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Re: Re: Needed Grid lines in Implicit Plot

On 13 Dec 2004, at 18:24, Bob Walker wrote:

> Remove ", {y, -4 , 4 }" iterator and it will work as you wish.
> ImplicitPlot accepts only one iterator.  You can inspect the
> ImplicitPlot signature by:
> ?ImplicitPlot
> I don't know why there is no error indication.
> Take care,
> Bob Walker
> Narasimham wrote:
>> << Graphics`ImplicitPlot`
>> ImplicitPlot[3 x + 5 y == 1, {x, -4 , 4}, {y, -4 , 4 },
>> GridLines ->{Range[-4, 4, 1], Range[-4, 4, 1]},AspectRatio ->
>> Automatic];

This is quite simply false.

There are two forms of Implicitplot. they are both described int he 
documentation (read till the end)

ImplicitPlot[eqn, {x, a, b}] draws a graph of the set of points that 
satisfy \
the equation eqn.  The variable x is associated with the horizontal 
axis and \
ranges from a to b.  The remaining variable in the equation is 
associated \
with the vertical axis. ImplicitPlot[eqn, {x, a, x1, x2, ..., b}] 
allows the \
user to specify values of x where special care must be exercised. \
ImplicitPlot[{eqn1, eqn2, ...}, {x, a, b}] allows more than one 
equation to \
be plotted, with PlotStyles set as in the Plot function. 
ImplicitPlot[eqn, \
{x, a, b}, {y, a, b}] uses a contour plot method of generating the 
plot.  \
This form does not allow specification of intermediate points

The form of ImplicitPLot with one iterator only works on equations that 
can be "solved" algebraically using Solve. it produces a Graphics 
object, which is why it accepts the GridLines option.

The other form of ImplicitPlot works much more generally, but it 
produces a CountourGraphics object, which needs to be converted to a 
Graphics object before using the GridLines option.

Andrzej Kozlowski
Chiba, Japan

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