well, let's first off, expand the expression, 3x(x-4) => 3x²-12x, which we can also write as
f(x) = 3x² - 12x + 0
now, this is a quadratic equation with a leading term with a positive coefficient, namely its graph is a parabola, or a Cup-Shaped figure, so the graph comes from the top, goes down down down, reaches a U-turn, then goes back up up up.
the U-turn, or vertex of the quadratic, is the lowest point in the "cup".
[tex]\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ \stackrel{f(x)}{y}=\stackrel{\stackrel{a}{\downarrow }}{3}x^2\stackrel{\stackrel{b}{\downarrow }}{-12}x\stackrel{\stackrel{c}{\downarrow }}{+0} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left( -\cfrac{(-12)}{2(3)}~~,~~0-\cfrac{(-12)^2}{4(3)} \right)\implies \left( \cfrac{12}{6}~,~0-\cfrac{144}{12} \right)\implies (2~,~-12)[/tex]
