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

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

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

tangent of

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

y = - 3 x

Solution steps

Find the tangent point:

(- 1, 3)

Find the slope of

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

Compute the slope of

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

x = - 1 : m = - 3

Find the line with slope m=

- 3

and passing through

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

y = - 3 x

h \to 0 lim (6 x + 3 h)

\frac{d}{d x} (1 + 2 x)

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

\frac{d y}{d x} - y x - y^{2} = 0

\int \frac{2}{v ^{2}} d v

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