3. Let $a$, $b$, and $c$ be the lengths of the sides of $\triangle ABC$, and the quadratic function $y=\left(a-\frac{b}{2}\right) x^{2}-c x-a-\frac{b}{2}$ has a minimum value of $-\frac{8}{5} b$ at $x=1$. Then $\triangle ABC$ is ( ).
(A) isosceles triangle
(B) acute triangle
(C) obtuse triangle
(D) right triangle
3. D.

From the problem, we have $\left\{\begin{array}{l}-\frac{-c}{2\left(a-\frac{b}{2}\right)}=1, \\ a-\frac{b}{2}-c-a-\frac{b}{2}=-\frac{8}{5} b .\end{array}\right.$
Therefore, $c=\frac{3}{5} b, a=\frac{4}{5} b$.
Thus, $a^{2}+c^{2}=b^{2}$.
Hence, $\triangle A B C$ is a right triangle.