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The general form of a quadratic For  a  > 1 (such For  a  < 1 (such
As you can see, as the Copyright There is a simple, if slightly
This can be useful information: Parabolas always have a Advertisement
If the quadratic is written (Note: The "a" Since the vertex is a useful For a given quadratic y = ax^{2} + bx + c,
To find the vertex, I h = ^{–b}/_{2a} = ^{–(1)}/_{2(3)} = ^{–1}/_{6} Then I can find k by evaluating y at h = ^{–1}/_{6}: k = 3( ^{–1}/_{6 })^{2} + ( ^{–1}/_{6 }) – 2 = ^{3}/_{36} – ^{1}/_{6} – 2 = ^{1}/_{12} – ^{2}/_{12} = ^{–25}/_{12} So now I know that the
When you write down the The only other consideration Helpful note: If your quadratic’s xintercepts
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Derivation of the formula for the vertex of a parabola
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I’m taking a course on Basic Conic Sections, and one of the ones we are discussing is of a parabola of the form
$$y = a x^2 + b x + c$$
My teacher gave me the formula:
$$x = \fracb2a$$
as the $x$ coordinate of the vertex.
I asked her why, and she told me not to tell her how to do her job.
My smart friend mumbled something about it involving calculus, but I’ve always found him a rather odd fellow and I doubt I’d be able to understand a solution involving calculus, because I have no background in it. If you use something you know from calculus, explain it to someone who has no background in it. Because I sure don’t.
Is there a purely algebraic or geometrical yet elegant derivation for the $x$ coordinate of a parabola of the above form?
 Which definition of parabola are you using? The conic section one or the locus one? If you use the second, you can derive the equation of the parabola and the coordinates of the vertex directly from it.
– zar
Aug 5 ’10 at 8:30  Suggestion: tag it also "planecurves"
– Américo Tavares
Aug 29 ’10 at 22:51
 25Offtopic: your teacher is terrible, and it’s great that you’re asking these kinds of questions. Don’t give up.
– Paul VanKoughnett
Oct 15 ’10 at 3:40  1Unfortunately we have a system that often encourages teachers and students to be idiots. In this case the teacher was in compliance. ${}\qquad{}$
– Michael Hardy
Sep 6 ’15 at 19:49  It seems I got here late, but my answer explains why the shape of the graph of $y=ax^2+bx+c$ is always the same regardless of which numbers are $a,b,c$ (as long as $a\ne0$. ${}\qquad{}$
– Michael Hardy
Sep 6 ’15 at 19:52
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13 Answers
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Already so many answers, but I haven’t seen my favorite one posted, so here’s another.
The vertex occurs on the vertical line of symmetry, which is not affected by shifting up or down. So subtract $c$ to obtain the parabola $y=ax^2+bx$ having the same axis of symmetry. Factoring $y=x(ax+b)$, we see that the $x$intercepts of this parabola occur at $x=0$ and $x=\fracba$, and hence the axis of symmetry lies halfway between, at $x=\fracb2a$.
 I see now that the idea of translating vertically is also used in Isaac’s answer, but it is different from my answer. math.stackexchange.com/questions/709/…
– Jonas Meyer
Oct 15 ’10 at 2:52  Simply subtracting $c$ is something I would have never thought about, and reveals the zeroes as plain as day. The biggest jump in intuition was knowing what to do with the zeroes…and realizing that the axis of symmetry is exactly in between them.
– Justin L.
Oct 16 ’10 at 4:45  3Very nice — shifting up and down is a worthy trick here, mathematically and pedagogically
– David Lewis
Feb 25 ’12 at 1:43  There’s no need for shifting. We need two points with the same second coordinate, so if we plug in $x=0$ and $x=b/2a$, the zeroes of $ax^2+bx$, we get the same second coordinate, namely $c$.
– Michael Hoppe
Aug 20 ’15 at 10:49  @MichaelHoppe Where you write $b/2a$: I believe you intended to write $b/a$.
– Benjamin Dickman
Oct 5 ’15 at 3:16

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By the vertex I assume you mean the minimum/maximum point of the parabola. Indeed, this result can be discovered easily through a bit of calculus, but there is also a simple purely algebraic way, which I will present here.
Let’s consider a generic quadratic expression:
$y = ax^2 + bx + c$
We now complete the square on this formula.
$y = a[x^2 + bx/a + c/a]$
$y = a[(x + b/2a)^2 – (b/2a)^2 + c/a]$
The expression $ (b/2a)^2 + c/a$ is a constant (it does not depend on x), so we can replace it with k
for the matter of discussion.
$y = a[(x + b/2a)^2 + k]$
Now, depending on whether a is positive or negative, the parabola given by y will either have a maximum or minimum. Since a and k are fixed, this must occur when $(x + b/2a)^2$ is zero (we know it cannot be less than zero, and it can extend to infinity).
Hence, we know that for $(x + b/2a)^2$ to be zero, $x = b/2a$. This in turn implies that the function y is at a minimum or a maximum when this is true. Q.E.D.
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Parabolas of the form you described (y = …) are symmetric over a vertical line through their vertex. Let’s call that line x = k. This means that if the graph crosses the xaxis (meaning that $ax^2+bx+c=0$ has real solution(s)), they must be equidistant from x = k, so (k,0) must be the midpoint of the segment with endpoints at the zeros of the quadratic or k is the average of the zeros. From the quadratic formula, the two zeros of the quadratic are $\fracb\pm\sqrtb^24ac2a$, so their sum is $\fracba$ and their average is $k=\fracb2a$. So, the xcoordinate of the vertex must be $\fracb2a$.
If the parabola does not cross the xaxis (no real solutions), there is another parabola with equation $y=ax^2+bx+c’$ for some $c’$ for which the graph is a vertical translation of the graph of the original quadratic, but crosses the xaxis. Its axis of symmetry is $x=\fracb2a$ and because it is a vertical translation of the original, the axis of symmetry of the original is also $x=\fracb2a$, so the vertices of both have xcoordinate $\fracb2a$.
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An alternative approach, which isn’t necessarily the best in and of itself, but stick with me …
Instead of using the fact that the vertex is the point where a vertical line cuts the parabola exactly in half, we’ll use the fact that it’s also the point where a horizontal line just brushes the curve, meeting it in a single point.
Consider how the horizontal line $y=d$ crosses the parabola as you adjust $d$. If $d$ places the line above or below the vertex, there are either 2 or 0 points of intersection (which is which depends upon whether the parabola points up or down); when $d$ places the line at the same height as the vertex, the parabola seems to sit comfortably on the line. (The line is tangent to the parabola the point they have in common, in the same way that lines can be tangent to circles.)
We can find the point of tangency (the vertex), then, by using algebra to determine when the intersection of our parabola and line equations have a single solution.
We begin with the standard findtheintersection strategy, setting our formulas for $y$ equal:
$d = a x^2 + b x + c$
When does this equation have a single solution for $x$? Let’s see. Rearranging and invoking the Quadratic Formula, we have:
$a x^2 + b x + c – d = 0$
$x = \frac12a\left( – b \pm \sqrt b^2 – 4 a (cd) \right)$
Now, when the stuff under the radical matters, then we have either 2 or 0 values of $x$: two values if the stuff is positive (the square root gives a quantity to add and subtract from $b$); no roots if the stuff is negative (the square root gives an imaginary number we can’t use here). We’ll have our soughtafter single value of $x$ when $d$ is whateverithastobe (its exact value doesn’t matter) to make the stuff under the radical becomes zero, making the radical term vanish; when that happens, our value of $x$ falls out as
$x = \fracb2a$
Just like with the other approaches.
With this approach, however, we can consider a more sophisticated situation. It may be somewhat advanced for your class, but –if you can make sense of my rambling– you can use it to blow your teacher’s mind. 🙂
(Before we begin: I wouldn’t necessarily recommend this exploration for every student just beginning to learn about parabola equations. However, I think having it “in the archive”, as it were, can be worthwhile for people who come across the question to see that the answer can lead to some interesting places.)
We ask ourselves this question:
What if we tilt the line?
That is: What can we say about the $x$coordinate of the point of tangency where the parabola is touched by a line with slope $m$?
Let the slanted line’s equation be $y = m x + d$, where $d$ is adjustable. We can make the same observations as in the horizontal case: if $d$ places the line above or below the point of tangency, the line will meet the parabola in either 2 or 0 places; for a special value of $d$, however, the line meets the parabola at the single point, the point of tangency.
As before, we set our formulas for $y$ equal to each other …
$m x + d = a x^2 + b x + c$
… then rearrange and invoke the Quadratic Formula …
$a x^2 + ( b – m ) x + ( c – d ) = 0$
$x = \frac12a\left( (bm) \pm \sqrtstuff \right) = \frac12a\left(mb\pm \sqrtstuff\right)$
And as before, we don’t really care about the complicated mess that “$stuff$” might be, since we’re only interested in the situation where it goes away anyhow.[*] Once the radical term vanishes and the dust settles, we have this:
For the parabola described by the equation $y = a x^2 + b x + c$, the $x$coordinate of the point where a line of slope $m$ lies tangent to the curve is given by
$x = \frac12a\left( m – b \right)$
Here’s the kicker that your smart friend[**] can appreciate: That statement really is a fact that most people don’t see until Calculus. Except there, the issue tends to be phrased a little differently: It’s not about finding the $x$coordinate of the pointoftangency based on a given slope, but about finding the slope of the line based on the $x$coordinate for the pointoftangency. So, we rephrase the above equation to yield $m$:
For the parabola described by the equation $y = a x^2 + b x + c$, the slope $m$ of the line tangent to the curve at a point with $x$coordinate $x$ is given by
$m = 2 a x + b$
You’ll be way ahead of the game if you can see a neat way to get directly from the parabola equation to the slope formula. I’ll give you a hint by showing how the formula fits in with a couple of facts you already know about the slopes of lines, and show you one step further, with a few unnecessary zeros thrown in:
$y = a$ slope = $0$.
$y = a x + b$ slope = $a + 0$.
$y = a x^2 + b x + c$ slope = $2 a x + b + 0$
$y = a x^3 + b x^2 + c x + d$ slope = $3 a x^2 + 2 b x + c + 0$
$y = a x^4 + b x^3 + c x^2 + d x + e$ slope = you tell me
Detect a pattern?
[*] Actually, you should care enough to convince yourself that you can find a value of $d$ that makes “$stuff$” go away, but for now you can trust me that such a $d$ exists.
[**] Keep in mind that “smart” doesn’t always mean “having the right answers”; just as often, it means “asking the right questions”. The fact that you chose not to accept a formula blindly and asked your teacher “why” –and then steadfastly continued your quest for an answer when she wouldn’t provide one– suggests to me that your smart friend has a smart friend, too.
 Your solution may not have exploited the special structure of the problem, but you do get an upvote for generality.
– J. M. is not a mathematician
Aug 9 ’10 at 1:21  1+1 for the "visual" solution. I didn’t realize you were talking about the derivative until you derived it.
– stackErr
Jun 17 ’14 at 10:29
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Here is the simplest way I know, using a bit of algebra. Starting with Isaac’s observation that the vertex lies on the axis of symmetry, the line x = k. Question is, what’s k? That’s your answer.
if f(x) = ax^{2} + bx + c
Then the symmetry around x = k means that, for any h
f(k+h) = f(kh)
From that, just multiply it all out and some simple additive cancellation gives
2akh – bh = 2akh + bh
h cancels out, and a bit more algebra gives
k = $\fracb2a$
To really complete this, however, we should show that x = k = $\fracb2a$ is a maximum or minimum point. Again, with more algebra:
f(k+h) – f(k) boils down to ah^{2}
So when a > 0, then f(k+h) > f(k) for any h != 0, establishing that x = k is a minimum point
And likewise when a < 0, x = k is a maximum. This also shows the symmetry around x = k, and how the coefficient a determines whether the curve is concave upward or downward.
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I like Americo’s answer using translation of y = ax^2. But to finish it, we should at least note that ax^2 has a maximum or minimum at <0,0>, which is quite obvious, and that translation preserves order and therefore maxima and minima, which is pretty easy.
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The vertex of a parabola is the point on the axis of symmetry that intersects it ( Wikipedia ).
The point $(x,y)=\left( 0,0\right) $ is the vertex of the parabola given by $y=ax^2$.
By making the change of coordinates
$X=x+\dfracb2a,Y=y+\dfracb^24ac4a$,
equation $y=ax^2+bx+c$ is transformed into $Y=aX^2$, the vertex of which is the point $(X,Y)=(0,0)$, i.e. $(x,y)=\left(\dfracb2a,\dfracb^24ac4a\right)$.
(This change of coordinates is a translation of both axes $x$, $y$, respectively, by $\dfracb2a,\dfracb^24ac4a$.)
Hence, the $x$coordinate of the vertex of the parabola is $\dfracb2a$.
Edit. Example. Plot of $y=2x^2x+4$; blue axes: $X=x1/4,Y=y31/8$
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Complete the square
\beginalign
y = ax^2 + bx + c & = a\left(x^2 + \frac b a x \right) + c \\[10pt]
& = a\left( x^2 + \frac b a x + \frac b^24a^2 \right) + c – \frac b^24a \\[10pt]
& = a\ \underbrace\left( x + \frac b 2a \right)^2_\texta square + \underbrace\frac4acb^24a_\text No $x$ appears here..
\endalign
The square is $0$ when $x = \fracb2a$ and is positive when $x\neq \fracb2a$. From that you can conclude things about the shape of the graph, including the location of the vertex.
In particular, this shows you why the shape of the graph is the same regardless of the values of $a,b,c$ as long as $a\ne0$.
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Echoing Americo’s and Isaac’s answers, but without appealing to the quadratic equation:
First treat the special case f(x)=ax²+d (i.e. no “x” term); by remembering that the line that divides the parabola symmetrically passes through the vertex, and is in fact the yaxis (x=0), we see that 0 is the abscissa of the vertex.
To treat the more general f(x)=ax²+bx+c, we try a substitution of the form x=x’h (geometrically, this corresponds to translating your coordinate system horizontally by h units). Expanding this, you will get something like a(x’)²+(2ah+b)x’+(constant term). To go back to the special case we first treated, we find the value of h such that the coefficient of x’ is 0. Thus you obtain the expression for the vertex’s xcoordinate.
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There are nice answers already but I would like to add another way to look at this:
The quadratic equation $a(xr_1)(xr_2) = ax^2 – a(r_1 + r_2)x + r_1 r_2 = 0$ defines a parabola, $r_1, r_2$ are the roots (intersections with the xaxis).
The middle point is between the two roots, $\fracr_1 + r_22 = \frac a(r_1 + r_2)2a$.
This leads to an interesting question about what happens when the parabola does not have real roots.
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alt text http://hotmath.com/images/gt/lessons/genericalg1/parabs.gif
$y = ax^2 + bx + c$
for parabola written in Vertex Form vertex is (h,k).
$y = m(xh)^2 + k = \underlinemx^2 + \underline(2mh)x+ \underlinemh^2 +k$
$\rightarrow$
$a = m$
$b = 2mh = 2ah \rightarrow h = b/2a$
$c = mh^2 + k = ah^2 + k \rightarrow k = c – b^2/4a$
source
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Lots of interesting answers here. Might as well make it an even dozen. 🙂
Actually, this can be derived directly from the definition of a parabola. Using the definition and the distance formula, you can derive the Standard Format for the equation of a parabola:
$$4p\left(yk\right) = \left(xh\right)^2$$
where $V=(h,k)$ is the vertex of the parabola and $p$ is the distance from the vertex to the focus (and also from the vertex to the directrix). If $p > 0$, the parabola opens upward; if $p < 0$, the parabola opens downward. I assume you’ve run across this standard form already and are familiar with it (there are plenty of online pages you can reference if not).
Now just expand the righthandside and solve for y. You end up with:
$$y = \left.\frac14p\right.x^2 – \left.\frach2p\right.x + \frach^24p + k$$
which is an equation of the form $y = ax^2 + bx + c$ with
$$a = \frac14p,\qquad
b = \frach2p,\qquad
c = \frach^24p + k$$
Now just go right down the line and solve for $p$, $h$, and $k$.
$$\beginalign
a & = \frac14p \\
p & = \frac14a \\
\endalign$$
[Here’s the one you’re interested in…]
$$\beginalign
b & = \frach2p \\
b & = \frach\frac12a\qquad \textsubstituting\; p = \frac14a\; \textabove \\
\fracb2a & = h \\
\endalign$$
and lastly,
$$\beginalign
c & = \frach^24p + k \\
c & = \frac\left(\fracb2a\right)^2\frac1a + k\qquad \textsubstituting for\; p\;\text&\;h\;\textabove \\
c & = \frac\fracb^24a^2\frac1a + k \\
c & = \fracb^24a + k \\
c – \fracb^24a & = k \\
\endalign$$
Now you can easily convert back & forth between Standard Format and the other.
 1Please see here for how to typeset common math expressions with LaTeX
– Zev Chonoles
Jun 4 ’13 at 4:23  I’ll take a look. Thanks.
– Bullwinkle
Jun 4 ’13 at 15:02
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It could be argued that the parabola is symetric respect to the vertical line passing through the vertex point. So, the vertex is located in the midpoint between the two roots of $y=0$.
As pointed out before, by completing the square, the parabola equation can be written as
$$ y = a(x – x_+)(x – x_)\,, $$
where $x_+$ and $x_$ are the roots mentioned above, given by
$$ x_\pm = \dfrac b \pm \sqrt b^2 – 4 a c 2a \,.$$
Then, the midpoint between $x_+$ and $x_$ is
$$ x_0 = \dfrac12 (x_+ + x_ ) = \fracb2a \,.$$
Note that this demonstration is also valid even when there is no real roots for $y=0$ (which is the case when $b^2 < 4a c$), since the complex terms in the last equation are mutually canceled.
Sketch of the parabola:
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Graphing Sections: Introduction ,
The general form of a quadratic For  a  > 1 (such For  a  < 1 (such
As you can see, as the Copyright There is a simple, if slightly
This can be useful information: Parabolas always have a Advertisement
If the quadratic is written (Note: The "a" Since the vertex is a useful For a given quadratic y = ax^{2} + bx + c,
To find the vertex, I h = ^{–b}/_{2a} = ^{–(1)}/_{2(3)} = ^{–1}/_{6} Then I can find k by evaluating y at h = ^{–1}/_{6}: k = 3( ^{–1}/_{6 })^{2} + ( ^{–1}/_{6 }) – 2 = ^{3}/_{36} – ^{1}/_{6} – 2 = ^{1}/_{12} – ^{2}/_{12} = ^{–25}/_{12} So now I know that the
When you write down the The only other consideration Helpful note: If your quadratic’s xintercepts
<< Previous Top  1  2  3  4 

 MathHelp.com Courses K12 Math 5th Grade Math College Math College PreAlgebra Standardized Test Prep ACCUPLACER Math more tests… 
Copyright � 20022014 Elizabeth 
 Feedback 


