first off, we know this equation is a quadratic equation, so it is of the form y = ax² + bx + c, where "a, b and c" are digits or constants, and we have no clue what they are.
well, let's take a peek at the table of values and let's make, hmmm usually we'd end up with a system of equations of 3 variables, but in this case we can cook it earlier by being wimpy and using the (0 , 0) point from the table, that says that y = 0 and x = 0, then we'll be using the point (-2 , 0), again being wimpy for the 0 and we know that y = 0 whilst x = -2.
[tex]y = ax^2+bx+c \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{using (0 , 0)}}{0 = a(0)^2+b(0)+c}\implies \boxed{0=c} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{using (-2 , 0)}}{0=a(-2)^2+b(-2)+0}\implies 0=4a-2b\implies 2b=4a \\\\\\ \cfrac{2b}{2}=2a\implies \underline{b=2a} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{using (2 , 4)}}{4=a(2)^2+b(2)+0}\implies 4=4a+2b\implies \stackrel{\textit{substituting 2b}}{4=4a+2(2a)} \\\\\\ 4=4a+4a\implies 4=8a\implies \cfrac{4}{8}=a\implies \boxed{\cfrac{1}{2}=a}[/tex]
[tex]~\dotfill\\\\ \stackrel{\textit{we know that}}{b=2a}\implies b=2\left( \cfrac{1}{2} \right)\implies \boxed{b=1} \\\\[-0.35em] ~\dotfill\\\\ y=\cfrac{1}{2}x^2+1x+0\implies \boxed{y=\cfrac{1}{2}x^2+x} \\\\[-0.35em] ~\dotfill\\\\ \textit{when x = 4, what's "y"?}\qquad y=\cfrac{1}{2}(4)^2+4\implies y=\cfrac{16}{2}+4\implies y=12[/tex]