Respuesta :
bearing in mind that 0.3 = 3/10
[tex]\bf ~\hfill \textit{parabola vertex form}~\hfill \\\\ \begin{array}{llll} y=a(x- h)^2+ k\qquad \leftarrow \textit{we'll use this one}\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{-\frac{1}{3}}{ h},\stackrel{\frac{3}{10}}{ k}) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ y=a\left[ x-\left( -\cfrac{1}{3} \right) \right]^2+\cfrac{3}{10}\implies y=a\left( x+\cfrac{1}{3} \right)^2+\cfrac{3}{10}[/tex]
[tex]\bf \textit{we also know a point } \begin{cases} x=-\frac{2}{15}\\\\ y=-\frac{1}{2} \end{cases}\implies -\cfrac{1}{2}=a\left( -\cfrac{2}{15}+\cfrac{1}{3} \right)^2+\cfrac{3}{10} \\\\\\ -\cfrac{1}{2}-\cfrac{3}{10}=a\left( \cfrac{1}{5} \right)^2\implies -\cfrac{4}{5}=a\left(\cfrac{1}{25} \right)\implies -\cfrac{4}{5}=\cfrac{a}{25}\implies \cfrac{-4\cdot 25}{5}=a \\\\\\ -4\cdot 5=a\implies \boxed{-20=a} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill y=-20\left( x+\cfrac{1}{3} \right)^2+\cfrac{3}{10}~\hfill[/tex]
