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

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

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tangent of y=x^2-2

tangent of

y = x^{2} - 2

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

Compute the slope of

y = x^{2} - 2

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

Find the line with slope m=

2 a_{0}

and passing through

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

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

x \to 3 lim (4)

15 x - 0.4 x^{2} - 80

\frac{\partial}{\partial x} (cos (5 x + 9 y + 7 z))

\int \frac{1}{x ^{2} - 2 x + 4} d x

f (x) = cos^{6} (x)