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

The tangent of f(x)=1+2x+x^2,\at x=-1 is y=0

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

tangent of

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

y = 0

Solution steps

Find the tangent point:

(- 1, 0)

Find the slope of

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

Compute the slope of

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

x = - 1 : m = 0

Find the line with slope m=

0

and passing through

(- 1, 0) : y = 0

y = 0

\int (sec (x))^{2} d x

x \to 0 + lim (\frac{5}{sin ( x )})

\int e^{7 x + e^{7 x}} d x

f (x) = 4 x^{3} + 3 x^{2} - 2

x \to \infty lim (x^{- 2})

What is the tangent of f(x)=1+2x+x^2,\at x=-1 ?
The tangent of f(x)=1+2x+x^2,\at x=-1 is y=0