Respuesta :
since the vertex is at 0,-4 and the zeros are above it, we'll use a vertical parabola equation, namely with the dependent variable to be the "y".
bear in mind that, a zeor or solution of say -1 and 1, simply means x-intercepts at -1, 0 and 1,0.
[tex]\bf \qquad \textit{parabola vertex form}\\\\ \begin{array}{llll} \boxed{y=a(x-{{ h}})^2+{{ k}}}\\\\ x=a(y-{{ k}})^2+{{ h}} \end{array} \qquad\qquad vertex\ ({{ h}},{{ k}})\\\\ -------------------------------\\\\ vertex\ (0,-4)\ \begin{cases} h=0\\ k=-4 \end{cases}\implies y=a(x-0)^2-4\implies y=ax^2-4 \\\\\\ \textit{we also know that } \begin{cases} y=0\\ x=-1 \end{cases}\implies 0=a(-1-0)^2-4 \\\\\\ 0=a(-1)^2-4\implies \boxed{4=a}\qquad thus\implies \boxed{y=4x^2-4}[/tex]
bear in mind that, a zeor or solution of say -1 and 1, simply means x-intercepts at -1, 0 and 1,0.
[tex]\bf \qquad \textit{parabola vertex form}\\\\ \begin{array}{llll} \boxed{y=a(x-{{ h}})^2+{{ k}}}\\\\ x=a(y-{{ k}})^2+{{ h}} \end{array} \qquad\qquad vertex\ ({{ h}},{{ k}})\\\\ -------------------------------\\\\ vertex\ (0,-4)\ \begin{cases} h=0\\ k=-4 \end{cases}\implies y=a(x-0)^2-4\implies y=ax^2-4 \\\\\\ \textit{we also know that } \begin{cases} y=0\\ x=-1 \end{cases}\implies 0=a(-1-0)^2-4 \\\\\\ 0=a(-1)^2-4\implies \boxed{4=a}\qquad thus\implies \boxed{y=4x^2-4}[/tex]
