The equation of the parabolas given will be found as follows: a] general form of the parabolas is: y=k(ax^2+bx+c) taking to points form the first graph say (2,-2) (3,2), thus y=k(x-2)(x-3) y=k(x^2-5x+6)
taking another point (-1,5) 5=k((-1)^2-5(-1)+6) 5=k(1+5+6) 5=12k k=5/12 thus the equation will be: y=5/12(x^2-5x+6)
b] Using the vertex form of the quadratic equations: y=a(x-h)^2+k where (h,k) is the vertex from the graph, the vertex is hence: (-2,1) thus the equation will be: y=a(x+2)^2+1 taking the point say (0,3) and solving for a 3=a(0+2)^2+1 3=4a+1 a=1/2 hence the equation will be: y=1/2(x+2)^2+1
