The sum of the [zeros](https://artofproblemsolving.com/wiki/index.php/Root), the product of the zeros, and the sum of the [coefficients](https://artofproblemsolving.com/wiki/index.php/Coefficient) of the [function](https://artofproblemsolving.com/wiki/index.php/Function) $f(x)=ax^{2}+bx+c$ are equal. Their common value must also be which of the following? 
$\textrm{(A)}\ \textrm{the\ coefficient\ of\ }x^{2}~~~ \textrm{(B)}\ \textrm{the\ coefficient\ of\ }x$
$\textrm{(C)}\ \textrm{the\ y-intercept\ of\ the\ graph\ of\ }y=f(x)$
$\textrm{(D)}\ \textrm{one\ of\ the\ x-intercepts\ of\ the\ graph\ of\ }y=f(x)$
$\textrm{(E)}\ \textrm{the\ mean\ of\ the\ x-intercepts\ of\ the\ graph\ of\ }y=f(x)$
By [Vieta's formulas](https://artofproblemsolving.com/wiki/index.php/Vieta%27s_formulas), the sum of the roots of a [quadratic equation](https://artofproblemsolving.com/wiki/index.php/Quadratic_equation) is $\frac {-b}a$,  the product of the zeros is $\frac ca$, and the sum of the coefficients is $a + b + c$. Setting equal the first two tells us that $\frac {-b}{a} = \frac ca \Rightarrow b = -c$. Thus, $a + b + c = a + b - b = a$, so the common value is also equal to the coefficient of $x^2 \Longrightarrow \fbox{\textrm{A}}$. 
To disprove the others, note that:

$\mathrm{B}$: then $b = \frac {-b}a$, which is not necessarily true.
$\mathrm{C}$: the y-intercept is $c$, so $c = \frac ca$, not necessarily true.
$\mathrm{D}$: an x-intercept of the graph is a root of the polynomial, but this excludes the other root.
$\mathrm{E}$: the mean of the x-intercepts will be the sum of the roots of the quadratic divided by 2.