To solve the problem of finding the quadratic function (f(x)) given the conditions (f(0)=1) and (f(x+1)-f(x)=2x), follow these steps:

Step 1: Assume the form of the quadratic function

Since (f(x)) is quadratic, let:
[f(x) = ax^2 + bx + c \quad (a \neq 0)]

Step 2: Use (f(0)=1) to find (c)

Substitute (x=0):
[f(0) = a(0)^2 + b(0) + c = c = 1]
Thus, (c=1).

Step 3: Compute (f(x+1)-f(x)) and equate coefficients

First, expand (f(x+1)):
[f(x+1) = a(x+1)^2 + b(x+1) + c = a(x^2 +2x+1) +b(x+1)+1]
[= ax^2 +2ax +a +bx +b +1]

Subtract (f(x)):
[f(x+1)-f(x) = [ax^2 +2ax +a +bx +b +1] - [ax^2 +bx +1]]
[=2ax + (a +b)]

Given this equals (2x), equate coefficients:

• Coefficient of (x): (2a =2 \implies a=1)
• Constant term: (a +b =0 \implies b=-1) (since (a=1))

Step 4: Write the final function

Substitute (a=1), (b=-1), (c=1):
[f(x) = x^2 -x +1]

Answer: (\boxed{x^2 -x +1})

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