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

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

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

tangent of

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

y = - 4 x - 1

Solution steps

Find the tangent point:

(- 1, 3)

Find the slope of

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

Compute the slope of

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

x = - 1 : m = - 4

Find the line with slope m=

- 4

and passing through

(- 1, 3) : y = - 4 x - 1

y = - 4 x - 1

y^{^{'}} + 5 y = 2 sin (e^{5 x})

\int_{2}^{6} t^{3} ln (5 x) d x

\frac{1}{( s ^{2} - 1 )}

\frac{\partial}{\partial x} (7 + x ln (x y - 5))

\frac{d}{d y} (2 y)

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