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if I have a plot let us say 

Plot[x^2 - 5 x - 10, {x, 0, 8}]

I want o check if there is a way to know the automatic plot range and to add or subtract from that range as I need.

I will assume something fictitious like this:

Plot[x^2 - 5 x - 10, {x, 0, 8}, PlotRange -> (Automatic + {0, 10})]

is there a way to do something like this?

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    $\begingroup$ Are you asking about y axis? Plot[x^2 - 5 x - 10, {x, 0, 8}, PlotRangePadding -> {Automatic, {Automatic, 8}}] ? $\endgroup$
    – Kuba
    Commented May 25, 2014 at 15:20
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    $\begingroup$ Related: "How to get the real PlotRange?" $\endgroup$
    – Alexey Popkov
    Commented May 25, 2014 at 15:31
  • $\begingroup$ @Kuba yes I am asking about y axis. thanks for the answer. can that be applied for both sides of y axis (+y and -y)? $\endgroup$
    – Basheer Algohi
    Commented May 25, 2014 at 15:37
  • $\begingroup$ @Algohi It's unclear what it is about. PlotRange padding will not create parts of plot that are not already there. Take as example Tan[x]. About x axis, isn't it the second argument of Plot? $\endgroup$
    – Kuba
    Commented May 25, 2014 at 16:08
  • $\begingroup$ pretty sure in general you need to generate the plot use AbsoluteOptions[] or some such to get the range then re-plot. $\endgroup$
    – george2079
    Commented May 25, 2014 at 17:32

2 Answers 2

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OP confirmed that PlotRangePadding is what solves the problem.

Plot[x^2 - 5 x - 10, {x, 0, 8},   PlotRangePadding -> {Automatic, {Automatic, 8}}]
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just for completeness here is the issue with padding, if automatic ranging clipped the function, PlotRangePadding does not recover the clipped portion:

 Plot[ 1/x , {x, -1, 1},
     PlotRangePadding -> {Automatic, {10, 10}}]

enter image description here

Fix by generating the plot first to get the auto generated range then again with manual padding:

 Plot[1/x, {x, -1, 1}, 
     PlotRange -> ({#[[1]], #[[2]] + {-10, 10}} &@
      (PlotRange /.  AbsoluteOptions[Plot[ 1/x , {x, -1, 1}]]))]

enter image description here

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  • $\begingroup$ you are right about this. I will consider this way also. great thanks. $\endgroup$
    – Basheer Algohi
    Commented May 27, 2014 at 17:10

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