It seems to me that and , your x-intercept is correct. Include y=0 in the range as it gives a corresponding x-value.
What about when ?
Okay. So our assignment is to get the domain, range, intercepts, etc. of the relation (also a function but not expressed as one)
. I didn't have a problem at all till I realize that I got
(I EXCLUDED ZERO because when I solved for x, y is in the DENOMINATOR) as the range and have an x-intercept
which is a contradiction with the range.
Now, it leaves me with a dilemma because I can't include zero in the range because x will be undefined (at least if you use quadratic formula by manipulating the given equation which I think alters its nature I presume), and I can't just simply remove the x-intercept because of the same reason.
Here's my question, do I include zero in the range OR remove the x-intercept OR just leave my answer as it is.
And please explain why.
I think you're confusing yourself . . .
Find the domain, range, intercepts, etc. of the relation: .
We have: .
Horizontal asymptote: .
Vertical asymptotes: .
. . . . . . .
Range: difficult to determine without using Calculus.
The graph looks something like this:
Code:| | : : | : :* |*: : | : : * | : : * oB: : * A | : : * - -------------------o+-:---------:-------------- * * | :1 :3 * * | : : o | : : D | : C : | : o : | : * * : | : * * : | : : | :* *:
Therefore, the range is: .