What is the tangent of f(x)=x^2-1,(-2,3) ?

The tangent of f(x)=x^2-1,(-2,3) is y=-4x-5

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tangent of f(x)=x^2-1,(-2,3)

tangent of

f (x) = x^{2} - 1, (- 2, 3)

y = - 4 x - 5

Solution steps

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

(- 2, 3) : m = - 4

Find the line with slope m=

- 4

and passing through

(- 2, 3) : y = - 4 x - 5

y = - 4 x - 5

x \to 2 lim (\frac{3}{2 - x})

\int (3 t + t^{2}) sin (3 t) d t

\int ln (1 + t^{2}) d t

\int \frac{tan ^{3} ( ln ( x ) )}{3 x} d x

2 w^{3 π}, at 7

What is the tangent of f(x)=x^2-1,(-2,3) ?
The tangent of f(x)=x^2-1,(-2,3) is y=-4x-5