I hope you can follow this vertex form is y= a(x-h)^2 +k where (h,k) are the coordinates of the vertex. Since the axis of symmetry is x=-2 the equation is a(x+2)^2 +k
We need another point on the graph. Since parabolas are symmetric around axis of symmetry another point would be (2,44)
Can now set up a system of equations for standard form of a parabola y=ax^2 +bx +c
Using these three points we have the following 3 equations with 3 unknowns
44= a(-6)^2 +b(-6) +c. 44=36a -6b +c
23 = a(1^2) +b(1) +c. 23= a+b+c
44 = a(2^) +b(2) +6. 44= 4a +2b +c
Solve this system by elimination and you will get a=3 you only need to solve for a
So vertex form is y=3(x+2)^2 +k. Use the point (1,23) to solve for k
23 = 3(1+2)^2 + k
k=-4 so the equation is y= 3(x+2)^2 -4
