# ListPlot with Joined set to True

Let

f[x_]=9x^2-2x-3;


I want to make a table of f at x ={-1, -0.5, 0, 0.5, 1, 1.5, 2}. Plot these values with ListPlot + option Joined set to True Make a "normal" plot of f on the interval [-1, 2]

Table[{f[x]}, {x, -1, 2, 0.5}]

Curves := Riffle[Table[{f[x]}, {x, -1, 2, 0.5}], Table[{f[x]}, {x, -1, 2, 0.5}]]
ListLinePlot[Curves]
Plot[f[x], {x, -1, 2}]


Is this ok for this task? Because i think with Joined the points dont get joined but are still separate, but I don't know if that's normal because of ListPlot. And with Plot I don't think I got the right function, but if I do:

Plot[f[x], {x, -1, 2}]


then I miss the points $-.5, .5, 1.5$. Or shouldn't that be a problem since the question is not that specific?

Thanks in advance!

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Is this what you want to do? f[x_] = 9 x^2 - 2 x - 3; Curves = Table[{x, f[x]}, {x, -1, 2, 0.5}]; ListLinePlot[Curves, InterpolationOrder -> 3] Plot[f[x], {x, -1, 2}] –  chris Oct 6 '12 at 13:05
Yes i think it is. –  Jaimy Oct 6 '12 at 13:08
But the next question was: describe the difference of the plots, where i expected listlineplot to be less accurate –  Jaimy Oct 6 '12 at 13:08
remove the InterpolationOrder->3 and it will be so. Interpolation was done precisely to make it look smooth! –  chris Oct 6 '12 at 13:11
ok thanks shouldve come up with that myself... –  Jaimy Oct 6 '12 at 13:12

## 1 Answer

It seems you want to do something like this:

f[x_] = 9 x^2 - 2 x - 3;
Curves = Table[{x, f[x]}, {x, -1, 2, 0.5}];
Show[{
ListLinePlot[Curves, PlotStyle -> Red,
Epilog -> {AbsolutePointSize[Large], Point /@ Curves}] ,
Plot[f[x], {x, -1, 2}, PlotStyle -> Dashed]
}]


which shows the difference between the Built in Plot and ListLinePlot routines.

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