Here is what this site states

All parabolas are the same shape, no matter how big they are. Although they are infinite, meaning that the arms will never close up, the arms will eventually become parallel.

Now, I have an argument against it. Let $f(x) = ax^2 + bx + c$ be a quadratic polynomial with $a , b$ and $c$ being real numbers and $a \ne 0$. So, its graph will give us a parabola ($\because$ graph of a quadratic polynomial is a parabola).


$$\dfrac{d (ax^2 + bx + c)}{dx}$$$$ = 2ax + b$$

i.e. the slope of a quadratic polynomial is given by $g(x) = 2ax + b$. Now, differentiating the equation for slope of the quadratic

$$\dfrac {d (2ax + b)}{d x}$$ $$ = 2a$$

So, if $a \gt 0$ then the slope of $g(x)$ will be increasing. This means that the slope of $f(x)$ will also be increasing. Similarly, if $a \lt 0$ then the slope of $f(x)$ will be decreasing.

This means that for all $x$ the slope of $f(x)$ will be different. So, this contradicts the fact (according to me) that the arms of a parabola will eventually be parallel. Where am I going wrong?

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    $\begingroup$ "eventually" is a big word and here means "at infinity": the direction of each arm each gets closer to the direction of the axis as you move towards infinity, but there are no asymptotes . $\endgroup$
    – Henry
    Sep 6, 2016 at 14:58
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    $\begingroup$ This is not true at all... You know, sites are like people: some of them are not reliable... You have to be cautious: the most reliable sites are university sites or very recognized sites (Wikipedia, Wolfram, Cut-the-link, etc...) or forums with efficient moderators... $\endgroup$
    – Jean Marie
    Sep 6, 2016 at 14:59
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    $\begingroup$ This is yet another example of one of my favourite mathematical errors: confusing the properties of a limit with the properties of a thing. You can come up with examples all day: 3, 3.1, 3.14, 3.1415, 3.14159, ... are all rational, the limit is pi, therefore pi is rational. And so on. Statement about the behaviour of mathematical objects in the limiting case should be made extremely carefully. $\endgroup$ Sep 6, 2016 at 16:24
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    $\begingroup$ As for "where are you going wrong?" You're not. Either the statement "the arms are nowhere parallel" is true or "there is a point where the arms are parallel" is true. You have correctly proved the first statement, and therefore the second statement is false. My advice: find a better web site. $\endgroup$ Sep 6, 2016 at 17:31
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    $\begingroup$ I'm peeved here. "Eventually" would mean "at some event," but there is no such event at which the arms will be parallel. If you're confused, it's because the claim almost says something correct, but as stated it's not even imprecise, just wrong. $\endgroup$
    – djechlin
    Sep 7, 2016 at 23:46

13 Answers 13


By "eventually" they mean "in the limit". As $a$ goes to positive infinity, the slope at $a$ goes to positive infinity, which means the arm is becoming more and more vertical. As $a$ goes to negative infinity, so does the slope at $a$ - but a slope of "negative infinity" also means that the line is vertical, so the left arm is also becoming closer and closer to vertical. If both arms are getting closer to closer to vertical, then they must be getting closer and closer to being parallel to one another.

Of course, your argument shows that they will never actually be parallel. Just very close.

  • $\begingroup$ I agree if you take "eventually"... But it seems sketchy to me. Moreover, the curve do not have any tangent... $\endgroup$
    – Martigan
    Sep 6, 2016 at 14:59
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    $\begingroup$ Ah, the political "eventually". i.e. $t=\infty$ $\endgroup$
    – Jam
    Sep 6, 2016 at 15:15
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    $\begingroup$ @Jam Also referred to as "eight to ten months". $\endgroup$
    – mbomb007
    Sep 6, 2016 at 19:58
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    $\begingroup$ I'm not inclined to be very generous toward the site and its phrasing. The sentence starts "Although they are infinite..." (emphasis mine), which implies that the statement somehow forms a contrast with the fact that the arms never come to an end--which, to me, would imply that the claim about the "eventual" behavior of the arms refers to an observable characteristic at a finite distance from the focal point. $\endgroup$ Sep 6, 2016 at 21:01
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    $\begingroup$ This answer is good and correct, but I'd add a disclaimer that this usage of 'eventually' is wrong or at least very misleading, and everyone reading this should know not to use the word to mean this. $\endgroup$
    – JiK
    Sep 8, 2016 at 11:12

The parabola (top picture), and hyperbola (bottom picture) viewed projectively, all lines $y=k$ are parallel to the line at infinity, $L_{\infty}$. Here we see all non-degenerate conic sections are ellipses in the projective plane.

Looking at the picture the arms of the parabola 'appear' to become parallel somewhere in the middle, before curving inwards to meet at $L_{\infty}$.

The Projective Parabola

The Projective Hyperbola

Here's a nice talk on Projective Geometry that explains this - get to about 36mins in for the parabola.

Edit: The appearance of parallel lines is a Euclidean one that the picture allows us to make, and indeed this notion of parallel changes with the perspective of the viewer, and was a nod to the initial question of what parallel actually means WRT the sites claim about parabolas:

Although they are infinite, meaning that the arms will never close up, the arms will eventually become parallel.

It's difficult to make claims like this without the setting of the appropriate geometry, as it has no meaning in Euclidean geometry and doing calculus won't help us see the bigger picture. To see what's happening we have to view things in the projective plane. Viewed projectively every non-degnerate conic is a ellipse that is in one piece. To understand this one needs the concept of points, and lines at infinity. The ellipse, parabola, and hyperbola have $0$, $1$, and $2$ points at infinity, respectively.

To see the points at infinity on the parabola, we tilt its perspective. Observe how the parabola will cut each ray at $0$ and one finite point, except for the $y$-axis, which it meets at $L_{\infty}$. Hence parabolas have just one point at infinity. (See top picture.)

The hyperbola's two points at infinity are where it meets its asymptotes, and the continuation of the hyperbola to form an ellipse comes from projecting the lower branch through the same centre of projection. (See bottom picture.)

Note also that when we look at the hyperbolic plane we look at transformations of $\mathbb{R}^2$ with a "point at infinity" added (the extended reals: $\hat{\mathbb{R}}=\mathbb{R}\cup\{\infty\}$), whose transformations are governed by the projective special linear group $\operatorname{PSL}_2(\mathbb{R})$. In hyperbolic geometry we get a notion of distance but not so in projective geometry, where the transformations are governed by $\operatorname{PSL}_3(\mathbb{R})$, which being bigger than $\operatorname{PSL}_2(\mathbb{R})$, means the projective plane comes with fewer geometrical properties but richer transformations.

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    $\begingroup$ In a projective geometry picture, lines that are parallel in the Euclidean sense intersect at a point at infinity (where the $y$ is written). In this case, the only time it approaches this point is as a tangent to the line at infinity, reflecting the fact that there are not two lines but one, since as @ttulinsky observed the two candidate lines are also being pushed away from each other, so that each line is pushed away from the origin to the limit $L_\infty$. $\endgroup$ Sep 7, 2016 at 7:48
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    $\begingroup$ @Mario Carneiro : Nicely put. Yes the appearance of parallelism I gave is a Euclidean idea that we see via the picture, and is in no way what is really going on in the projective plane, merely a nod to the site's claim that a parabola's arms eventually become parallel which I'm not quite sure what they mean by that, but it could be they were hinting at the projective plane. The lecture does a better job at explaining all this, and I agree with the lecturer that aliens would be asking us "Why is projective geometry not taught anymore?" $\endgroup$ Sep 7, 2016 at 11:10

I also disagree with the statement "the arms will eventually become parallel."

I don't think the analysis of $f''$ as you've done is really necessary. I think it's enough to stop with $f'(x) = 2ax + b$ and note that $$2ax_1 + b = 2a x_2 + b \iff x_1 = x_2.$$

That is, there can't be two distinct points on the graph of $f(x)$ that have the same slope.

I wonder if perhaps they were thinking something like this when they wrote that: $$\lim_{x \to +\infty} (2ax + b) = (\operatorname{sgn}a)\infty$$ and $$\lim_{x \to -\infty} (2ax + b) = -(\operatorname{sgn}a)\infty,$$ where $\operatorname{sgn}a$ is $1$ if $a > 0$ and $\operatorname{sgn}a$ is $-1$ if $a < 0$. Basically the "limit" of each arm is a vertical line. That's the only thing I can think they'd be getting at, but that's an awful and pointless thing to explain at a precalculus level.

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    $\begingroup$ It's worth noting that slopes are measured with projective real numbers, not extended real numbers; in particular, we should have $(-1) \cdot \infty = \infty$ here. $\endgroup$
    – user14972
    Sep 6, 2016 at 16:22
  • $\begingroup$ The lines can be parallel if they slope 180 degrees opposite to each other as well as if they have identical slopes - i.e. x1 = -x2. $\endgroup$
    – DeadMG
    Sep 7, 2016 at 7:14
  • $\begingroup$ @DeadMG, I know, but that never happens for any $x$ on any parabola. $\endgroup$
    – user307169
    Sep 7, 2016 at 11:01
  • $\begingroup$ @Hurkyl : Yes, with the extended real numbers we get hyperbolic geometry, along with a notion of distance. But we need projective geometry to understand what is going on here, and hence projective real numbers, which also dispenses with any notion of distance. $\endgroup$ Sep 7, 2016 at 11:15

The site is wrong. Not only do the arms of a parabola not eventually become parallel, they don't even approach being parallel as a limit as y goes to infinity.

The parabola is defined for all real numbers--there is no x beyond which the parabola "will not go". A function whose graph does approach being parallel as y goes to infinity is

y = 1/abs(x)

where the arms become arbitrarily close to the y axis and never reach it (undefined at x=0).

By contrast the arms of a parabola keep spreading further and apart as y tends to infinity which is to say the distance between them is unbounded.

It is interesting that the slopes of the arms do become arbitrarily close, and one definition of parallel lines is that they have the same slope. I don't know how to account for that. But I am sure that the fact that the distance between the arms is unbounded "takes precedence" over the slopes becoming closer.

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    $\begingroup$ Perhaps you are saying that there is no way to judge whether two lines are parallel if they are infinitely far apart. $\endgroup$
    – amI
    Sep 6, 2016 at 18:28
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    $\begingroup$ I would encourage looking at @DanielBuck's picture, which suggests the "right" interpretation of a limit of lines that recede from the origin. $\endgroup$ Sep 7, 2016 at 7:57

What they mean by "eventually the arms of the parabola become vertical" is that when taking the limit $$\lim_{x\rightarrow\infty}[2ax+b]=\cases{\infty \quad \;\;, \;a>0\\ -\infty \;\;\;,\;a<0}$$ the slope is vertical, although "eventually" and infinity are not to used synonymously, so their choice of words is unfortunate.

  • $\begingroup$ Limits do not show that a thing happens "eventually." $\endgroup$ Sep 6, 2016 at 20:55
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    $\begingroup$ @KyleStrand What.. that could be said to be precisely what they do! Of course, it all depends on how to interpret "eventually", which is not a mathematical concept, although you seem to know its true definition. $\endgroup$ Sep 6, 2016 at 21:04
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    $\begingroup$ Of course. "Eventually" means that an event actually occurs in some finite-but-unspecified amount of time. But mathematically, limiting behavior is not guaranteed to happen in finite <distance/time/etc>, and in this case (as proven by the OP!), it is guaranteed not to happen, regardless of the value (however large) of $x$. $\endgroup$ Sep 6, 2016 at 21:07
  • $\begingroup$ @KyleStrand Well that's reasonable, but you have to consider the context: "As proven by the accepted answer" the authors clearly mean "in the limit".. we can all agree that it was an unfortunate choice of words to say "eventually" instead, but I'm sure we can all also agree on what the intention was. But let's not go any further, these kinds of conversations tend not to be very productive! (I have added some quotation marks to humor you, I hope you like them :)) $\endgroup$ Sep 6, 2016 at 21:13
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    $\begingroup$ I'd still disagree that that's a reasonable use of "eventually", but that's not on you; if you explained what's meant by the word, even if the usage is questionable, you'd be clarifying rather than obscuring matters. $\endgroup$ Sep 6, 2016 at 21:51

What they probably meant to say that the tangents at both arms asymptotically tend to vertical, because, as you correctly state, the slope is $2ax + b$, which goes to $\pm\infty$ if $x\to\pm\infty$, and since the angle between the horizontal and the tangent line is equal to the arctangent of the slope, for $x\pm\infty$ you get that the angle goes to $\pm\pi/2$.

Lines with that direction angle are simply perpendicular to the $x$ axis, and so they are parallel. But the statement is somewhat flawed, since the arms will never be truly parallel. Just really close to that.


The reason behind this comes from projective geometry : a parabol is an ellipse tangent to the infinity circle. With that view it seems correct to say that tangents tend toward an intersection on the circle.

  • $\begingroup$ The "line at infinity", rather than "circle". It is the support of a degenerate circle, but only if "counted twice". $\endgroup$
    – egreg
    Sep 6, 2016 at 15:04
  • $\begingroup$ "Circular points at infinity" in a way... To be fair, the good way to vulgarize projective geometry implies a lot of poetry. $\endgroup$
    – marmouset
    Sep 6, 2016 at 16:20
  • $\begingroup$ Poetry is not helpful for the target audience of that site, which is obviously people without a ton of experience in mathematics. $\endgroup$ Sep 6, 2016 at 21:04
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    $\begingroup$ I wouldn't say it is correct to say that the tangents tend to an intersection on the circle, because the tangents actually coincide in the limit. $\endgroup$ Sep 7, 2016 at 7:52

One thing about comparing parabolas is that any two are "similar" in that a rescaling of both axes by the same factor can bring one parabolaa into any other. To see this consider $y=ax^2$ and replace $y$ by $ky,$ and $x$ by $kx.$ Then we have $(ky)=[a/k]\cdot (kx)^2.$ So in the rescaled system the constant, originally $k,$ has become $(a/k)$ in front of the square of the input. By this means two parabolas with a common vertex are seen to be "similar", and using a shift one can in the same way compare two other parabolas having different vertices. I think this similarity is what the first sentence in the quote refers to.

Compare this to cubics, for example $y=x^3-x$ is not similar to $y=x^3-2x.$


There is no contradiction. As you said, the slope keeps increasing, this means that as you go towards $x\rightarrow\pm\infty$, the slope also goes to $\pm\infty$, i.e the graph of the function gets more and more "vertical".

  • $\begingroup$ "More and more vertical" $\neq$ "parallel" or even "parallel to the vertical access". The word "eventually" does not clarify that the site is describing behavior that occurs nowhere. $\endgroup$ Sep 6, 2016 at 21:02

The first sentence is 'All parabolas are the same shape, no matter how big they are'. What is meant by this is: Given any two parabolas, it is possible to rescale and translate one so that it matches the other (note: 'rescale' includes 'flip over' via negative scale factors). [Caveat: This assumes both are oriented vertically; otherwise add 'rotate' to the action set]

The claim 'the arms eventually become parallel' is correctly addressed by @Reese.


The precise sense in which the statement is correct is that when viewed from a great distance above the plane of the parabola, thought of as a dynamic process of "zooming out to infinity", two things will happen (not necessarily at the same speed, but both processes will become visible given enough distance of the viewer from the plane).

  1. The two arms start to look flat and linear, like two straight lines with the parabola's axis of symmetry bisecting the angle between them.

  2. The angle between those two lines tends to 0, until the arm "lines" look indistinguishable from the axis of symmetry.

If there are several parabolas in the plane with parallel axes of symmetry then their "arms" will all appear to collapse into one line when seen from a great enough perpendicular distance to the plane.

  • $\begingroup$ Technically the arms become half-lines (rays) but I think calling them "lines" does not lead to any confusion in the question or answers. Also the symmetry axes toward which the half-lines collapse are also half lines, not the whole line of symmetry. $\endgroup$
    – zyx
    Sep 7, 2016 at 4:24

To find inclination you need to differentiate once only, you did it twice.

What he meant is

at $ x = + \infty $ slope $x \rightarrow +\infty $


at $ x = - \infty $ slope $x \rightarrow -\infty $

So the final inclination to x-axis is $\pm 90^0 $ and they remain parallel.

  • $\begingroup$ The double differentiation showed that the slope kept on varying $\endgroup$
    – Parth
    Sep 7, 2016 at 11:32
  • $\begingroup$ Single differentiation shows slope as infinite except for sign. There are only two points at $ \infty$ you are talking about. Not two thousand of them. Double differentiation is for deciding max/min or curvatures. $\endgroup$
    – Narasimham
    Sep 7, 2016 at 11:44

You cannot just take the derivative of a function until you get to a value $\neq \pm\infty$, since the difference between two infinite values is undefined. You have to check the behaviour of the functions as $x$ approaches infinity. If you do this, you can prove the tangents become parallel.

Let's define $t$ as the direction of the tangent, i.e.

$$t(x) = \begin{pmatrix}\frac{d(f(x))}{dx}\\\\1\end{pmatrix}$$

since $\lim\limits_{x\to\infty} | t(x) | = +\infty$ we need the normalized direction, i.e.

$$\begin{align} t_{norm}(x) &=&\frac{t(x)}{|t(x)|}\\\\ &=& \frac{1}{\sqrt{(2ax+b)^2+1}}\cdot\begin{pmatrix} 2ax+b\\1 \end{pmatrix} \end{align}$$ which gives us

$$\begin{align} \lim\limits_{x\to+\infty}t_{norm}(x) &=& \begin{pmatrix} 1\\0 \end{pmatrix}\\\\ \lim\limits_{x\to-\infty}t_{norm}(x) &=& \begin{pmatrix} -1\\0 \end{pmatrix}\\\\ &=& - \lim\limits_{x\rightarrow+\infty}t_{norm}(x) \end{align}$$ Which means the tangents are parallel.


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