Answer
$${y^2} = 4\left( {x + 1} \right)$$
Work Step by Step
$$\eqalign{
& {\text{From the graph we got the following data:}} \cr
& {\text{Directrix }}x = - 2,\,\,\,\,{\text{vertex}}\left( { - 1,0} \right) \cr
& {\text{The axis of symmetry is }}x,{\text{ then the equation of the parabola is}} \cr
& {\text{of the form}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\left( {y - k} \right)^2} = 4p\left( {x - h} \right) \cr
& {\text{vertex }}\left( {h,k} \right) \cr
& {\text{directrix }}x = - p + h \cr
& \cr
& {\text{then,}} \cr
& {\text{vertex}}\left( { - 1,0} \right) \cr
& {\text{vertex }}\left( {h,k} \right) \Rightarrow \,\,\,\,\,h = - 1,\,\,\,\,k = 0 \cr
& {\text{directrix }}x = - p + h \Rightarrow \,\, - p - 1 = - 2 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,p = 1 \cr
& \cr
& {\text{An equation of the parabola is }}{\left( {y - k} \right)^2} = 4p\left( {x - h} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\left( {y - 0} \right)^2} = 4\left( 1 \right)\left( {x - \left( { - 1} \right)} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{y^2} = 4\left( {x + 1} \right) \cr} $$
