Re: graphs and AxesOrigin
- To: mathgroup at smc.vnet.net
- Subject: [mg43096] Re: [mg43088] graphs and AxesOrigin
- From: Tomas Garza <tgarza01 at prodigy.net.mx>
- Date: Tue, 12 Aug 2003 04:43:08 -0400 (EDT)
- References: <200308110616.CAA05144@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Try the option PlotRange->{{0, 320},{0, 220}} in your Plot. This will
include the origin in the graph, but I wonder if what you get is
interesting.
Tomas Garza
Mexico City
----- Original Message -----
From: "Emmanuel Dechenaux" <dechenau at purdue.edu>
To: mathgroup at smc.vnet.net
Subject: [mg43096] [mg43088] graphs and AxesOrigin
> Hi,
> I have what I assume is a simple question. I'm trying to make the graph
> below but if I let mathemtica determine the origin it puts the vertical
> axis in the middle of the curve. I don't want this to happen, so I use
> AxesOrigin. Now unfortunately it doesn't show the axes all the way to
> the origin, it doesn't go beyong the range I specified for the variable
> in the function I want to plot. I was wondering if there's a way for
> this not to happen (the plot range can't possibly be different). Thanks.
> Emmanuel
>
> \!\(c = 0.5\[IndentingNewLine]
> M = 1000\[IndentingNewLine]
> k1 = 950\[IndentingNewLine]
> d = 0.52\[IndentingNewLine]
> k2 = 750\[IndentingNewLine]
> p2 = \((1 - c)\) \(M - d \((M - k2)\)\)\/\(2 \((1 - d)\) k1\) +
> c\[IndentingNewLine]
> S1 = \((1 - c)\) \((\((1 - d)\) k1 + d \((M -
k2)\))\)\[IndentingNewLine]
> L1 = \((1 -
> c)\) \((\(\((1 - d)\) \((p2 - c)\) k1 + d \((1 - c)\) \((M - \
> k2)\)\)\/\(1 - c\))\)\[IndentingNewLine]
> U1 = \((1 -
> c)\) \((d \((k1)\) + \((1 - d)\) \((M -
> k2)\))\)\[IndentingNewLine]
> Plot[Min[\((1 - c)\) M -
> v, \((\(v - d \((1 - c)\) \((M - k2)\)\)\/\(\((1 - d)\) k1\))\)
> \((M \
> - v\/\((1 - c)\))\)], {v, L1, U1}]\)
>
>
>
- References:
- graphs and AxesOrigin
- From: Emmanuel Dechenaux <dechenau@purdue.edu>
- graphs and AxesOrigin