6. The parabola $y=a x^{2}+b x+c$ has vertex at $\left(\frac{1}{4},-\frac{9}{8}\right)$. If $a>0$ and $a+b+c$ is an integer, find the minimum possible value of $a$.

The parabola $y=a x^{2}+b x+c$ has its vertex at $\left(\frac{1}{4},-\frac{9}{8}\right)$. If $a>0$ and $a+b+c$ is an integer, find the minimum possible value of $a$.
6. $\frac{2}{9}$
6. We have $y=a\left(x-\frac{1}{4}\right)^{2}-\frac{9}{8}$. Note that $a+b+c$ is equal to the value of $y$ when $x=1$, i.e. $a+b+c=a\left(1-\frac{1}{4}\right)^{2}-\frac{9}{8}=\frac{9 a-18}{16}$. Hence we need to look for the smallest positive $a$ for which $9 a-18$ is divisible by 16. This corresponds to $9 a-18=-16$, or $a=\frac{2}{9}$.