### Instruction: Suppose that a parabola has vertex $\left(\frac{1}{4},-\frac{9}{8}\right)$ and equation $y = ax^2 + bx + c$, where $a > 0$ and $a + b + c$ is an integer.  Find the smallest possible value of $a.$.### Response: The first thing I notice is that the vertex of the parabola is given, and I recall that the vertex form of a parabola is $y = a(x-h)^2 + k$, where $(h,k)$ is the vertex.
So I can rewrite the equation of the parabola as $y = a\left(x-\frac{1}{4}\right)^2 - \frac{9}{8}$.
Comparing this with the standard form $y = ax^2 + bx + c$, I can see that $b = -\frac{a}{2}$ and $c = -\frac{9}{8} + \frac{a}{16}$.
Now I want to use the fact that $a + b + c$ is an integer.
Substituting the expressions for $b$ and $c$, I get $a - \frac{a}{2} - \frac{9}{8} + \frac{a}{16}$ is an integer.
Simplifying this, I get $\frac{9a}{16} - \frac{9}{8}$ is an integer.
Multiplying both sides by $\frac{16}{9}$, I get $a - 2$ is an integer.
Since $a > 0$, the smallest possible value of $a$ that makes $a - 2$ an integer is $a = 3$.
I can check that this works by plugging it back into the equation of the parabola and verifying that the vertex is still $\left(\frac{1}{4},-\frac{9}{8}\right)$.