And if that's just part of the definition of a parabola, I guess my question becomes why is the graph of any quadratic a parabola?

My attempt at explaining:

The way I understand it after some thought is that any quadratic can be written by completing the square as a perfect square + a constant, for ex:

$f(x) = x^2 + x$ can be written as $f(x) = (x+\frac{1}{2})^2 - \frac{1}{4}$.

So essentially, any quadratic is a displaced version of $f(x) = x^2$, and it's pretty obvious why $f(x) = x^2$ has an axis of symmetry and why it's at the vertex.

Is my reasoning correct? And if you have a different way to think about it and explain it, whether geometric, algebraic or other, I would love to see it.

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    $\begingroup$ Cf. en.wikipedia.org/wiki/… $\endgroup$
    – Simon S
    May 30, 2015 at 15:14
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    $\begingroup$ If I would ask this question and someone would give me this answer, then my next question would be: "pretty obvious..., please explain". It is a good thing however to bring it back to one specific case. $\endgroup$
    – drhab
    May 30, 2015 at 15:14
  • $\begingroup$ @SimonS, Thanks, just got done reading that, very interesting. $\endgroup$ May 30, 2015 at 16:45

4 Answers 4


You have a valid argument that the graph of a quadratic function always has a vertical axis of symmetry.

If we look at a parabola geometrically it is defined as the intersection between a plane and a cone in a certain orientation with respect to the plane. In that case we can show generally that every conic section must have an axis of symmetry:

Consider the axis of the cone. If it is perpendicular to the plane, the conic section is a circle, and we know that a circle has lots of axes of symmetry -- any diameter will do.

Otherwise, project each point on the cone's axis perpendicularly onto the plane. This creates a line in the plane. The projected line is either parallel to the cone's axis, or it intersects it. In each case the two lines lie in a common plane, which is perpendicular to the image plane.

Reflecting every point in space in this common plane takes the cone to itself and the image plane to itself (and therefore takes the conic section to itself). This reflecting within the image plane is a reflection through the projected cone axis; and we have just seen that the conic section is symmetric under this reflection.

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    $\begingroup$ My brain is hurting is from trying to visualize all of this, but I can finally see it. It's a very nice way to think about it, I hope one day I can have similar insights. Thank you. $\endgroup$ May 30, 2015 at 16:10

Certainly completing the square explains why all graphs of equations of the form $y=ax^2+bx+c$ have the same shape (where $a\ne0$).

But as for why all are "parabolas", perhaps that depends on which definition of "parabola" you use.

A parabola is either

  • A certain kind of section of a cone; or
  • The set of points equidistant from a focus and a directix; or
  • Perhaps any of several other characterizations?

Why are all of these the same thing? That is interesting but I'm not sure that was the intent of your question.

  • $\begingroup$ I'm guessing the definition the OP had in mind is that a parabola is the shape produced by a second degree polynomial. $\endgroup$
    – kasperd
    May 31, 2015 at 14:07

I love Henning's conic section answer above..... But you can also think about it physically.

A parabola is a good approximation of the path of a physical object that is undergoing constant acceleration. Like a ball thrown on a planet under somewhat constant acceleration from Gravity.

How does this relate to quadratics? First forget about horizontal position.

Imagine as time steps by, we update position and velocity of a particle in one dimension, a particle that you throw down off a cliff on a planet with no atmosphere. y = vertical distance from your position on the cliff. At each time step:

y_position = y_position + y_velocity
y_velocity = y_velocity + y_acceleration
y_acceleration = constant

imagine time steps, with initial position 0, velocity 1, and acceleration of 2

time    p   v   a
0-step: 0,  1,  2
1-step: 1,  3,  2
2-step: 4,  5,  2
3-step: 9,  7,  2
4-step: 16, 9,  2
5-step: 25, 11, 2
6-step: 36, 13, 2
7-step: 49, 15, 2

as u can see, the position is increasing as a square of time, even though we are only using addition to update our particle information. Position turns out to be Time Squared. Position is f(x)=x^2, and time is x.

Why is this quadratic? Because at each step, you add the velocity... but you have previously added the other velocities. At time 4, position is 9+7, but where did the 9 come from? from previously adding 4 and 5. position 16 is actually the sum of all previous velocities - 7 + 5 + 3 + 1 = 16. It becomes sort of like the equation for the sum of numbers between 1 and n, from the famous story about Guass' childhood. That sum is a square. (quadratic). Multiplication is a form of repeated summing, and that's what we are doing here.

What about the non x^2 terms? Like, what if we have f(x) = x^2 + x? Well, the situation turns out to be the same - acceleration is constant. Let's imagine that Position = Time Squared + Time, or f(x) = x^2 + x:

time    p      v   a
0-step: 0+0 , ?,  ?
1-step: 1+1 , ?,  ?
2-step: 4+2 , ?,  ?
3-step: 9+3 , ?,  ?
4-step: 16+4, ?,  ?
5-step: 25+5, ?,  ?
6-step: 36+6, ?,  ?
7-step: 49+7, ?,  ?

We have no idea what velocity and acceleration are now.... but we can figure them out. First, velocity. Look at the pattern of the position numbers. We know that at each stage, the new position is the old position + velocity. So we can find velocity by looking at the differences.

time    p    v   a
0-step: 0 ,  2,  ?
1-step: 2 ,  4,  ?
2-step: 6 ,  6,  ?
3-step: 12,  8,  ?
4-step: 20, 10,  ?
5-step: 30, 12,  ?
6-step: 42, 14,  ?
7-step: 56,   ,   

We also know acceleration is the difference in velocities across the steps, so we can find a.

time    p    v   a
0-step: 0 ,  2,  2
1-step: 2 ,  4,  2
2-step: 6 ,  6,  2
3-step: 12,  8,  2
4-step: 20, 10,  2
5-step: 30, 12,  2
6-step: 42, 14,  2
7-step: 56,   ,  

Ohhh... so... by adding in that extra term to our quadratic... f(x)=x^2+x, all we really did was change our initial velocity and our position also changed faster! Acceleration is still constant 2.

All that extra 'x' in the equation did was scale the velocity up in a linear fashion. It didn't make acceleration itself increase.

Adding a constant number to the equation, like f(x) = x^2+x+2 would have a similar result. Acceleration would still be constant.

And if we graph position vs time, we would still get a parabola.

But.... what about the ball thrown with horizontal velocity? Ok... the horizontal speed of the throw doesnt change... the vertical position is what we care about. And the path is a parabola, we are simply stretching time out over a horizontal distance. For a simple example, imagine a 2 dimensional coordinate for a particle, x,y, and imagine it goes like this: x = 0 1 2 3 4, y = 0 1 4 9 16, time = 0 1 2 3 4. f(x) = y = time^2. x = time.

Now if you want to mess around with other stuff and make it more complicated, you can. Time = 2x? Sure. Why not. But the principles are the same. Constant acceleration of the particle produces a quadratic.

This works not only for f(x)=x^2 type situations but for all the ways you can rotate a parabola in space too. (like with the general conic equation linked above)

(Note all this depends on some big assumptions.... for example, that space is everywhere uniform and flat (which is what using Euclidean geometry and Cartesian coordinates imply), gravity is constant (it's not), etc etc. But inverse square laws are found quite often in physics as good approximations of reality.)

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    $\begingroup$ "as u can see, the position is increasing as a square of time". You do not "see" anything. Either you measure it and you will not get these exact figures (for many reasons, measurement uncertainty, air friction, ...) or you know that these are the numbers you should get because you assume a physical model where F=mg, from which you derive a quadratic relation between time and position. $\endgroup$
    – WoJ
    May 31, 2015 at 8:28
  • $\begingroup$ Thank you for taking the time to write all this. It took me a while to get through but it was very helpful. $\endgroup$ Jun 2, 2015 at 14:17
  • $\begingroup$ you are most welcome. $\endgroup$
    – don bright
    Jun 2, 2015 at 23:38

Suppose, $f(x)=(x+a)^2+b$

Then, we have $$f(-a-x)=f(-a+x)=x^2+b$$ for all $x\in \mathbb R$


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