Werid Algebra/Calculus?

Question

A quadratic function, p(x)= Ax^2+Bx+C, is called a quadratic approximation of the function F for values x values near a, provided that the following three conditions are met;

F(a)=p(a)
F prime(a)=p prime(a)
F double prime(a)=p double prime(a)

For values of x near a=0, find a quadratic polynomial,
p(x)= Ax^2+Bx+C, that is a quadratic approximation for the function F(x)= e^(-x^2).

Answer 1

F(a) = p(a)
F'(a) = p'(a)
F''(a) = p''(a)

p(x) = Ax^2 + Bx + C
F(x) = e^(-x^2)
p'(x) = 2Ax + B
F'(x) = (-2x)(e^(-x^2))
p''(x) = 2A
F''(x) = (4x^2)(e^(-x^2))

Using the equalities:
Ax^2 + Bx + C = e^(-x^2)
2Ax + B = (-2x)(e^(-x^2))
2A = (4x^2)(e^(-x^2))

And plugging in a=0 for x, we have:
A(0)^2 + B(0) + C = e^(-(0)^2)
2A(0) + B = (-2(0))(e^(-(0)^2))
2A = (4(0)^2)(e^(-(0)^2))

Simplifying:
C = e^0
B = 0 * e^0
2A = 0 * e^0

So
C = 1
B = 0
A = 0

Therefore the quadratic approximation of F(x) = e^(-x/2) around a=0 is:
p(x) = 1

Answer 2

f(x) =f(0) +x*f´(0) +x^2/2! f´´(0)
f(0)=1
f´(x) = e^-x^2(-2x) so f´(0)=0
f´´(x) = -2(( e^-x^2+x*e^-x^2*(-2x)) and f´´ (0) =-2
so
p(x) = 1-x^2