derivitive of parabola (parabolization)

[matt944]

In calc, we are studying parabolization. It is the linerazation of a parabola. The linerazation of a standard line is L(x) = b0+b1(x-a) when f(x) is at x=a

b0 = f(a)
b1=f'(a)

The parabolization of f(x) at x=a is given by the equation
P(x)=c0+c1(x-a)+c2(x-a)^2.

f(a) = P(a)
F'(a) = P'(a)
f''(a) = P''(a)

I need to find a formula for c0, c1, and c2 in terms of f(a), f'(a) and f"(a)

Im sure that I need to find the first and second derivitive of the equation c0+c1(x-a)+c2(x-a)^2. Im just not sure where to start...

Thanks!

Matt

 

[tkhunny]

"Parabolization"? What a silly term. It is a second degree approximation. Why does it need a name? I can hardly wait for the authors name for a 5th degree approximation.

How about a quick example of what you ahve written?

f(x) = sin(x) at x = 0 -- sin(0) = 0
f'(x) = cos(x) -- cos(0) = 1
f"(x) = -sin(x) -- -sin(x) = 0

P(0) = c0+c1(0)+c2(0)^2 = c0 = 0
P'(0) = c1+2*c2(0) = c1 = 1
P"(0) = 2*c2 = 0 ==> c2 = 0

P(x) = 0+1(x)+0(x)^2 = x