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

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

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

tangent of

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

y = - 3 x + 1

Solution steps

Find the tangent point:

(- 1, 4)

Find the slope of

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

Compute the slope of

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

x = - 1 : m = - 3

Find the line with slope m=

- 3

and passing through

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

y = - 3 x + 1

9 y^{^{''}} + 18 y^{^{'}} + 17 y = 0

n = 1 \sum \infty \frac{2 n}{5 n + 1}

\frac{\partial}{\partial y} (\frac{12 x}{x + y})

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

sin (3 x + 1)

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