Answer
$${\text{The equation describes a parabola}}{\text{.}}$$
Work Step by Step
$$\eqalign{
& y = 8{x^2} + 16x + 8 \cr
& {\text{Factor}} \cr
& y = 8\left( {{x^2} + 2x + 1} \right) \cr
& y = 8{\left( {x + 1} \right)^2} \cr
& {\left( {x + 1} \right)^2} = \frac{1}{8}y \cr
& \left( a \right) \cr
& {\text{The equation is written in the form }}{\left( {x - h} \right)^2} = 4p\left( {y - k} \right), \cr
& {\text{then, the equation describes a parabola}}{\text{.}} \cr
& \underbrace {{{\left( {x - h} \right)}^2} = 4p\left( {y - k} \right)}_{{{\left( {x + 1} \right)}^2} = \frac{1}{8}y} \to h = - 1,{\text{ }}k = 0,{\text{ }}p = \frac{1}{{32}} \cr
& \cr
& \left( b \right) \cr
& {\text{Characteristics:}} \cr
& {\text{Focus: }}\left( {h,p + k} \right) \cr
& {\text{Focus: }}\left( { - 1,\frac{1}{{32}}} \right) \cr
& {\text{Vertex: }}\left( {h,k} \right) \cr
& {\text{ Vertex: }}\left( { - 1,0} \right) \cr
& {\text{Directrix: }}y = - p + k \cr
& {\text{Directrix: }}y = - \frac{1}{{32}} \cr
& \cr
& \left( c \right) \cr
& {\text{The eccentricity of a parabola is 1}} \cr
& e = 1 \cr
& \cr
& \left( d \right) \cr
& {\text{Graph}} \cr} $$
