What is the tangent of x^2-6 ?

The tangent of x^2-6 is y=2a_{0}x-a_{0}^2-6

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tangent of x^2-6

tangent of

x^{2} - 6

y = 2 a_{0} x - a_{0}^{2} - 6

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{2} - 6)

Find the slope of

y = x^{2} - 6 : \frac{d y}{d x} = 2 x

Compute the slope of

y = x^{2} - 6

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

Find the line with slope m=

2 a_{0}

and passing through

(a_{0}, a_{0}^{2} - 6) : y = 2 a_{0} x - a_{0}^{2} - 6

y = 2 a_{0} x - a_{0}^{2} - 6

(2 + x) (\frac{d y}{d x}) = 3 y

f (x) = \frac{3 x}{x}, x = 8

3 x^{3} - 3 x

\frac{\partial}{\partial y} (\frac{4 ln ( y + 2 )}{x e ^{x}})

\int 7 x d x