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Someone else should weigh in

How did you do the second problem? i don't get it.

Well my solution isn't a general solution. It does not really answer the general case.

Can you do the 2nd problem please?

*Solving for d where (-d/2)^2+d(-d/2)+3(-d/2) = 0 will yield d = 0 and d = -6

The vertex for the second equation is not on the x axis. I think it is at 3.

If d is what y is when x = 0 then d would equal 3.

I think d is we are looking for when y = 0, not necessarily when x = 0.

I am unfamiliar with these types of questions but y never touches 0 in the second equation.

Is the second formula equal to f(x) = x^2 + 3?