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

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

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

tangent of

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

y = - 2 x + 1

Solution steps

Find the tangent point:

(- 1, 3)

Find the slope of

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

Compute the slope of

f (x) = 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

\frac{d}{d t} (\frac{1}{1 + 4 t ^{2}})

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

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

\int \frac{x + 6}{( x + 2 ) ^{2}} d x

\frac{d}{d x} (in sin (x))

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