What is the tangent of f(x)=x^2-8 ?

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

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tangent of f(x)=x^2-8

tangent of

f (x) = 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

f (x) = x^{2} - 8 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = 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} (\frac{x + 3}{∣ x + 3 ∣})

x \to 0 + lim (\frac{( ln ( x ) ) ^{2}}{x})

\frac{\partial}{\partial x} (sin^{4} (x))

(- 4, 6), (3, - 8)

\int (4 x^{3} - 2) x^{2} d x

What is the tangent of f(x)=x^2-8 ?
The tangent of f(x)=x^2-8 is y=2a_{0}x-a_{0}^2-8