Hello,
I’ve been trying to plot this rational function and illustrate the asymptotes of it, but I don’t know how to get both the asymptotes plotted in together with the function.
f(x_):=(3x-1)/(2x+6)
You get a graph like that with
f[x_] := (3 x - 1)/(2 x + 6);
Plot[{f[x], 3/2}, {x, -11, 7}, Exclusions -> {x == -3},
ExclusionsStyle -> Red, PlotStyle -> {Black, Red}, PlotRange -> 10,
AspectRatio -> Automatic]
You can insert the labels with Epilog:
f[x_] := (3 x - 1)/(2 x + 6);
Plot[{f[x], 3/2}, {x, -11, 7}, Exclusions -> {x == -3},
ExclusionsStyle -> Red, PlotStyle -> {Black, Red}, PlotRange -> 10,
AspectRatio -> Automatic,
Epilog -> {Text[x == -3, {-5, -8}], Text[y == 1.5, {5, 2}]}]
Hi Torjus,
I had a quick look, and there are a few ways of doing this. When you just plot the function like this:
Plot[(3 x - 1)/(2 x + 6), {x, -10, 6}]
you get an output with the vertical asymptote:
You can then just add another function to plot of 1.5, like this:
Plot[{(3 x - 1)/(2 x + 6), 1.5}, {x, -10, 6}]
This is fine, but if you want both asymptotes to be a different graphics primitive we will have to try something else. You can also use GridLines, though I find this a bit clunky:
Plot[(3 x - 1)/(2 x + 6), {x, -10, 6},GridLines->{{-3,0},{0,1.5}}]
OK, so now we have both asymptotes, but still the vertical one from the original plot. To stop this from showing up use Exclusions (an option in plot):
Plot[(3 x - 1)/(2 x + 6), {x, -10, 6},Exclusions->{x==-3},GridLines->{{-3,0},{0,1.5}}]
Another way, and my preferred method, is to use the option Epilog to add a Line primitive onto the plot. You can then style the asymptotes separately to make them more clear:
Plot[{(3 x - 1)/(2 x + 6), 1.5}, {x, -10, 6},
PlotStyle -> {Black, Red}, Exclusions -> {x == -3},
Epilog -> {Red, Line[{{-3, -100}, {-3, 100}}]}]
From here, you can use labelling options to add in your asymptote labels.
I hope this helps!
Nia Knibbs Vaughan