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We're required to take 30 hours of math courses. We end in a minor of math. I'm in my third year of university, though haven't really taken much of the math yet. We're allowed to pretty much make our own schedule and are responsible for knowing material prior to a class. I've taken one (out of three required) 5 hour calculus course, one (out of two required) discrete math course, and college algebra and trig my first year, which I don't really count as those are pretty simple. But yea, I haven't gotten much of my math in yet and this particular topic isn't one that has like ever come up. At least not quite this way. So kind of both.

This seems different than how other schools (such as Helios') handle this, and start you off with a lot of math. And my university had a more "practical" approach to CS than some schools which are more research and theory oriented. But we are still a certified computer science program by whatever organization handles those standards (I wanna say IEEE but I think that's wrong).

This seems different than how other schools (such as Helios') handle this, and start you off with a lot of math. And my university had a more "practical" approach to CS than some schools which are more research and theory oriented. But we are still a certified computer science program by whatever organization handles those standards (I wanna say IEEE but I think that's wrong).

If I delete this post (I have actually tried), I see this:

http://i.imgur.com/D0YBAIh.png

I think it's because quarkonium and ResidentBiscuit replied at the same time and confused the forum software. What an appropriate name, quarkonium.

http://i.imgur.com/D0YBAIh.png

I think it's because quarkonium and ResidentBiscuit replied at the same time and confused the forum software. What an appropriate name, quarkonium.

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@L B Excuse my ignorance, but what does the graph tell me? |

I don't think anyone answered this question, so without further ado:

The x-intercepts on a graph are the zero's or "solutions" to the equation. For example x^2 - 1 factors into (x+1)(x-1). If f(x) = (x+1)(x-1) then f(1) = 0, f(-1) = 0. It should be clear from this that the zeros of a function are the x-intercepts, or the values that make the function output equal zero.

If you look at the graph of the function in question you'll see that it clearly never intercepts the x-axis and therefore has no solutions (and this makes sense because the minimum value the function can output is 3 so therefore it will never be equal to zero).

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