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

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

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

tangent of

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

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 1

Find the line with slope m=

- 1

and passing through

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

y = - x - 1

\int_{- 2}^{1} (x^{2} - 6 x + 12) d x

\frac{d}{d x} (\frac{x}{( 3 x - 2 ) ^{4}})

\int (sec^{2} (x) + 7) d x

f (x) = - 3 x^{4}, at x = - 1

\int_{- 2}^{3} (x^{2} + 4) - (- x^{2} - 1) d x

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