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

Guest Mar 15, 2020

#1**+4 **

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*1^{2} + b*1 + c = 9

a*0^{2} + b*0 + c = 1

You should be able to obtain a, b and c from these three equations.

Alan Mar 15, 2020