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

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

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

tangent of

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

y = - x - 1

Solution steps

Find the tangent point:

(- 1, 0)

Find the slope of

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

Compute the slope of

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

x = - 1 : m = - 1

Find the line with slope m=

- 1

and passing through

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

y = - x - 1

\frac{d}{d x} (\frac{2901}{1 + 4.76 e ^{- 0.3 x}})

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

\frac{d}{d x} (ln (x^{5} - 3 x^{3}))

n = 0 \sum \infty \frac{1}{6 ^{n}}

x y^{^{'}} + 2 y = 12 x^{2}

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