Question 42088: Identify the vertex, axis of symmetry, and direction of opening for:
y=1/2(x-8)^2+2.
Thanks
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! Identify the vertex, axis of symmetry, and direction of opening for:
y=1/2(x-8)^2+2.
(X-8)^2=2(Y-2)
STD EQN.IS
(X-H)^2=4A(Y-K)
WHERE H,K IS VERTEX..SO..H=8...K=2...SO..(8,2) IS VERTEX
AXIS OF SYMMETRY IS X=H...OR....X=8
SINCE Y IS MINIMUM AT X=8,WE HAVE VERTEX AT THE BOTTOM AS A TROUGH.SO DIRECTION OF OPENING IS UPWARDS
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