Need help finding smallest value of $x^2 + y^2$ I need to find the smallest value of $x^2 + y^2$ with the restriction $2x + 3y = 6$. This chapter focuses on the vertex formula. 

Let's solve for $y$ in the equation $2x + 3y = 6$. This gives $y = 2 - \frac{2x}{3}$.
After rewriting $x^2 + y^2$ in terms of $x$, we have 
$x^2 + y^2 = x^2 + (2 - \frac{2x}{3})^2 = x^2 + (4 - 2 \cdot \frac{4x}{3} +\frac{4x^2}{9})$
This is a quadratic "in $x$" that we would like to minimize. You can write it in standard form $ax^2 + bx + c$ and then use the methods you have learned. 
