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for f(x)= x^2 = vertex is always 0,0. always 9,4,1,0,1,4,9
for f(x)= ax^2 = vertex is always 0,0. the y is 9(a),4(a),1(a),0,1(a),4(a) 9(a)
for f(x)= (x-h)^2 = the x of the vertex is equal to h and the y is 0. the y is 9,4,1,0,1,4,9
for f(x)= x^2+k = vertex is always 0,k (no pun intended). the y is 9+k,4+k,1+k,k,1+k,4+k,9+k
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finding the roots of x
the quadratic formula
-b(sqrt b^2 - 4ac) /2
completing the sqaure
transpose the third term to the right side
find the value of b. divide it by two then square it. itll become the new third term
square root both sides
transpose the second term to the right
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