= Solution
The true quadratic has $g(x)=c+Hx$ and the exact expansion
$$
f(x+s)=f(x)+g(x)^Ts+\frac12s^THs.
$$
Subtracting the model with Hessian $I$ yields $f(x+s)-m(s)=\tfrac12s^T(H-I)s$. Since $H$ is symmetric positive definite and $\|H\|_2=1$, its <eigenvalues> lie in $(0,1]$, so $\|H-I\|_2\leq1$. Therefore
$$
\boxed{|f(x+s)-m(s)|\leq\frac12\|s\|_2^2\leq\frac12\Delta^2\leq\Delta^2.}
$$
In fact the stated assumptions give a factor one-half better than the requested bound.
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