Chapter 3 - Quadratic Functions - 3.1 Introduction to the Family of Quadratic Functions - Exercises and Problems for Section 3.1 - Exercises and Problems - Page 121: 35

$ y=\frac{1}{3} x^2-\frac{2}{3} x-1$

We know there are zeros at $x=-1$ and $x=3$, so we use the factored form $$ y=a(x+1)(x-3) $$ We solve for $a$ by substituting $x=0, y=-1$,this gives $$ \begin{aligned} & -1=a(1)(-3) \\ &a= \frac{1}{3} . \end{aligned} $$ Hence $$ \begin{aligned} & y=\frac{1}{3}(x+1)(x-3) \\ & y=\frac{1}{3} x^2-\frac{2}{3} x-1 \end{aligned} $$