 Quadratic minimization along a line

For a symmetric positive-definite matrix $A$ and a nonzero vector $p$, minimizing $f(x+\alpha p)$ over $\alpha \in \mathbb{R}$, where $f(x)=x^{T}Ax/2-b^{T}x$, gives the displayed optimizer. Put $r=b-Ax$. Expansion gives $f(x+\alpha p)=f(x)-\alpha p^{T}r+{\alpha }^2{p}^{T}Ap/2$, a strictly convex quadratic function of $\alpha$. At its minimum,Strict decrease is therefore equivalent to $p^{T}r\ne 0$. This is minimization over a full line, so the optimizer can be negative. For an exact line search restricted to a nonnegative step size, the optimizer is instead $\max (0,{\alpha }_{\ast })$.