On the same set of axes are drawn the graph of $y=ax^2+bx+c$ and the graph of the equation obtained by replacing $x$ by $-x$ in the given equation. 
If $b \neq 0$ and $c \neq 0$ these two graphs intersect: 
$\textbf{(A)}\ \text{in two points, one on the x-axis and one on the y-axis}\\ \textbf{(B)}\ \text{in one point located on neither axis}\\ \textbf{(C)}\ \text{only at the origin}\\ \textbf{(D)}\ \text{in one point on the x-axis}\\ \textbf{(E)}\ \text{in one point on the y-axis}$
Replacing $x$ with $-x,$ we have the graph of $y = a(-x)^2 + b(-x) + c = ax^2 - bx+c.$
We can plug in simple values of $a, b,$ and $c$ for convenient drawing. For example, we can graph $x^2+2x+1$ and $x^2-2x+1.$ We see that the parabolas intersect at $(0, 1),$ which is on the y-axis.
The answer is $\boxed{\textbf{(E)}}.$