What is the tangent of x^2-9 ?

The tangent of x^2-9 is y=2a_{0}x-a_{0}^2-9

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tangent of x^2-9

tangent of

x^{2} - 9

y = 2 a_{0} x - a_{0}^{2} - 9

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{2} - 9)

Find the slope of

y = x^{2} - 9 : \frac{d y}{d x} = 2 x

Compute the slope of

y = x^{2} - 9

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

Find the line with slope m=

2 a_{0}

and passing through

(a_{0}, a_{0}^{2} - 9) : y = 2 a_{0} x - a_{0}^{2} - 9

y = 2 a_{0} x - a_{0}^{2} - 9

\frac{\partial}{\partial x} (\frac{3 y ^{2}}{( 5 x + 3 y ) ^{2}})

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = - 2, y^{^{'}} (0) = 1

\int \frac{1}{x} - x^{2} + sin (x) - 3^{x} d x

\frac{d y}{d x} + 2 y = 9 e^{x}