Double discontinuity on a graph (1 Viewer)

leehuan · Jan 29, 2016

I'm too lazy to figure out anything myself so I'm asking you guys this time.

Is there any difference between the graphs of:

$\\ y=\frac { { x }^{ 2 }-1 }{ { x }^{ 2 }-2x+1 } \\ y=\frac { { x }^{ 3 }-{ x }^{ 2 }-x+1 }{ { x }^{ 3 }-3{ x }^{ 2 }+3x-1 }$

InteGrand · Jan 29, 2016

leehuan said:
I'm too lazy to figure out anything myself so I'm asking you guys this time.

Is there any difference between the graphs of:

$\\ y=\frac { { x }^{ 2 }-1 }{ { x }^{ 2 }-2x+1 } \\ y=\frac { { x }^{ 3 }-{ x }^{ 2 }+x+1 }{ { x }^{ 3 }-3{ x }^{ 2 }+3x-1 }$

$\noindent Based on your query, not sure whether you typed those both correctly (mainly the second one), but yes, the graphs are different. I think you meant to have a different numerator for the second one. Something like $(x+1)(x-1)^2$ so that both equations reduce to $\frac{x+1}{x-1}$ for $x\neq 1$. In this case, both graphs are the same, with a vertical asymptote at $x=1$ (unless we define a value there).$

leehuan · Jan 30, 2016

InteGrand said:
Based on your query, not sure whether you typed those both correctly (mainly the second one), but yes, the graphs are different. E.g. the first one has a y-intercept of -1 whilst the second has a y-intercept of +1.

Oh. That should be a - on the x

InteGrand · Jan 30, 2016

leehuan said:
Oh. That should be a - on the x

Yeah, the equations have the same graph.

leehuan · Jan 30, 2016

InteGrand said:
Yeah, the equations have the same graph.

Cool. All I needed

Obviously y=(x+1)/(x-1) itself won't have the discontinuity

InteGrand · Jan 30, 2016

leehuan said:
Cool. All I needed

Obviously y=(x+1)/(x-1) itself won't have the discontinuity

$\noindent It would have a vertical asymptote at $x=1$, since as $x\to 1$, the numerator approaches a non-zero number whilst the denominator approaches 0. This is the same as the graph of the equations you posted in the Original Post.$

leehuan · Jan 30, 2016

Actually.

Now I'm beginning to wonder if I accidentally inverted the numerator and denominator when I was writing the question.

InteGrand · Jan 30, 2016

leehuan said:
Actually.

Now I'm beginning to wonder if I accidentally inverted the numerator and denominator when I was writing the question.

$\noindent The reciprocal function would have a vertical asymptote at $x=-1$ (in fact, the whole graph gets flipped about the $y$-axis. In other words, that function $f$ satisfies $\frac{1}{f(x)}=f(-x)$ for all $x$ for which this equation makes sense (i.e. no 0 denominators etc.).$

leehuan · Jan 30, 2016

Ok, now that I'm more awake.

At the time I was thinking about the distinguishing factors between:

y=(x+1)/(x-1)

y=(x+1)^2/((x+1)(x-1)) which means yes I must've mixed up the numerator with denominator

and

y=(x+1)^3/((x+1)^2(x-1))

Double discontinuity on a graph (1 Viewer)

leehuan

Well-Known Member

InteGrand

Well-Known Member

leehuan

Well-Known Member

InteGrand

Well-Known Member

leehuan

Well-Known Member

InteGrand

Well-Known Member

leehuan

Well-Known Member

InteGrand

Well-Known Member

leehuan

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)