What is the tangent of f(x)=x^2-4,\at x=0 ?

The tangent of f(x)=x^2-4,\at x=0 is y=-4

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tangent of f(x)=x^2-4,\at x=0

tangent of

f (x) = x^{2} - 4, at x = 0

y = - 4

Solution steps

Find the tangent point:

(0, - 4)

Find the slope of

f (x) = x^{2} - 4 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = x^{2} - 4

x = 0 : m = 0

Find the line with slope m=

0

and passing through

(0, - 4) : y = - 4

y = - 4

\frac{d}{d x} (\frac{3 x ^{4} + 1}{x ^{3}})

\frac{d}{d x} (2 - \frac{x ^{2} - 2 x + 4}{x - 2})

\int \frac{ln ( x )}{x} d x

x^{2} \frac{d ^{2} y}{d x ^{2}} - 5 x y^{^{'}} + 8 y = 4 x^{2}

\int_{0}^{1} 2 π (x + 1) (x^{2}) d x

What is the tangent of f(x)=x^2-4,\at x=0 ?
The tangent of f(x)=x^2-4,\at x=0 is y=-4