Chapter 10 - Section 10.5 - Conic Sections - 10.5 Exercises - Page 680: 29

Parabola Focus: $( 1,-\frac{11}{6})$ Vertex: $(1,-2)$

$3x^2-6x-2y=1$ $3(x^2-2x)=2y$ $3(x-1)^2-4=2y$ $y=\frac{3}{2}(x-1)^2-2$, the equation of an upwards opening parabola. $(h,k+p)=(1,-2+1/6)=( 1,-\frac{11}{6})$ Focus: $( 1,-\frac{11}{6})$ Vertex: $(1,-2)$