The graph of any parabola has the form

y = f(x) = Ax² + Bx + C

(-1,10),(2,-5)and(3,-18). 

Since it goes through (-1,10), substitute x=-1 and y=f(x)=10

f(x) = Ax² + Bx + C
  10 = A(-1)² + B(-1) + C
  10 = 1A - B + C
A - B + C = 10

Since it goes through (2,-5), substitute x=2 and y=f(x)=-5

f(x) = Ax² + Bx + C
  -5 = A(2)² + B(2) + C
  -5 = 4A + 2B + C
4A + 2B + C = -5

Since it goes through (3,-18), substitute x=3 and y=f(x)=-18

f(x) = Ax² + Bx + C
 -18 = A(3)² + B(3) + C
 -18 = 9A + 3B + C
9A + 3B + C = -18

So we have this system of 3 equations and 3 unknowns:



Solve that system either by elimination or substitution
and get A=-2, B=-3, C=9
 
So the equation  
f(x) = Ax² + Bx + C
becomes
f(x) = -2x² - 3x + 9

The vertex is the point 






So the vertex is 

Since the coefficient of A is negative the parabola opens
downward, and the vertex is a MAXIMUM. The focus is BELOW
the vertex.

Since A=-2, the focus is 2 units BELOW the vertex, so we add
A=-2 units to the y-coordinate of the vertex:



For the focus, we use the same x-coordinate as the vertex

So the vertex is  

The x-intercepts are found by setting y = 0:
 
         f(x) = -2x² - 3x + 9
            0 = -2x² - 3x + 9
 2x² + 3x - 9 = 0
  (2x-3)(x+3) = 0
  2x-3=0;  x+3=0
    2x=3;    x=-3
     x=

So the x-intercepts are the points

 and 

Here is the graph.  The points marked are the vertex, focus,
and x-intercepts.



Edwin