bdsams Veteran Share Posted October 25, 2004 if some one could explain how to solve this, i know the formulas ( i think) but cant seem to get the correct solution...thx y = ax^2+bx+C that satisfies the conditions: 1.Vertex (3,-2) a=-1 thats problem one 2. Vertex (-1,-1) point on graph (0,4) prob 2 i know that you have to use y=a(x-h)^2+k formula... and that you should substitue the x and y values in? any help would be appreciated Link to post Share on other sites More sharing options...

atomicski Share Posted October 25, 2004 you start with f(x)=a(x-h)^2+k you put opposite of x of vertex in for h and y of vertex in for k. Now you put y of point in for f(x) and x of point in for x. problem 1 would be y=-l(x-3)^2+-2 problem 2 would be 0=a(4+1)^2+-1 i think thats right, but not positive Link to post Share on other sites More sharing options...

## Recommended Posts