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

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

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tangent of f(x)=1-x-x^2,\at x=0

tangent of

f (x) = 1 - x - x^{2}, at x = 0

y = - x + 1

Solution steps

Find the tangent point:

(0, 1)

Find the slope of

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

Compute the slope of

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

x = 0 : m = - 1

Find the line with slope m=

- 1

and passing through

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

y = - x + 1

x (29 - 2 x) (16 - 2 x)

lo g_{7} (x)

ln (x + 2), - 1

2 x y^{^{'}} - y = 3 x^{2}

\int 9 tan (x) sec (3) (x) d x

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