Why do parabolas' arms eventually become parallel?

Question

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).

Now,

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

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.

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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.

Answer 4

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".

Answer 5

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.

Answer 6

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.

Answer 7

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 $y=x^3-2x.$

Answer 8

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.

Answer 9

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.

Answer 10

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 $L_{\infty}$.

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.

Answer 11

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.

Answer 12

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

What he meant is

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

and

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

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

Answer 13

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.