^{4} + 5 x^{2} - 3, (1, 0)

What is the tangent of 5-9+8x^2 ?
The tangent of 5-9+8x^2 is y=16a_{0}x-8a_{0}^2-4