# Calculus

posted by
**Matt** on
.

Great! I understand now that

C0=f(a)

c1=f'(a)

c2=f"(a)/2

Now, If I am to find the parabolization of the equation x^2-x at x=2, then

c0=x^2-x=2^2-2=2

c1=2x-1=2(2)-1=3

c2=2/2=1

So, the equation (taken from c0+c1(x-a)+c2(x-a)^2) is >>

2+3(x-2)+1(x-2)^2??

Is this correct?

Thanks

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

Actually, linearization is approxiamtion by a straight line while Parabolization is approximation of a function by a parabola.

You start from:

f(a) = P(a)

f'(a) = P'(a)

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

Then you insert

P(x)=c0+c1(x-a)+c2(x-a)^2

in here.

So,

P(a) = C0

P'(x) = C1+2c2(x-a) --->

P'(a) = C1

P''(x) = 2C2 -->

P''(a) = 2C2

This means that:

f(a) = P(a) = C0

f'(a) = P'(a) = C1

f''(a) = P''(a) = 2C2

And it follows that:

P(x) = C0+C1(x-a)+c2(x-a)^2 =

f(a) + f'(a)(x-a) + f''(a)/2 (x-a)^2

There is an easy way to check. If you parabolize a parabola, you should get the same thing back. So, let's see:

2+3(x-2)+ (x-2)^2 =

2 + 3x-6 + (x^2 - 4x + 4) =

2-6+4 + 3x-4x + x^2 =

x^2 - x.

So, it's indeed the same!