Question 558697
The equation {{{(y+1)^2=-8(x-3)}}} is the equation of a parabola in vertex form. In that form, the equation shows you the coordinates of the vertex. They are subtracted from the {{{x}}} and the {{{y}}}). The location of the directrix and the focus point can be calculated from the coefficient {{{-8}}} in the equation.
Your book tells you all about it, but you can figure it all out without the book, and without memorizing any formulas.
VERTEX AND AXIS OF SYMMETRY
What happens when {{{y+1=0}}} <---> {{{y=-1}}} ?
{{{0=-8(x-3)}}} ---> {{{x-3=0}}} ---> {{{x=3}}}
The fact that both parenteses are zero tells you that it is a very special point.
For any other value of {{{y}}} , {{{-8(x-3)=(y+1)^(2)>0}}} ---> {{{x-3<0}}} ---> {{{x<3}}}
So, {{{x<=3}}}, no matter what value y takes. It's a maximum value for x. Point (3,-1) is the vertex, and {{{y=-1}}} is the axis of symmetry.
TWO RANDOM POINTS 
Let's try {{{y+1=2}}} <---> {{{y=1}}}
{{{(y+1)^2=2^2=4=-8(x-3)}}} ---> {{{x-3=-4/8=-1/2}}} ---> {{{x=31/2=5/2=2&1/2}}}
Let's try {{{y+1=8}}} <---> {{{y=7}}}
{{{(y+1)^2=8^2=-8(x-3)}}} ---> {{{x-3=-8}}} ---> {{{x=-5}}}
DIRECTRIX AND FOCUS POINT
The directrix will be {{{y=3+c}}} and the focus will be at (3-c,-1)
The points on the parabola for {{{x=3-c}}} will be at a distance
{{{abs(y-(-1))=abs(y+1)}}} , measured vertically, above and below the focus point.
They will also be at a distance {{{2c}}} measured horizontally from the directrix.
By the definition of parabola, those distances are the same, so
{{{abs(y+1)=2c}}} --> {{{(y+1)^2=(2c)^2}}} --> {{{(y+1)^2=4c^2}}}
On the other hand, from the equation given, {{{(y+1)^(2)=-8(x-3)}}} , we know that for {{{x=3-c}}} 
{{{(y+1)^2=-8(3-c-3)}}} --> {{{(y+1)^2=8c}}}
Substituting that expression for {{{(y+1)^2}}} in {{{(y+1)^2=4c^2}}} we get
{{{8c=4c^2}}} --> {{{8c/4c=4c^2/4c}}} --> {{{2=c}}} is the solution we need.
The directrix is {{{x=3+2}}} --> {{{x=5}}}
Thee focus point has {{{x=3-2=1}}}, so it's the point (1,-1).