Chapter 3 - Quadratic Functions - 3.2 The Vertex of a Parabola - Exercises and Problems for Section 3.2 - Exercises and Problems - Page 128: 28

$ y=\frac{2}{3} x^2-2 $

Substituting the coordinates of the vertex gives $$ -2=a(0)^2+k\implies k=-2 $$ Substituting the coordinates of the point, $(3,4)$ gives $$ \begin{aligned} & 4=a(3)^2-2\\ & 6=9a \\ & a=\frac{2}{3} \end{aligned} $$ Hence $$ y=\frac{2}{3} x^2-2 $$