Answer:
something noteworthy is that the independent and squared variable in this case will be the "x", namely the graph of that quadratic is a vertical parabola.
\begin{gathered}\bf f(x) = (x+2)(x-4)\implies 0=(x+2)(x-4)\implies x = \begin{cases} -2\\ 4 \end{cases} \\\\\\ \boxed{-2}\rule[0.35em]{7em}{0.25pt}0\rule[0.35em]{3em}{0.25pt}\stackrel{\downarrow }{1}\rule[0.35em]{10em}{0.25pt}\boxed{4}\end{gathered}
f(x)=(x+2)(x−4)⟹0=(x+2)(x−4)⟹x={
−2
4
−2
0
1
↓
4
so the parabola has solutions at x = -2 and x = 4, and its vertex will be half-way between those two guys, namely at x = 1.
since this is a vertical parabola, its axis of symmetry, the line that splits its into twin sides, will be a vertical line, and it'll be the x-coordinate of the vertex, since the vertex hasa a coordinate of x = 1, then the axis of symmetry is the vertical line of x = 1.
