What is the tangent of y=x^2-8 ?

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

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tangent of y=x^2-8

tangent of

y = 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

\int_{- \infty}^{1} x e^{2 x} d x

\int x^{2} - a^{2} d x

- \frac{1}{4 s}

f (x) = x^{π^{2}} + (π^{2})^{x}

n = 1 \sum \infty ln (\frac{1}{1 + 1})