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

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

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

tangent of

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

y = - 3 x - 1

Solution steps

Find the tangent point:

(- 1, 2)

Find the slope of

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

Compute the slope of

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

x = - 1 : m = - 3

Find the line with slope m=

- 3

and passing through

(- 1, 2) : y = - 3 x - 1

y = - 3 x - 1

\int (a^{2} + x^{2})^{\frac{1}{2}} d x

y^{^{''}} - 8 y^{^{'}} + 9 y = x e^{x}

sin (2 (t - \frac{π}{2}))

x^{2} - 4 x, x - 4

\frac{\partial}{\partial y} (2 x^{2} - 5 y^{2})

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