Solving MatLab "for" Loops With Vectors

[sirclash]

MatLab "for" loops

Homework Statement

Write an m-file to evaluate y(x)= x^2 - 3x +2 for all values of x between -1 and 3, in steps of .1 . Do this twice, once with a for loop and once with vectors. Plot the resulting functions.

Homework Equations

The Attempt at a Solution

count=0;
for x= -1:.1:3
count = count+1;
y(count)= x.^2 -3*x +2;

end
plot(x,y)

z=-1:.1:3;
u= z.^2 -3*z + 2;
plot(z,u)

My vector is correct, however i have no clue why my "for" loop won't show the correct graph.

 

[tedbradly]

the problem is that x is a scalar, so you're plotting (3, {y_1,y_2,...y_n}) which makes no sense.

Instead, try this: plot(-1:.1:3,y)

also, inside your for loop, you don't need ".^". You can just use "^" since x is scalar there.

 

[lorenb]

well, when you use the "for" part and say for x = -1:.1:3 . MATLAB only uses this for the loop. it does not make a vector out of it. well, i think it just keeps changing the value of x so that the value of x is the value for your last loop
try putting x = -1:.1:3 after your loop and keeping the plot(x,y)

 

[viscousflow]

sirclash, here check this code. It can clean up your code and make it a lot more efficient

%plot x, y
x= -1:.1:3;

for i=1:length(x)
y(i)= x(i)^2 -3*x(i) +2;
end

plot(x,y)

%plot z, u
z=-1:.1:3;
u= z.^2 -3*z + 2; 
plot(z,u)

 

[DeeNos]

Is there something wrong with my code or is there a better way to solve this problem?
Your code for the "for" loop looks correct, so there may be an issue with how you are plotting the results. Make sure you are plotting the correct variables (x and y) and that you are using the "hold on" command to plot both the "for" loop and vector results on the same graph. Also, double check that you are using the correct syntax for the plot command. A possible alternative solution could be to use the "linspace" function to create a vector of evenly spaced values between -1 and 3, and then use element-wise operations to evaluate the function for each value in the vector. This may be a more efficient way to solve the problem.