Hello
If i had the following
clear
clc
x=0:10
y=-2*x+6
y1=3*x
plot(x,y)
hold on
plot(x,y1)
how would i just plot the graph with the limits being the x and y intercept

7 comentarios

Geoff Hayes
Geoff Hayes el 25 de Abr. de 2020
Elliot - plaese clarify what you mean by with the limits being the x and y intercept. What do you want the minimum and maximum limits for the x and y axes?
Elliott Cameron
Elliott Cameron el 25 de Abr. de 2020
i want the plot to show no lower than where y1 intercepts the x axis and no higher than where y1 intercepts the y axis sorry i realise that wasnt very clear
Star Strider
Star Strider el 25 de Abr. de 2020
Note that ‘y1’ intercepts both the x and y axes at (0,0).
Elliott Cameron
Elliott Cameron el 25 de Abr. de 2020
sorry y and y1 are the wrong way round
Tommy
Tommy el 25 de Abr. de 2020
You can solve for the intercepts using the equation for y1 and update the axes limits accordingly:
clear
clc
x=0:10;
y1=-2*x+6;
y=3*x;
plot(x,y)
hold on
plot(x,y1)
yint = 6; % -2*0 + 6 => 6
xint = 3; % (0 - 6)/(-2) => 3
ax = gca;
ax.XLim(2) = xint;
ax.YLim(2) = yint;
dpb
dpb el 25 de Abr. de 2020
@Tommy...you have/are developing my habit of putting Answers in Comments... :)
Tommy
Tommy el 25 de Abr. de 2020
Ah! Yes I agree... I do often question if I really understand what's being asked, and whether my "answer" is a full answer, but too often I default to the comment section

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 Respuesta aceptada

Tommy
Tommy el 25 de Abr. de 2020

0 votos

You can solve for the intercepts using the equation for y1 and update the axes limits accordingly:
clear
clc
x=0:10;
y1=-2*x+6;
y=3*x;
plot(x,y)
hold on
plot(x,y1)
yint = 6; % -2*0 + 6 => 6
xint = 3; % (0 - 6)/(-2) => 3
ax = gca;
ax.XLim(2) = xint;
ax.YLim(2) = yint;

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