## Algebra 2 (1st Edition)

$(x+4)^2=-4(y+3)$; Parabola
Given: $x^2+8x+4y+28=0$ $(x+4)^2 +4y+28-16=0$ This gives: $(x+4)^2=-(4y+12)$ Thus, we have $(x+4)^2=-4(y+3)$ This is the standard form of a Parabola.