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

The tangent of x^2+2,\at x=-1 is y=-2x+1

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

tangent of

x^{2} + 2, at x = - 1

y = - 2 x + 1

Solution steps

Find the tangent point:

(- 1, 3)

Find the slope of

y = x^{2} + 2 : \frac{d y}{d x} = 2 x

Compute the slope of

y = x^{2} + 2

x = - 1 : m = - 2

Find the line with slope m=

- 2

and passing through

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

y = - 2 x + 1

x \to 0 lim (\frac{1 - 2 cos ( x ) + cos ( 2 x )}{x ^{2}})

y = - \frac{2}{5} x - 3

\frac{d}{d x} (e^{c o s (2 x)} sin (2 x))

\int - 6 x^{2} d x

x \to \infty lim (\frac{2 e ^{2 x}}{2 x})

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