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The functions are: $y=\frac{x^2-4}{x+2}$ and $(x+2)y=x^2-4$.

I've seen this problem some time ago, and the official answer was that they are not. My question is: Is that really true?

The functions obviously misbehave when $x = -2$, but aren't both of them indeterminate forms at that point? Why are they different?

Deathkamp Drone
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3 Answers3

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$(1)$The first function is undefined at $x = -2$,

$(2)$ the second equation is defined at $x = -2$:

$$(x + 2) y = x^2 - 4 \iff xy + 2y = x^2 - 4\tag{2}$$ It's graph includes the entire line $x = -2$. At $x = -2$, all values of y are defined, so every point lying on the line $x = -2$: each of the form $(-2, y)$ are included in the graph of function (2). Not so with the first equation.

ADDED:

Just to see how well Wolfram Alpha took on the challenge:

Graph of Equation $(1)$: (It fails to show the omission at $x = -2$) But it does add:

enter image description here

enter image description here

Graph of Equation $(2)$:

enter image description here

Note: The pair of graphs included here do not match in terms of the scaling of the axes, so the line $y = x - 2$ looks sloped differently in one graph than in the other.

amWhy
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  • I'm sorry if this looks dumb, but I've been told that if a number is equal to $0/0$, then it can assume any value. I can't see why there isn't a asymptote on the first function too. – Deathkamp Drone May 17 '13 at 02:56
  • Carpe: But the second equation does not evaluate to $0/0$. When $x = -2$, y can be any value whatsoever, so the line containing all points $(-2, y)$ defined by $x = - 2$ is part of its graph. – amWhy May 17 '13 at 02:59
  • There is no asymptote at $x = -2$ because the function described in the first equation has values that approach $-4$ as $x$ approaches 2 "from above or below". A vertical asymptote would be due to the function tending to infinity as $x$ approaches 2, which does not happen here. – colormegone May 17 '13 at 03:00
  • Alas, not before I started typing, which I guess I don't do fast enough... (I saw your last edit appear as I was still finishing.) – colormegone May 17 '13 at 03:04
  • Is this clear now, Carpe? Sorry you experienced confusion. I hope we've cleared things up a bit! ;-) $\quad0/0;$ IS indeed, undefined. And particularly, division of any number by zero is undefined!! But that's not a problem for the second equation. So the functions differ only that in the first, there is no point at $x = -2$, whereas, in the second, there are infinitely many points, $(-2, y)$, at $x = -2$ – amWhy May 17 '13 at 03:07
  • @amWhy: you are always conscientious! +1 – Amzoti May 17 '13 at 03:10
  • @amWhy: But I am missed to make +'s here. Missed that. :-( – Mikasa May 17 '13 at 04:51
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In the second equation, when $x=-2$, there is no restriction whatsoever on $y$. The graph of the second equation includes the entire vertical line $x=-2$.

Jonathan
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  • Why isn't there a line on the first one too? $y = 0/0$ isn't true for every $y$? – Deathkamp Drone May 17 '13 at 02:53
  • Thus, the second equation does not describe a function... – colormegone May 17 '13 at 02:53
  • $y$ isn't defined at $x=-2$ in the first equation. The graph of the first equation is like that of $y=x-2$ but with a hole at $x=-2$. – Jonathan May 17 '13 at 02:54
  • Khan Academy lied to me? – Deathkamp Drone May 17 '13 at 02:55
  • I'm not sure how to respond to that – Jonathan May 17 '13 at 02:55
  • @Carpe Noctem The distinction is that in the first equation, $y = f(x)$ has no definition at $ x = -2$ ($0/0$ is undefined because it is indeterminate), while in the second equation, the graph is actually a composite of two "curves", one of which is not the graph of a function (because $x = -2$ works regardless of the value of $y$). [And sorry, but not even KhanAcademy knows everything... ;) ] – colormegone May 17 '13 at 02:57
  • I'm sorry about that. I've watched this video a long time ago: https://www.khanacademy.org/math/trigonometry/functions_and_graphs/undefined_indeterminate/v/undefined-and-indeterminate

    I thought $0/0$ was defined (to be every possible value, for some reason), but now I see that it is undefined too.

     – Deathkamp Drone May 17 '13 at 03:05
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Forget functions for a moment. An equation is a sentence, and the graph of an equation (in two variables) is the set of pairs $(x,y)$ which satisfy the sentence. So $(1, -\frac{1}{3})$ is in the graph of the equation $y = \frac{x^2 -4}{x+2}$ and $(x+2)y = x^2-4$. You can see this by plugging in $x = 1$ and $y = -1/3$ -- each sentence is true.

The point $(-2, 0)$ lies in the graph of the second equation, but not in the first, because $\frac{0}{0}$ is a nonsense phrase (i.e. "undefined"). It doesn't make sense to call nonsense like $y = \frac{0}{0}$ true or false. Moreover, $(-2, 3)$ and $(-2, 4)$ also satisfy the second sentence. In fact, ANY pair $(-2, y)$ satisfies the second equation, so all those points lie on the graph of the second equation.

Now let's return to functions. The first equation is the graph of a function. The second equation is not, because it fails the vertical line test. To be really nit-picky, a function is a rule which assigns to each element of the domain a unique element of the range. So an equation is not a function.

Finally, "indeterminate forms" concern limits, which don't seem immediately relevant here.

Adam Saltz
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