If in applying the [quadratic formula](https://artofproblemsolving.com/wiki/index.php/Quadratic_formula) to a [quadratic equation](https://artofproblemsolving.com/wiki/index.php/Quadratic_equation)
\[f(x) \equiv ax^2 + bx + c = 0,\]
it happens that $c = \frac{b^2}{4a}$, then the graph of $y = f(x)$ will certainly:
$\mathrm{(A) \ have\ a\ maximum  } \qquad \mathrm{(B) \ have\ a\ minimum} \qquad$ $\mathrm{(C) \ be\ tangent\ to\ the\ x-axis} \qquad$ $\mathrm{(D) \ be\ tangent\ to\ the\ y-axis} \qquad$ $\mathrm{(E) \ lie\ in\ one\ quadrant\ only}$
The [discriminant](https://artofproblemsolving.com/wiki/index.php/Discriminant) of the quadratic equation is $b^2 - 4ac = b^2 - 4a\left(\frac{b^2}{4a}\right) = 0$. This indicates that the equation has only one root (applying the quadratic formula, we get $x = \frac{-b + \sqrt{0}}{2a} = -b/2a$). Thus it follows that $f(x)$ touches the x-axis exactly once, and hence is tangent to the x-axis $\Rightarrow \mathrm{(C)}$.