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Question

Answer

px2 + qx + r = p (x – h)2 + k
= p(x2 -2hx + h2 ) + k
=px2 -2phx + (ph2 + k)
So comparing the coefficient of x,
q = -2ph => h = -(q/2p)
Comparing the constant,
r = ph2 + k but we know h = -(q/2p) from above,
So r = p[ -(q/2p) ]2 + k
r = p [ (q2 /4p2 ) ] + k
Therefore, k = r – (q2 /4p)