If $y=f(x)$ be the equation of the parabola which is touched by the line $y=x$

Question

If $y=f(x)$ be the equation of the parabola which is touched by the line $y=x$ at the point where $x=1$.Prove that $f'(1)=1$ and $2f(0)=1-f'(0)$

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Let $f(x)=ax^2+bx+c$ be the parabola.Since the parabola is touched by the line $y=x$ at $x=1$.So $f'(1)=1$,because slope of $y=x$ is $1$.
But i cannot prove the second part of the question.

Answer 1

We have the point of intersection as $(1,1)$ . From here $1=a+b+c $ also $\frac {dy }{dx}=1$ thus $1=2a+b $ thus $a=\frac {1-b}{2}$ putting this in earlier equation. We have $1=\frac {1-b}{2}+b+c $ thus $2=1+b+2c $ hence $2c=1-b $ now $2f (0)=2c,1-f'(0)=1-b $. Hence the proof.

Answer 2

The tangent at $x=1$ is given by $$ y-f(1) = f'(1)(x-1), $$ so for it to be $y=x$, $f'(1)=1$ and $f(1)=1$. This gives us two equations for the coefficients: \begin{align} f(1) = a+b+c = 1 \tag{1} \\ f'(1) = 2a+b = 1. \tag{2} \end{align} Meanwhile, $$ f(0) = c, \\ f'(0) = b. $$ $2(1)-(2)$ gives $$ b+2c=1 \implies f'(0) + 2f(0) = 1, $$ as required.