quadratic equation plot investigation Let $f(x) =-x^2-4x+18 $ so i plot it like this :

But my imagination created the following:
$-x^2-4x+18=0 ->
x^2+4x = 18->
x^2+4x-18 = 0$
Which yields the parabola upside down. Where's the mistake I made?
 A: The trouble with your logic is that you are setting the expression to zero when you reverse signs.
Note that the graph of your function was $y=f(x)$. If you reverse signs for $f(x)$, you reverse signs for $y$.
By setting $f(x)=0$ and reversing signs, you have shown that both $f(x)$ and $-f(x)$ have the same $x$-intercepts.
A: The points of the graph have two coordinates $(x,y)$ where $y=-x^2-4x+18$.
So, if you take $y'=x^2+4x-18=-y\;$ you find  a graph that is the symmetric of the previous one with respect to the $x$ axis. 
For $y=0$ you find, if they exists, the points where the graph intersects the $x$ axis and the abscissas of these points are the solutions of the equation $-x^2-4x+18=0$.
 It is true that these solutions are the same as the solutions of $x^2+4x-18=0$ but this simply means that the points of the two graphs that stay on the $x$ axis are fixed points of the symmetry $ (x,y)\rightarrow (x,-y)$.
A: $ y=-x^2-4x+18 $ and $y=x^2+4x-18$ have same roots, but are mirrored from one to the other about the x-axis.
