How to justify why a negative sign reflects a parabola over its vertex (x value)

Question

Sorry if this is a really dumb question but I have to justify the above and just cannot think of how to.

Answer 1

Graph $x^2$ and $-x^2$ side by side.

Answer 2

Hint:

From the way the question has been posed, I assume you are talking about functions. Otherwise, leave a comment so as to make the appropriate edits.

Let $f:\mathbb{R}\to\mathbb{R}$ be a function such that its graph is a parabola. Then, by definition, its graph is the set: $$G(f):=\{(x,f(x))|x\in\mathbb{R}\}$$ Now, the graph of $-f$ is the set: $$G(-f)=\{(x,-f(x))|x\in\mathbb{R}\}.$$ Can you continue now?

Answer 3

Because if a parabola, say $f(x)$, has absolute maximum at point $x_0$, then $-f(x)$ should have absolute minimum at $x_0$ and vice versa. But when we multiply $f(x)$ by $(-1)$, we don't flip the parabola over it's turning point. We are taking the symmetry of it with respect to the $x-$axis.