Suppose a parabola has vertex $(\frac{1}{4},\frac{-9}{8})$ and equation $ax^2+bx+c=y$ where $a>0$ and $a+b+c $ is an integer. Find the minimum possible value of $a$ under the given condition.
My approach $(x-\frac{1}{4})^2=4a'(y+\frac{9}{8})$
$\frac{x^2-\frac{x}{2}+\frac{1}{16}}{4a'}-\frac{9}{8}=y$
$\frac{1}{4a'}-\frac{1}{4a'}+\frac{1}{64a'}-\frac{9}{8}=Z$, where $Z$ is an integer
$\frac{9}{64a'}-\frac{9}{8}=Z$, from here I am confused
On
We know that the vertex x coordinates is $\frac{-b}{2a} = \frac{1}{4}$ solve for $b = \frac{-a}{2}$ now the point $\frac{1}{4} , \frac{-9}{8}$ satisfies the parabola so
$\frac{a}{16} - \frac{a}{8} + c = \frac{-9}{8}$
$a-2a + 16c = -18 $
$c = \frac{-18 +a}{16}$ finally
$a+b+c = a + \frac{-a}{2} +\frac{-18+a}{16} \in \mathbb{Z}$ find the solution and pick the smallest one
we will have $16 \mid 9a-18 $
The equation of the parabola is:
$$a \left(x-\frac{1}{4}\right)^2 - \frac{9}{8}$$ $$=a \left(x^2-\frac{1}{2}x+\frac{1}{16} \right)- \frac{9}{8}$$ $$=ax^2-\frac{a}{2}x+ \left(\frac{a}{16}-\frac{9}{8} \right)$$
Then we find that: $$a+b+c = \frac{9}{16}(a-2).$$
When $a=0$, $a+b+c = -\frac{9}{8}$. Since the gradient of the line is positive, if we try $a+b+c=-1$, we will get a positive value of $a$.
Therefore we have that: $-1 = \frac{9}{16}(a-2) \Rightarrow a = \frac{2}{9}$.