What is the tangent of x^2-8 ?

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

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tangent of x^2-8

tangent of

x^{2} - 8

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

Compute the slope of

y = x^{2} - 8

x = a_{0} : m = 2 a_{0}

Find the line with slope m=

2 a_{0}

and passing through

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

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

\frac{d}{d x} (sin (2 x) cos (2 x))

f (x) = x^{3}, at x = 6

\int - 9 x d x

x \to 2 lim (51 - x - 7)

f (x) = x - 1, 1, 2