jmc

If these two graphs intersect then the points of intersection occur when $$x^2+a=ax,$$ or $$x^2-ax+a=0.$$ This quadratic has solutions exactly when the discriminant is nonnegative: $$(-a{)}^2-4\cdot 1\cdot a\ge 0.$$ This simplifies to $$a(a-4)\ge 0.$$ This quadratic (in $a$) is nonnegative when $a$ and $a-4$ are either both $\ge 0$ or both $\le 0$. This is true for $a$ in $$(-\infty ,0]\cup [4,\infty ).$$ Therefore the line and quadratic intersect exactly when $a$ is in $\boxed{(-\infty ,0]\cup [4,\infty )}$.