Consider the equation $x^2 - 5x + 6 = 0$. By factorising I get $(x-3)(x-2) = 0$. Which means it represents a pair of straight lines, namely $x-2 =0 $ and $x- 3 = 0$, but when I plot $x^2 - 5x + 6 = 0$, I get a parabola, not a pair of straight lines. Why?
Plotting: x^2 - 5x + 6 = 0
at Wolfram Alpha, I get the result:
On
This is because Wolframalpha is plotting $y=(x-2)(x-3)$, which is a parabola.
As you have entered $(x-2)(x-3)=0$, it is merely indicating where the intersection is between $y=(x-2)(x-3)$ and $y=0$, which is why there are red dots on the points where the $x$-coordinates are $2$ and $3$.
This is not interpreted as a function by Wolframalpha as it contains only one variable, and thus it is commonly interpreted as mentioned above, but given that $x $ is some function of $y $, it will be, as you have suggested, two lines parallel to the $y $-axis.
On
I get $(x-3)(x-2) = 0$. which means it represent a pair of striaght lines namely $x-2 =0 $ and $x- 3 = 0$
When we just write $$ (x-3)(x-2) = 0 $$ and ask for $x$ we usually mean the set of solutions $$ S = \{ x \mid (x-3)(x-2) = 0 \} $$
Two vertical lines in two dimensions we can model as: $$ L_1 = \{ (x,y) \mid (x-3) = 0 \} \\ L_2 = \{ (x,y) \mid (x-2) = 0 \} $$ These are different sets. We sloppily write $x-3=0$ and $x-2=0$ but mean the above.
On
After much discussion in comments, I have decided to interpret this as a WolframAlpha question. Many people would not plot an equation in one variable. The solution could be plotted on a number line. In $2$ dimensions, the plot would in fact be $2$ lines parallel to the $y$-axis.
WolframAlpha does not interpret it in this fashion. It seems to interpret each side of the equation as a function in $1$ variable. $f(x)=x^2-5x+6,g(x)=0$. It graphs both and highlights points of intersection.
On
To get from the equation $$x^2 - 5x + 6 = 0$$ to a set of solutions, one must do some interpretation. Namely, in what universe or "base set" do we search the solutions?
If this is to be interpreted as $\{ (x,y)\in\mathbb{R}^2 \mid x^2 - 5x + 6 = 0 \}$, then the solution set is indeed two vertical lines: $$\{ (x,y)\in\mathbb{R}^2 \mid x=2 \quad\mathrm{or}\quad x=3 \} = \{ (x,y)\in\mathbb{R}^2 \mid x=2 \} \cup \{ (x,y)\in\mathbb{R}^2 \mid x=3 \}$$
However, one could also interprete as the set $\{ x\in\mathbb{R} \mid x^2 - 5x + 6 = 0 \}$ and then your solution set is: $$\{ x\in\mathbb{R} \mid x=2 \quad\mathrm{or}\quad x=3 \} = \{ 2,3 \}$$
Wolfram Alpha took the last approach and showed with two red dots on the $x$ axis (abscissa) the two-point set. Then, it also plotted the difference between the left-hand side and the right-hand side of the equation $x^2 - 5x + 6 = 0$ with another color, along a vertical coordinate axis (ordinate). In a sense, this shows how "far" from being a solution other $x$ values (than $2$ or $3$) are.
In fact, the blue "helpful" addition to the two-point solution set, is the same as $y = (x^2 - 5x + 6) - 0$ or just $y=x^2 - 5x + 6$.
On
I believe the other answers are correct in the interpretation of the question and in the way that they have addressed the issue. But I would like to look at the question a bit differently. Whether this is of any use or interest is up to the OP.
$y = (x -2)$ and $y = (x-3)$ are lines. So why is their product a parabola?
Consider $y = (x-2)(x -3) = (x-2)x - 3(x-2)$. If you compare this to the slope-intercept form of a line, $y = mx + b$, then you get $$m = (x-2)\\b=-3(x-2)$$
In particular, it is the $m = (x-2)$ part that makes this a parabola instead of a line. In a line, the slope is constant. But in this parabola, the slope is changing in a linear fashion itself. This is not just true of this parabola, but of any parabola. That is, a parabola is just a "line" whose slope changes linearly as you go along - a relation that is more explicit when examined using calculus.
So the parabola is indeed a combination of two lines. Only the combination is by one of the lines modifying the slope of the other "line".
On
As has been pointed out $x^2-5x+6=0\quad$ is not the equation of the parabola.
The equation of the parabola is:
$y_p = x^2-5x+6 = (x-2)(x-3)$
The equations of the two straight lines within the equation are is $y_1 = x-2$ and $ y_2 = x - 3$ and both lines can be plotted as a function of the same variable $x$. But if we multiple the two lines together we get:
$y_1y_2 = (x-2)(x-3)$
and $y_2= y_1 - 1$ so
$y_1(y_1-1) = (x-2)(x-3)$
$y_1^2 -y_1 = (x-2)(x-3)$
and thus $y_p = y_1^2 - y_1$ which isn't terribly useful.
But as $x$ goes to infinity the constant term 6 in the equation becomes negligible. Thus the parabola approaches the curve $y_3 = x^2-5x$ as $x$ approaches $+\infty$ or as $x$ approaches $-\infty$.
Now if we translate the axis to the point ($\dfrac{5}{2}$, $-\dfrac{1}{4}$), we can define:
$y' = y + \dfrac{1}{4}$ and $x' = x - \dfrac{5}{2}$
so that
$y' = x'^2$
and the quandary would seem to disappear.
On
You're asking about two different things in your question and in its title.
As for the title:
Why does "$x^2 - 5x + 6 = 0$", which is the same as "$(x-3)(x-2) = 0$", represent a parabola?
the answer is: it doesn't.
The equation $x^2 - 5x + 6 = 0$ is just an equation, and it does not represent anything unless you define the way of interpreting.
For example, if you say 'it's an equation of one real variable $x$' then it will mean just that: equality of two algebraic expressions, which holds for two real values only: for $x=2$ or $x=3$.
But if you say 'it's an equation of a figure in 3D spherical coordinates, with $x$ denoting a distance from origin', the equation will represent two concentric spherical surfaces.
And for 'it's an equation of a figure in planar Cartesian coordinates XY' it will represent two parallel lines.
Any way it won't represent a parabola.
As for the question itself:
when I plot $x^2 - 5x + 6 = 0$ (at Wolfram Alpha), I get a parabola, not a pair of straight lines. Why?
The shortest answer is: because you plot it at Wolfram Alpha.
The site performs plenty of pre-processing of users' queries, so that it can usually make a mathematical sense of questions asked in nearly everyday's prose. That applies also to equations. And your equation has a form $f(x)=0$ which often appears in questions like "what are zero points of a function $f(x)$?" So WAlpha tries to explain the problem to you by plotting $y=f(x)$ (which is a parabola!) and marking its zero points.
But the parabola here is not the 'answer' – it is just an illustration of the nature of the problem (as Wolfram Alpha 'thinks' you have). The actual answer is just a pair of red dots on the OX axis.
You can make this way of iterpreting invalid to WAlpha by adding explicit $y$ term, which renders any attempt to extract $y(x)$ doomed. That adds another line to your plot, but allows you to see the two lines you expected:
plot y(x^2 - 5x +6) = 0
results in this plot:
$x^2-5x+6=0\quad$ is not the equation of the parabola.
The equation of the parabola is $\quad x^2-5x+6=y(x)$
$(x-3)(x-2)=0\quad$ is not the equation of two straight lines.
The equations of the two straight lines is $\quad x-3=y(x)\quad\text{and}\quad x-6=y(x)\quad$ or : $$(x-2-y)(x-3-y)=0$$
Do not confuse the "equation" of a curve with the "equation" to be solved for an unknown $x$.
The meaning of the word "equation" isn't the same. In the first case, it means a relationship between two variables $y$ and $x$. In the second case, it means an equality not for any values of $x$, but only for some particular values of $x$. Then, solving for $x$ means finding those particular values.
In addition :
In the very different case $\quad x^2-4xy-y^2=0\quad$ there is $y$ in the equation. This is a relationship between $y$ and $x$. So, it is valid for various values of $x$ and the related values of $y$. This allows to draw a curve $$y(x)=(2\pm \sqrt{5})\:x$$ So, two straight lines : $\quad y(x)=(2+ \sqrt{5})\:x \quad$ and $\quad y(x)=(2- \sqrt{5})\:x$
Further addition :
$x^2-5x+6=0\quad$ is commonly understood as to be solve for $x$, that is, to find the roots of the equation. The answer is two constant values : $x=2$ and $x=3$.
If one want to make understand that the question is not to find the roots of the equation on the common sens, but is to find some unknown relationship between $x$ and $y$ satisfying $x^2-5x+6=0$ , in order to avoid the ambiguity, the equation should be written as : $$\left( x(y)\right)^2-5x(y)+6=0$$ because, this specifies that $y$ exists and that the equation have to be solved for a function $x(y)$.
Solving it leads to $$x(y)-2=0 \quad\to\quad x(y)=2$$ $$x(y)-3=0 \quad\to\quad x(y)=3$$ that is two lines parallel to the y-axis.