quadratic equation plot investigation

Question

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?

Answer 1

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)$.

Answer 2

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.

Answer 3

$ 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.