Is the graph of this problem statement incorrect?
In solving the quadratic equation, x= $\frac{(a+3)\pm\sqrt{(a+3)^2-12a}}{2a}$ =$\frac{(a+3)\pm|a-3|}{2a}$
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when a$\geq$3, $x_1$=1>$x_2$=$\frac{3}{a}$;x2 is the smaller root. In this case, the area of $\Delta$ AOC has many values pending on value of a.
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when a$\leq$3, $x_1$=1<$x_2$=$\frac{3}{a}$;1 is the small root. In this case, the area of $\Delta$ AOC is $\frac{3}{2}$.
Hope this clarify the picture.
We have $a+b+c=f(1)=0$, hence clearly $1$ is a root. The question is whether it is the smallest root.
Also $f(0)=c=3$.
From vieta's formula, $\frac{c}{a}=1\cdot x_2$, that the other root is $\frac{c}{a}=\frac3a$.
Hence if $\frac3a > 1$, then the answer is $\frac32$.
However, it is also possible that $\frac3a < 1$ as well. Then the answer is $\frac{3}{2}\cdot \frac{3}{a}=\frac{9}{2a}$ which is a function of $a$.
For example, consider the graph of $$f(x)=6x^2-9x+3=3(2x^2-3x+1)=3(2x-1)(x-1).$$