= Solution
The convex <Legendre transform>, or <convex conjugate>, is
$$
\boxed{f^*(p)=\sup_{x\in\mathbb R^n}\{p^Tx-f(x)\}.}
$$
For a differentiable strictly <convex function>, an attained maximum is characterized by $p=\nabla f(x)$, and then $f^*(p)=p^Tx-f(x)$ at that maximizing $x$.
For the <affine covariance of the convex conjugate>, assume $\lambda>0$ and substitute $y=x-x_0$:
$$
\begin{aligned}
g^*(p)&=\sup_y\{p^T(y+x_0)-\lambda f(y)+\mu\}\\
&=p^Tx_0+\mu+\lambda\sup_y\{(p/\lambda)^Ty-f(y)\}.
\end{aligned}
$$
Hence
$$
\boxed{g^*(p)=\lambda f^*(p/\lambda)+p^Tx_0+\mu.}
$$
Positive scaling is necessary for this supremum definition, although the printed formula does not specify it. For example, $f(x)=x^2/2$ and $\lambda=-1$ give $g^*(p)=+\infty$, not the finite expression $-p^2/2$. If the transform is instead defined as a stationary value using a locally invertible gradient, the same algebraic identity holds for any $\lambda\ne0$: the critical-point relation is $p=\lambda\nabla f(y)$, and substitution of that stationary $y$ gives the displayed formula. That extension does not assert a supremum for negative scaling.
For the <positive-definite quadratic form>, symmetry and positive <eigenvalues> of $A$ give
$$
p^Tx-\tfrac12x^TAx=\tfrac12p^TA^{-1}p-\tfrac12(x-A^{-1}p)^TA(x-A^{-1}p).
$$
The second term is nonpositive and vanishes exactly at $x=A^{-1}p$. Therefore
$$
\boxed{f^*(p)=\tfrac12p^TA^{-1}p.}
$$
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