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 件のコメント

Geoff Hayes
Geoff Hayes 2020 年 4 月 25 日
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 2020 年 4 月 25 日
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 2020 年 4 月 25 日
Note that ‘y1’ intercepts both the x and y axes at (0,0).
Elliott Cameron
Elliott Cameron 2020 年 4 月 25 日
sorry y and y1 are the wrong way round
Tommy
Tommy 2020 年 4 月 25 日
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 2020 年 4 月 25 日
@Tommy...you have/are developing my habit of putting Answers in Comments... :)
Tommy
Tommy 2020 年 4 月 25 日
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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 採用された回答

Tommy
Tommy 2020 年 4 月 25 日

0 投票

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;

1 件のコメント

Elliott Cameron
Elliott Cameron 2020 年 4 月 26 日
brilliant thats perfect, thank you

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