Re: graphs and AxesOrigin
- To: mathgroup at smc.vnet.net
- Subject: [mg43103] Re: graphs and AxesOrigin
- From: "Allan Hayes" <hay at haystack.demon.co.uk>
- Date: Tue, 12 Aug 2003 04:43:14 -0400 (EDT)
- References: <bh7d59$54k$1@smc.vnet.net>
- Reply-to: "Allan Hayes" <hay at haystack.demon.co.uk>
- Sender: owner-wri-mathgroup at wolfram.com
"Emmanuel" <dechenau at purdue.edu> wrote in message news:bh7d59$54k$1 at smc.vnet.net... > 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}]\) > Emmanuel, Try the option PlotRange -> {{250, 310}, {180, 210}} This does not affect the range that you construct the graphics objects over, but does change the amount of "canvas" that you show, and it determines the axes. Allan --------------- Allan Hayes hay at haystack.demon.co.uk Voice: +44 (0)116 241 8747 Fax: +44 (0)870 164 0565