math help


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bdsams

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

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atomicski

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

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