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 commentaires

Geoff Hayes
Geoff Hayes le 25 Avr 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 le 25 Avr 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 le 25 Avr 2020
Note that ‘y1’ intercepts both the x and y axes at (0,0).
Elliott Cameron
Elliott Cameron le 25 Avr 2020
sorry y and y1 are the wrong way round
Tommy
Tommy le 25 Avr 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 le 25 Avr 2020
@Tommy...you have/are developing my habit of putting Answers in Comments... :)
Tommy
Tommy le 25 Avr 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

Connectez-vous pour commenter.

 Réponse acceptée

Tommy
Tommy le 25 Avr 2020

0 votes

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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