Confusing range of a Relation.

**Kaloda** · June 28th 2012, 02:56 PM

Okay. So our assignment is to get the domain, range, intercepts, etc. of the relation (also a function but not expressed as one)
$5y(x-3)(x-1)=2(5x+3)$ . I didn't have a problem at all till I realize that I got $(-\infty,-5]\cup[-1/5,0)\cup(0,+\infty)$
(I EXCLUDED ZERO because when I solved for x, y is in the DENOMINATOR) as the range and have an x-intercept $(-3/5,0)$
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.

**pickslides** · June 28th 2012, 03:45 PM

It seems to me that $y \in \mathbb{R}$ and $x \in (-\infty, 1) \cup ( 1,3) \cup (3,\infty)$ , your x-intercept is correct. Include y=0 in the range as it gives a corresponding x-value.

What about when $x=0$ ?

**Kaloda** · June 28th 2012, 04:37 PM

Originally Posted by pickslides

It seems to me that $y \in \mathbb{R}$

Huh? Do you mean that the range is all real numbers? I think you must recheck it (or maybe I'm the one who is wrong).

**Soroban** · June 28th 2012, 04:38 PM

Hello, Kaloda!

I think you're confusing yourself . . .

Find the domain, range, intercepts, etc. of the relation: . $5y(x-3)(x-1)\:=\:2(5x+3)$

We have: . $y \;=\;\frac{2}{5}\cdot\frac{5x+3}{(x-3)(x-1)}$

If $y = 0\!:\;x = \text{-}\tfrac{3}{5}\;\hdots\;x\text{-intercept: }\left(\text{-}\tfrac{3}{5},\:0\right)$

If $x = 0\!:\;y = \tfrac{2}{5}\;\hdots\;y\text{-intercept: }\left(0,\:\tfrac{2}{5}\right)$

Horizontal asymptote: . $y = 0$
Vertical asymptotes: . $x = 1,\;x = 3$

Domain: . $x \ne 1,\:x \ne 3$
. . . . . . . $(\text{-}\infty,\,1)\,\cup\,(1,\,3)\,\cup\,(3,\,\infty)$

Range: difficult to determine without using Calculus.

The graph looks something like this:

Code:

                        |
                        | :         :
                        | :         :*
                        |*:         :
                        | :         : *
                        | :         :  *
                        oB:         :     *
                      A | :         :          *
  - -------------------o+-:---------:--------------
        *             * | :1        :3
              *     *   | :         :
                 o      | :         :
                 D      | :    C    :
                        | :    o    :
                        | :  *   *  :
                        | : *     * :
                        | :         :
                        | :*        *:

$A\text{ and }B\text{ are the }x\text{- and }y\text{-intercepts, respectively.}$

$C\text{ has coordinates: }\,\left(\tfrac{9}{5},\text{-}5\right)$
$D\text{ has coordinates: }\,\left(\text{-}3,\,\text{-}\tfrac{1}{5}\right)$

Therefore, the range is: . $\left(\text{-}\infty,\,\text{-}5)\,\cup\,\leeft(\text{-}\tfrac{1}{5},\,\infty\right)$

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