bdsams Veteran Posted October 25, 2004 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 comment Share on other sites More sharing options...
atomicski Posted October 25, 2004 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 comment Share on other sites More sharing options...
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