Why does "$x^2 - 5x + 6 = 0$", which is the same as "$(x-3)(x-2) = 0$", represent a parabola? 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:


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. 
