What is the tangent of f(x)=x^2-1,\at x=0 ?

The tangent of f(x)=x^2-1,\at x=0 is y=-1

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tangent of f(x)=x^2-1,\at x=0

tangent of

f (x) = x^{2} - 1, at x = 0

y = - 1

Solution steps

Find the tangent point:

(0, - 1)

Find the slope of

f (x) = x^{2} - 1 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = x^{2} - 1

x = 0 : m = 0

Find the line with slope m=

0

and passing through

(0, - 1) : y = - 1

y = - 1

x \to - 2 lim (\frac{1}{( x + 2 ) ^{2}})

\int j

\int (x) sin (x) d x

\frac{d}{d x} (ln (8 - x^{2}))

y = arcsin (5 x + 1)

What is the tangent of f(x)=x^2-1,\at x=0 ?
The tangent of f(x)=x^2-1,\at x=0 is y=-1