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The equation $y(x-1)=x^2-1$ can be graphed on Cartesian plane by inserting values in for $y$ and then solving for $x$

For example, if $y=3$ then:

$3(x-1)=x^2-1$

$3x-3=x^2-1$

$3x=x^2+2$

$0=x^2-3x+2$

$0=(x-2)(x+1)$

Thus $x=1$ or $2$ and so co-ordinates (1,3) and (2,3) can be plotted.

Repeating this process with different values for $y$ produces this graph with a vertical line.

Is the vertical line that appears in the graph a removable singularity or something else?

Mr Indeterminate
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  • Your equation is the one of a conical section. Other than ellipse, parabola and hyperbole there are some degenerate conics. In this case the conic section degenerates to a pair of lines. – N74 Mar 16 '19 at 10:36

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Let’s rewrite your equation: $$x^2-1-y(x-1) = (x-1)(x-y+1) = 0,$$ so it’s the product of the equations $x-1=0$ and $x-y+1=0$. The graph of your original implicit equation is therefore the union of the graphs of these two equations, i.e., it’s a pair of lines. One of them happens to be vertical, but that doesn’t mean that there’s a singularity per se. It simply means that $y$ can’t be expressed as a function of $x$ everywhere on the plane.

amd
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  • The reason I was pondering if the vertical line is a singularity is because it is a very similar equality to $y=(x^2-1)/(x-1)$ which is typically graphed with a hole at $x=1$. – Mr Indeterminate Mar 16 '19 at 22:03
  • The two equations are not equivalent, since the latter is undefined at $x=1$, nor are their graphs similar: the graph of the latter is a single line with a removable singularity, as you noted. Effectively, by dividing through by $x-1$ you both eliminate one of the components of the graph and introduce a hole in the line. This is an illustration of the usual caveat that dividing by a quantity that is potentially zero can remove solutions to the original equation. – amd Mar 16 '19 at 22:11
  • One equation is typically depicted with hole at $x=1$ and the other can be shown to have a vertical line at $x=1$ otherwise the graphs look the same. These two features appear to be complete opposites as nothing is potted $x=1$ in one instance and in the other the whole of $x=1$ is plotted. The hole is known as a removable singularity. What is the vertical line known as? – Mr Indeterminate Mar 16 '19 at 23:11
  • A vertical line. – amd Mar 16 '19 at 23:11
  • Lol, thanks for the answer – Mr Indeterminate Mar 17 '19 at 02:14