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Plotting with different color for a single curve
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
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
How to plot a function $f(x)=frac3(4+x)3(2-x)-16$ (say $x in [-15,15]$ ) with the condition that i want to give different color for each of the following cases
(i) when $fracx+43x+10>0$ and $fracx^2+8x+123x+10>0$
(ii) when $fracx+43x+10>0$ and $fracx^2+8x+123x+10<0$
(iii) when $fracx+43x+10<0$ and $fracx^2+8x+123x+10>0$
plotting
asked 8 hours ago
WomWom
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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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add a comment |
$begingroup$
How to plot a function $f(x)=frac3(4+x)3(2-x)-16$ (say $x in [-15,15]$ ) with the condition that i want to give different color for each of the following cases
(i) when $fracx+43x+10>0$ and $fracx^2+8x+123x+10>0$
(ii) when $fracx+43x+10>0$ and $fracx^2+8x+123x+10<0$
(iii) when $fracx+43x+10<0$ and $fracx^2+8x+123x+10>0$
plotting
asked 8 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 $f(x)=frac3(4+x)3(2-x)-16$ (say $x in [-15,15]$ ) with the condition that i want to give different color for each of the following cases
(i) when $fracx+43x+10>0$ and $fracx^2+8x+123x+10>0$
(ii) when $fracx+43x+10>0$ and $fracx^2+8x+123x+10<0$
(iii) when $fracx+43x+10<0$ and $fracx^2+8x+123x+10>0$
plotting
asked 8 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 $f(x)=frac3(4+x)3(2-x)-16$ (say $x in [-15,15]$ ) with the condition that i want to give different color for each of the following cases
(i) when $fracx+43x+10>0$ and $fracx^2+8x+123x+10>0$
(ii) when $fracx+43x+10>0$ and $fracx^2+8x+123x+10<0$
(iii) when $fracx+43x+10<0$ and $fracx^2+8x+123x+10>0$
plotting
plotting
asked 8 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 8 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 8 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 8 hours ago
WomWom
162 bronze badges
asked 8 hours ago
WomWom
162 bronze badges
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 |
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3 Answers
3
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 8 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 7 hours ago
Carl WollCarl Woll
85k3 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 8 hours ago
AccidentalFourierTransformAccidentalFourierTransform
5,6051 gold badge11 silver badges42 bronze badges
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3 Answers
3
active
oldest
votes
3 Answers
3
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 8 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 8 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 8 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 8 hours ago
mjwmjw
1,37510 bronze badges
answered 8 hours ago
mjwmjw
1,37510 bronze badges
answered 8 hours ago
mjwmjw
1,37510 bronze badges
answered 8 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 7 hours ago
Carl WollCarl Woll
85k3 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 7 hours ago
Carl WollCarl Woll
85k3 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 7 hours ago
Carl WollCarl Woll
85k3 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 7 hours ago
Carl WollCarl Woll
85k3 gold badges109 silver badges220 bronze badges
answered 7 hours ago
Carl WollCarl Woll
85k3 gold badges109 silver badges220 bronze badges
answered 7 hours ago
Carl WollCarl Woll
85k3 gold badges109 silver badges220 bronze badges
answered 7 hours ago
Carl WollCarl Woll
85k3 gold badges109 silver badges220 bronze badges
85k3 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 8 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 8 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 8 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 8 hours ago
AccidentalFourierTransformAccidentalFourierTransform
5,6051 gold badge11 silver badges42 bronze badges
answered 8 hours ago
AccidentalFourierTransformAccidentalFourierTransform
5,6051 gold badge11 silver badges42 bronze badges
answered 8 hours ago
AccidentalFourierTransformAccidentalFourierTransform
5,6051 gold badge11 silver badges42 bronze badges
answered 8 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 |
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.
Wom is a new contributor. Be nice, and check out our Code of Conduct.
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