All lines with a slope have a y-intercept?

I still disagree with the statement, What if you had a line of the equation (x*(x-3))/x? At x = 0 there’s a hole, leaving the idea of a y-intercept impossible, but otherwise it’s a completely normal linear function.

With the usual definition of a line, if you take a point out of the line it’s no longer a line.

I always understood the it to be " a line that is discontinuous at the interval -1<x<1 (or -infinity<x<infinity)" but I guess I’m wrong, alright then.

And for those who are interested, “m” is used for slope because it is from the French word “monter” which was used by René Descartes.

Well, the curve you are talking about is “a line that’s undefined (with a removable discontinuity) at 0”, if you like. But “a line that’s undefined (with a removable discontinuity) at 0” is not strictly covered by what’s usually meant when one says simply “a line”, just as “a square with rounded edges” is not strictly covered by what’s usually meant when one says simply “a square”.

This was mentioned earlier, but I want to clarify something about this. I’m assuming that you’re referring to a parabola which opens vertically, in which case it will always have a y-intercept, regardless of how far you shift it horizontally. Parabolas always keep widening and never become truly vertical. Of course, after a while they become nearly vertical (the slope approaches infinity) but they continue to widen. Eventually, and it might be way the blork up or down, the parabola will hit the axis.

True, but the poster admitted later he meant hyperbola, not parabola. :wink: