The graph of the function f(x)=ax^2+bx+c has its vertex at (0,1) and passes through the point (1,9) find a b and c
+33614 The quadratic is symmetrical about its vertex, so, if it passes through (1, 9) it also passes through (-1,9), since the x-value of the vertex is at 0.
So you have three equations:
a*(-1)2 +b*(-1) + c = 9
a*12 + b*1 + c = 9
a*02 + b*0 + c = 1
You should be able to obtain a, b and c from these three equations.
