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Eliminate the parameter. x = t - 3, y = t2 + 5
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in the first equation solve for t, then replace it into the second equation.
y = x2 + 14?
NO. from the first equation \[\large t=x+3 \] so \[\large y=t^2+5=(x+3)^2+5 \]
This has to be it: y = x2 + 6x + 14
yes. but there's (usually) no need to expand the square: u lose information.
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(i mean u hide it, u don't lose it, it is still there)
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