What is the tangent of f(x)=x^2-2,\at x=0 ?

The tangent of f(x)=x^2-2,\at x=0 is y=-2

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tangent of f(x)=x^2-2,\at x=0

tangent of

f (x) = x^{2} - 2, at x = 0

y = - 2

Solution steps

Find the tangent point:

(0, - 2)

Find the slope of

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

Compute the slope of

f (x) = x^{2} - 2

x = 0 : m = 0

Find the line with slope m=

0

and passing through

(0, - 2) : y = - 2

y = - 2

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\int_{0}^{\frac{π}{4}} sin^{4} (x) d x

2 y^{^{'}} = 4 t cos (t) y^{2}, y (0) = 1

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What is the tangent of f(x)=x^2-2,\at x=0 ?
The tangent of f(x)=x^2-2,\at x=0 is y=-2