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

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

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

y = - x + 1

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