What is the tangent of f(x)=-1/2 x^2-2x+3,\at x=0 ?

The tangent of f(x)=-1/2 x^2-2x+3,\at x=0 is y=-2x+3

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tangent of f(x)=-1/2 x^2-2x+3,\at x=0

tangent of

f (x) = - \frac{1}{2} x^{2} - 2 x + 3, at x = 0

y = - 2 x + 3

Solution steps

Find the tangent point:

(0, 3)

Find the slope of

f (x) = - \frac{1}{2} x^{2} - 2 x + 3 : \frac{df ( x )}{d x} = - x - 2

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 = - 2

Find the line with slope m=

- 2

and passing through

(0, 3) : y = - 2 x + 3

y = - 2 x + 3

f (t) = t - sin (t)

\frac{d}{d y} (\frac{t ^{2}}{4 y ^{4}})

\frac{d}{d x} (10 x e^{x^{2}})

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

f (x) = 12 x

What is the tangent of f(x)=-1/2 x^2-2x+3,\at x=0 ?
The tangent of f(x)=-1/2 x^2-2x+3,\at x=0 is y=-2x+3