Since P = (x,y) is on the graph of y = x2 - 6, P can be expressed as (x, x2-6).
So, d = √[(x - 2)2 + (x2 - 6 - 0)2] = √[x2 - 4x + 4 + x2 - 6]
d = √[2x2 - 4x - 2]
Carlee P.
asked 10/31/19Since P = (x,y) is on the graph of y = x2 - 6, P can be expressed as (x, x2-6).
So, d = √[(x - 2)2 + (x2 - 6 - 0)2] = √[x2 - 4x + 4 + x2 - 6]
d = √[2x2 - 4x - 2]
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