Plotting with different color for a single curvePlotting piecewise function with distinct colors in each...
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Plotting with different color for a single curve
Plotting piecewise function with distinct colors in each sectionSubset of edges with a different colorNeed 4D plot (3D + color for function)Question about plotting one function with different colorsWant a different color for each curve displayed with ShowHow to plot data with different colors (or symbols) depending on a conditionListPlot with different color optionsListPlot with different color options part IIPlotting: every point in different colorPlotting a function with different parameters sets
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$begingroup$
How to plot a function $f(x)=frac{3(4+x)}{3(2-x)-16}$ (say $x in [-15,15]$ ) with the condition that i want to give different color for each of the following cases
(i) when $frac{x+4}{3x+10}>0$ and $frac{x^2+8x+12}{3x+10}>0$
(ii) when $frac{x+4}{3x+10}>0$ and $frac{x^2+8x+12}{3x+10}<0$
(iii) when $frac{x+4}{3x+10}<0$ and $frac{x^2+8x+12}{3x+10}>0$
plotting
asked 11 hours ago
WomWom
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add a comment |
$begingroup$
How to plot a function $f(x)=frac{3(4+x)}{3(2-x)-16}$ (say $x in [-15,15]$ ) with the condition that i want to give different color for each of the following cases
(i) when $frac{x+4}{3x+10}>0$ and $frac{x^2+8x+12}{3x+10}>0$
(ii) when $frac{x+4}{3x+10}>0$ and $frac{x^2+8x+12}{3x+10}<0$
(iii) when $frac{x+4}{3x+10}<0$ and $frac{x^2+8x+12}{3x+10}>0$
plotting
asked 11 hours ago
WomWom
162 bronze badges
New contributor
Wom is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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add a comment |
$begingroup$
How to plot a function $f(x)=frac{3(4+x)}{3(2-x)-16}$ (say $x in [-15,15]$ ) with the condition that i want to give different color for each of the following cases
(i) when $frac{x+4}{3x+10}>0$ and $frac{x^2+8x+12}{3x+10}>0$
(ii) when $frac{x+4}{3x+10}>0$ and $frac{x^2+8x+12}{3x+10}<0$
(iii) when $frac{x+4}{3x+10}<0$ and $frac{x^2+8x+12}{3x+10}>0$
plotting
asked 11 hours ago
WomWom
162 bronze badges
New contributor
Wom is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
How to plot a function $f(x)=frac{3(4+x)}{3(2-x)-16}$ (say $x in [-15,15]$ ) with the condition that i want to give different color for each of the following cases
(i) when $frac{x+4}{3x+10}>0$ and $frac{x^2+8x+12}{3x+10}>0$
(ii) when $frac{x+4}{3x+10}>0$ and $frac{x^2+8x+12}{3x+10}<0$
(iii) when $frac{x+4}{3x+10}<0$ and $frac{x^2+8x+12}{3x+10}>0$
plotting
plotting
asked 11 hours ago
WomWom
162 bronze badges
New contributor
Wom is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 11 hours ago
WomWom
162 bronze badges
New contributor
Wom is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 11 hours ago
WomWom
162 bronze badges
New contributor
Wom is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 11 hours ago
WomWom
162 bronze badges
asked 11 hours ago
WomWom
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162 bronze badges
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Wom is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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4 Answers
4
active
oldest
votes
$begingroup$
Admittedly, not very different from the naïve answer already posted, just avoided repeating the function definitions.
f[x_] := (3 (x + 4))/(3 (2 - x) - 16);
g[x_] := (x + 4)/(3 x + 10);
h[x_] := (x^2 + 8 x + 12)/(3 x + 10);
a[x_] := (g[x] > 0 && h[x] > 0);
b[x_] := (g[x] > 0 && h[x] < 0);
c[x_] := (g[x] < 0 && h[x] > 0);
Plot[{f[x] && a[x], f[x] && b[x], f[x] && c[x]}, {x, -15, 15},
PlotRange -> {-3, 3}, PlotStyle -> Thickness[.01], Frame -> True,
Axes -> False]
answered 10 hours ago
mjwmjw
1,37510 bronze badges
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add a comment |
$begingroup$
You could use the option ColorFunction with ColorFunctionScaling->False. First your conditions:
cond1[x_] := (x+4)/(3x+10)>0 && (x^2+8x+12)/(3x+10)>0
cond2[x_] := (x+4)/(3x+10)>0 && (x^2+8x+12)/(3x+10)<0
cond3[x_] := (x+4)/(3x+10)<0 && (x^2+8x+12)/(3x+10)>0
And your function:
f[x_] := (3(4+x))/(3(2-x)-16)
Then:
Plot[f[x], {x, -15, 15},
PlotRange -> {All, {-3, 3}},
ColorFunctionScaling -> False,
ColorFunction -> Function @ Piecewise[
{
{ColorData[97][1], cond1[#]},
{ColorData[97][2], cond2[#]},
{ColorData[97][3], cond3[#]}
},
ColorData[97][4]
]
]
answered 10 hours ago
Carl WollCarl Woll
85.1k3 gold badges109 silver badges220 bronze badges
$endgroup$
add a comment |
$begingroup$
Naïve solution:
f1[x_] /; And[(x + 4)/(3 x + 10) > 0, (x^2 + 8 x + 12)/(3 x + 10) > 0] := (3 (4 + x))/(3 (2 - x) - 16)
f2[x_] /; And[(x + 4)/(3 x + 10) > 0, (x^2 + 8 x + 12)/(3 x + 10) < 0] := (3 (4 + x))/(3 (2 - x) - 16)
f3[x_] /; And[(x + 4)/(3 x + 10) < 0, (x^2 + 8 x + 12)/(3 x + 10) > 0] := (3 (4 + x))/(3 (2 - x) - 16)
Plot[{f1[x], f2[x], f3[x]}, {x, -15, 15}, PlotRange -> {All, {-10, 10}}]
answered 11 hours ago
AccidentalFourierTransformAccidentalFourierTransform
5,6051 gold badge11 silver badges42 bronze badges
$endgroup$
add a comment |
$begingroup$
An alternative way to specify a color function:
f[x_] := (3 (4 + x))/(3 (2 - x) - 16)
cf = ColorData[97]@(1 + {1, 2}.UnitStep[{(# + 4)/(3 # + 10), (#^2 + 8 # + 12)/(3 # + 10)}]) &;
Plot[f[x], {x, -15, 15},
PlotRange -> {All, {-3, 3}},
BaseStyle -> AbsoluteThickness[5],
ColorFunction -> cf,
ColorFunctionScaling -> False]
answered 1 hour ago
kglrkglr
200k10 gold badges229 silver badges455 bronze badges
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add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Admittedly, not very different from the naïve answer already posted, just avoided repeating the function definitions.
f[x_] := (3 (x + 4))/(3 (2 - x) - 16);
g[x_] := (x + 4)/(3 x + 10);
h[x_] := (x^2 + 8 x + 12)/(3 x + 10);
a[x_] := (g[x] > 0 && h[x] > 0);
b[x_] := (g[x] > 0 && h[x] < 0);
c[x_] := (g[x] < 0 && h[x] > 0);
Plot[{f[x] && a[x], f[x] && b[x], f[x] && c[x]}, {x, -15, 15},
PlotRange -> {-3, 3}, PlotStyle -> Thickness[.01], Frame -> True,
Axes -> False]
answered 10 hours ago
mjwmjw
1,37510 bronze badges
$endgroup$
add a comment |
$begingroup$
Admittedly, not very different from the naïve answer already posted, just avoided repeating the function definitions.
f[x_] := (3 (x + 4))/(3 (2 - x) - 16);
g[x_] := (x + 4)/(3 x + 10);
h[x_] := (x^2 + 8 x + 12)/(3 x + 10);
a[x_] := (g[x] > 0 && h[x] > 0);
b[x_] := (g[x] > 0 && h[x] < 0);
c[x_] := (g[x] < 0 && h[x] > 0);
Plot[{f[x] && a[x], f[x] && b[x], f[x] && c[x]}, {x, -15, 15},
PlotRange -> {-3, 3}, PlotStyle -> Thickness[.01], Frame -> True,
Axes -> False]
answered 10 hours ago
mjwmjw
1,37510 bronze badges
$endgroup$
add a comment |
$begingroup$
Admittedly, not very different from the naïve answer already posted, just avoided repeating the function definitions.
f[x_] := (3 (x + 4))/(3 (2 - x) - 16);
g[x_] := (x + 4)/(3 x + 10);
h[x_] := (x^2 + 8 x + 12)/(3 x + 10);
a[x_] := (g[x] > 0 && h[x] > 0);
b[x_] := (g[x] > 0 && h[x] < 0);
c[x_] := (g[x] < 0 && h[x] > 0);
Plot[{f[x] && a[x], f[x] && b[x], f[x] && c[x]}, {x, -15, 15},
PlotRange -> {-3, 3}, PlotStyle -> Thickness[.01], Frame -> True,
Axes -> False]
answered 10 hours ago
mjwmjw
1,37510 bronze badges
$endgroup$
Admittedly, not very different from the naïve answer already posted, just avoided repeating the function definitions.
f[x_] := (3 (x + 4))/(3 (2 - x) - 16);
g[x_] := (x + 4)/(3 x + 10);
h[x_] := (x^2 + 8 x + 12)/(3 x + 10);
a[x_] := (g[x] > 0 && h[x] > 0);
b[x_] := (g[x] > 0 && h[x] < 0);
c[x_] := (g[x] < 0 && h[x] > 0);
Plot[{f[x] && a[x], f[x] && b[x], f[x] && c[x]}, {x, -15, 15},
PlotRange -> {-3, 3}, PlotStyle -> Thickness[.01], Frame -> True,
Axes -> False]
answered 10 hours ago
mjwmjw
1,37510 bronze badges
answered 10 hours ago
mjwmjw
1,37510 bronze badges
answered 10 hours ago
mjwmjw
1,37510 bronze badges
answered 10 hours ago
mjwmjw
1,37510 bronze badges
1,37510 bronze badges
add a comment |
add a comment |
$begingroup$
You could use the option ColorFunction with ColorFunctionScaling->False. First your conditions:
cond1[x_] := (x+4)/(3x+10)>0 && (x^2+8x+12)/(3x+10)>0
cond2[x_] := (x+4)/(3x+10)>0 && (x^2+8x+12)/(3x+10)<0
cond3[x_] := (x+4)/(3x+10)<0 && (x^2+8x+12)/(3x+10)>0
And your function:
f[x_] := (3(4+x))/(3(2-x)-16)
Then:
Plot[f[x], {x, -15, 15},
PlotRange -> {All, {-3, 3}},
ColorFunctionScaling -> False,
ColorFunction -> Function @ Piecewise[
{
{ColorData[97][1], cond1[#]},
{ColorData[97][2], cond2[#]},
{ColorData[97][3], cond3[#]}
},
ColorData[97][4]
]
]
answered 10 hours ago
Carl WollCarl Woll
85.1k3 gold badges109 silver badges220 bronze badges
$endgroup$
add a comment |
$begingroup$
You could use the option ColorFunction with ColorFunctionScaling->False. First your conditions:
cond1[x_] := (x+4)/(3x+10)>0 && (x^2+8x+12)/(3x+10)>0
cond2[x_] := (x+4)/(3x+10)>0 && (x^2+8x+12)/(3x+10)<0
cond3[x_] := (x+4)/(3x+10)<0 && (x^2+8x+12)/(3x+10)>0
And your function:
f[x_] := (3(4+x))/(3(2-x)-16)
Then:
Plot[f[x], {x, -15, 15},
PlotRange -> {All, {-3, 3}},
ColorFunctionScaling -> False,
ColorFunction -> Function @ Piecewise[
{
{ColorData[97][1], cond1[#]},
{ColorData[97][2], cond2[#]},
{ColorData[97][3], cond3[#]}
},
ColorData[97][4]
]
]
answered 10 hours ago
Carl WollCarl Woll
85.1k3 gold badges109 silver badges220 bronze badges
$endgroup$
add a comment |
$begingroup$
You could use the option ColorFunction with ColorFunctionScaling->False. First your conditions:
cond1[x_] := (x+4)/(3x+10)>0 && (x^2+8x+12)/(3x+10)>0
cond2[x_] := (x+4)/(3x+10)>0 && (x^2+8x+12)/(3x+10)<0
cond3[x_] := (x+4)/(3x+10)<0 && (x^2+8x+12)/(3x+10)>0
And your function:
f[x_] := (3(4+x))/(3(2-x)-16)
Then:
Plot[f[x], {x, -15, 15},
PlotRange -> {All, {-3, 3}},
ColorFunctionScaling -> False,
ColorFunction -> Function @ Piecewise[
{
{ColorData[97][1], cond1[#]},
{ColorData[97][2], cond2[#]},
{ColorData[97][3], cond3[#]}
},
ColorData[97][4]
]
]
answered 10 hours ago
Carl WollCarl Woll
85.1k3 gold badges109 silver badges220 bronze badges
$endgroup$
You could use the option ColorFunction with ColorFunctionScaling->False. First your conditions:
cond1[x_] := (x+4)/(3x+10)>0 && (x^2+8x+12)/(3x+10)>0
cond2[x_] := (x+4)/(3x+10)>0 && (x^2+8x+12)/(3x+10)<0
cond3[x_] := (x+4)/(3x+10)<0 && (x^2+8x+12)/(3x+10)>0
And your function:
f[x_] := (3(4+x))/(3(2-x)-16)
Then:
Plot[f[x], {x, -15, 15},
PlotRange -> {All, {-3, 3}},
ColorFunctionScaling -> False,
ColorFunction -> Function @ Piecewise[
{
{ColorData[97][1], cond1[#]},
{ColorData[97][2], cond2[#]},
{ColorData[97][3], cond3[#]}
},
ColorData[97][4]
]
]
answered 10 hours ago
Carl WollCarl Woll
85.1k3 gold badges109 silver badges220 bronze badges
answered 10 hours ago
Carl WollCarl Woll
85.1k3 gold badges109 silver badges220 bronze badges
answered 10 hours ago
Carl WollCarl Woll
85.1k3 gold badges109 silver badges220 bronze badges
answered 10 hours ago
Carl WollCarl Woll
85.1k3 gold badges109 silver badges220 bronze badges
85.1k3 gold badges109 silver badges220 bronze badges
add a comment |
add a comment |
$begingroup$
Naïve solution:
f1[x_] /; And[(x + 4)/(3 x + 10) > 0, (x^2 + 8 x + 12)/(3 x + 10) > 0] := (3 (4 + x))/(3 (2 - x) - 16)
f2[x_] /; And[(x + 4)/(3 x + 10) > 0, (x^2 + 8 x + 12)/(3 x + 10) < 0] := (3 (4 + x))/(3 (2 - x) - 16)
f3[x_] /; And[(x + 4)/(3 x + 10) < 0, (x^2 + 8 x + 12)/(3 x + 10) > 0] := (3 (4 + x))/(3 (2 - x) - 16)
Plot[{f1[x], f2[x], f3[x]}, {x, -15, 15}, PlotRange -> {All, {-10, 10}}]
answered 11 hours ago
AccidentalFourierTransformAccidentalFourierTransform
5,6051 gold badge11 silver badges42 bronze badges
$endgroup$
add a comment |
$begingroup$
Naïve solution:
f1[x_] /; And[(x + 4)/(3 x + 10) > 0, (x^2 + 8 x + 12)/(3 x + 10) > 0] := (3 (4 + x))/(3 (2 - x) - 16)
f2[x_] /; And[(x + 4)/(3 x + 10) > 0, (x^2 + 8 x + 12)/(3 x + 10) < 0] := (3 (4 + x))/(3 (2 - x) - 16)
f3[x_] /; And[(x + 4)/(3 x + 10) < 0, (x^2 + 8 x + 12)/(3 x + 10) > 0] := (3 (4 + x))/(3 (2 - x) - 16)
Plot[{f1[x], f2[x], f3[x]}, {x, -15, 15}, PlotRange -> {All, {-10, 10}}]
answered 11 hours ago
AccidentalFourierTransformAccidentalFourierTransform
5,6051 gold badge11 silver badges42 bronze badges
$endgroup$
add a comment |
$begingroup$
Naïve solution:
f1[x_] /; And[(x + 4)/(3 x + 10) > 0, (x^2 + 8 x + 12)/(3 x + 10) > 0] := (3 (4 + x))/(3 (2 - x) - 16)
f2[x_] /; And[(x + 4)/(3 x + 10) > 0, (x^2 + 8 x + 12)/(3 x + 10) < 0] := (3 (4 + x))/(3 (2 - x) - 16)
f3[x_] /; And[(x + 4)/(3 x + 10) < 0, (x^2 + 8 x + 12)/(3 x + 10) > 0] := (3 (4 + x))/(3 (2 - x) - 16)
Plot[{f1[x], f2[x], f3[x]}, {x, -15, 15}, PlotRange -> {All, {-10, 10}}]
answered 11 hours ago
AccidentalFourierTransformAccidentalFourierTransform
5,6051 gold badge11 silver badges42 bronze badges
$endgroup$
Naïve solution:
f1[x_] /; And[(x + 4)/(3 x + 10) > 0, (x^2 + 8 x + 12)/(3 x + 10) > 0] := (3 (4 + x))/(3 (2 - x) - 16)
f2[x_] /; And[(x + 4)/(3 x + 10) > 0, (x^2 + 8 x + 12)/(3 x + 10) < 0] := (3 (4 + x))/(3 (2 - x) - 16)
f3[x_] /; And[(x + 4)/(3 x + 10) < 0, (x^2 + 8 x + 12)/(3 x + 10) > 0] := (3 (4 + x))/(3 (2 - x) - 16)
Plot[{f1[x], f2[x], f3[x]}, {x, -15, 15}, PlotRange -> {All, {-10, 10}}]
answered 11 hours ago
AccidentalFourierTransformAccidentalFourierTransform
5,6051 gold badge11 silver badges42 bronze badges
answered 11 hours ago
AccidentalFourierTransformAccidentalFourierTransform
5,6051 gold badge11 silver badges42 bronze badges
answered 11 hours ago
AccidentalFourierTransformAccidentalFourierTransform
5,6051 gold badge11 silver badges42 bronze badges
answered 11 hours ago
AccidentalFourierTransformAccidentalFourierTransform
5,6051 gold badge11 silver badges42 bronze badges
5,6051 gold badge11 silver badges42 bronze badges
add a comment |
add a comment |
$begingroup$
An alternative way to specify a color function:
f[x_] := (3 (4 + x))/(3 (2 - x) - 16)
cf = ColorData[97]@(1 + {1, 2}.UnitStep[{(# + 4)/(3 # + 10), (#^2 + 8 # + 12)/(3 # + 10)}]) &;
Plot[f[x], {x, -15, 15},
PlotRange -> {All, {-3, 3}},
BaseStyle -> AbsoluteThickness[5],
ColorFunction -> cf,
ColorFunctionScaling -> False]
answered 1 hour ago
kglrkglr
200k10 gold badges229 silver badges455 bronze badges
$endgroup$
add a comment |
$begingroup$
An alternative way to specify a color function:
f[x_] := (3 (4 + x))/(3 (2 - x) - 16)
cf = ColorData[97]@(1 + {1, 2}.UnitStep[{(# + 4)/(3 # + 10), (#^2 + 8 # + 12)/(3 # + 10)}]) &;
Plot[f[x], {x, -15, 15},
PlotRange -> {All, {-3, 3}},
BaseStyle -> AbsoluteThickness[5],
ColorFunction -> cf,
ColorFunctionScaling -> False]
answered 1 hour ago
kglrkglr
200k10 gold badges229 silver badges455 bronze badges
$endgroup$
add a comment |
$begingroup$
An alternative way to specify a color function:
f[x_] := (3 (4 + x))/(3 (2 - x) - 16)
cf = ColorData[97]@(1 + {1, 2}.UnitStep[{(# + 4)/(3 # + 10), (#^2 + 8 # + 12)/(3 # + 10)}]) &;
Plot[f[x], {x, -15, 15},
PlotRange -> {All, {-3, 3}},
BaseStyle -> AbsoluteThickness[5],
ColorFunction -> cf,
ColorFunctionScaling -> False]
answered 1 hour ago
kglrkglr
200k10 gold badges229 silver badges455 bronze badges
$endgroup$
An alternative way to specify a color function:
f[x_] := (3 (4 + x))/(3 (2 - x) - 16)
cf = ColorData[97]@(1 + {1, 2}.UnitStep[{(# + 4)/(3 # + 10), (#^2 + 8 # + 12)/(3 # + 10)}]) &;
Plot[f[x], {x, -15, 15},
PlotRange -> {All, {-3, 3}},
BaseStyle -> AbsoluteThickness[5],
ColorFunction -> cf,
ColorFunctionScaling -> False]
answered 1 hour ago
kglrkglr
200k10 gold badges229 silver badges455 bronze badges
answered 1 hour ago
kglrkglr
200k10 gold badges229 silver badges455 bronze badges
answered 1 hour ago
kglrkglr
200k10 gold badges229 silver badges455 bronze badges
answered 1 hour ago
kglrkglr
200k10 gold badges229 silver badges455 bronze badges
200k10 gold badges229 silver badges455 bronze badges
add a comment |
add a comment |
Wom is a new contributor. Be nice, and check out our Code of Conduct.
Wom is a new contributor. Be nice, and check out our Code of Conduct.
Wom is a new contributor. Be nice, and check out our Code of Conduct.
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