Calculator use is allowed on these questions, so use it wisely.
The graph of $y=g(x)$ is shown in the figure above. If $g(x)=a{x}^2+bx+c$ for constants $a,b$, and $c$, and if $abc\ne 0$, then which of the following must be true?

Remember your transformation rules. Whenever a parabola faces down, the quadratic equation has a negative sign in front of $x^2$ term. It always helps to Plug In! For example, if your original equation was $(x$ $-2{)}^2$, putting a negative sign in front would make the parabola open downward, so you'll have $-(x-2{)}^2$. If you expand it out, you get $-x^2$ $+4x-4$. Notice that the value of $a$ in this equation is $-1$. Also notice that the value of $c$ is $-4$. This allows you to eliminate (B) and (D). Now you must plug in differently to distinguish between (A) and (C). Be warned: You must use fractions to help discern which is correct. Say the $x$-intercepts take place at $x=\frac{1}{2}$ and $x=\frac{3}{4}$. Rewriting those two expressions means that the factors are $\left(x-\frac{1}{2}\right)$ and $\left(x-\frac{3}{4}\right)$. If you FOIL out the terms, you end up with $x^2-\frac{5}{4}x+\frac{3}{8}$. Remember, the parabola opens downward, so you must multiply by $-1$ to each term to yield $-x^2+\frac{5}{4}x-\frac{3}{8}$. Your values of $a$ and $c$ are now $-1$ and $-\frac{3}{8}$, respectively. Multiply the two values and you get $\frac{3}{8}$, which allows you to eliminate (A) and confidently choose (C). (And that was worth only 1 point! Embrace the POOD!)