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

The tangent of f(x)=x^2-1,\at x=-3 is y=-6x-10

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

tangent of

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

y = - 6 x - 10

Solution steps

Find the tangent point:

(- 3, 8)

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 = - 3 : m = - 6

Find the line with slope m=

- 6

and passing through

(- 3, 8) : y = - 6 x - 10

y = - 6 x - 10

\frac{\partial}{\partial x} (3 y x^{2})

\frac{d}{d x} (5 - 8 x)

\frac{\partial}{\partial y} (- \frac{y ^{3}}{3} - 8 x)

n \to \infty lim (\frac{n}{10 + n})

\frac{( s + 2 ) ^{2}}{s ^{3}}

What is the tangent of f(x)=x^2-1,\at x=-3 ?
The tangent of f(x)=x^2-1,\at x=-3 is y=-6x-10