Let the quadratic curve passing through the point (-1, 0) and touching the line $$y = x$$ at (1, 1) be $$y = f(x)$$. Then the $$x$$-intercept of the normal to the curve at the point $$(\alpha, \alpha + 1)$$ in the first quadrant is _______.

Quadratic $$f(x) = ax^2+bx+c$$ passes through (-1,0) and touches y=x at (1,1).

f(-1)=0: a-b+c=0. f(1)=1: a+b+c=1. f'(1)=1: 2a+b=1.

From first two: 2a+2c=1. From third: b=1-2a. From first: a-(1-2a)+c=0, 3a+c=1.

Also 2a+2c=1, so c=(1-2a)/2. Substituting: 3a+(1-2a)/2=1, (6a+1-2a)/2=1, 4a=1, a=1/4.

b=1/2, c=1/4. So f(x)=(x²+2x+1)/4=(x+1)²/4.

f(α)=α+1: (α+1)²/4=α+1, α+1=4 (since α+1≠0 in first quadrant), α=3.

f'(3)=(3+1)/2=2. Normal slope=-1/2. Normal at (3,4): y-4=-1/2(x-3).

x-intercept: y=0: -4=-1/2(x-3), x-3=8, x=11.

The answer is 11.