Create a vector 't' and vector 'f' where 'f' is a mathematical function to plot.

Create a vector 't' which consists of 100 numbers uniformly spread between 0.01 and 1. Also create a vector 'f' which includes the corresponding 100 values of the mathematical function t^(2) + 3t – 15. Plot 'f' versus 't'. This is what I have so far:
% Creating a vector 't' which consists of 100 numbers uniformly spread between 0.01 and 1
t=0.01+(1-0.01).*rand(1,100);
% Creating a vector 'f' which includes the corresponding 100 values of the mathematical function t^(2)+3t-15
f=t.^2+3*t-15
%add code
% Plotting 'f' versus 't'
figure(1),plot(f,t)

1 Comment

You have made the same mistake again. You need to plot(t,f) instead of plot(f,t).

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 Accepted Answer

How about this:
t = 0:.01:1;
f = t.^2+3.*t-15;
figure;
plot(t,f,'b');
grid;

7 Comments

I'm really confused on this part of the question: "Also create a vector 'f' which includes the corresponding 100 values of the mathematical function t^(2) + 3t – 15."
First of all, the question calls for a series of values that are "uniformly spaced". This means that they have the same spacing from one to the next. It has nothing to do with randomness. so the line:
t=0: 0.01 : 1
creates a vector t starting at 0, with increments of 0.01 and ending at 1, so you have a vector from 0 to 1 in 100 equal increments.
the statement "create a vector 'f' which includes the corresponding 100 values of the mathematical function t^(2) + 3t – 15" means to create a vector f, using the 100 values of t, according to the function t^(2) + 3t - 15. This is accomplished in the line of code:
f=t.^2+3.*t-15
note that the syntax ".^" and ".*" (using the dot) means to perform this operation on each element of the vector.
The instruction says you want a uniform spread from .01 to 1, so 1 is the highest value. The expression 1: .01 : 100 says start at one and go to 100 in steps of .01. This is not what you want. You want to start at 0 and go to 1 in steps of 0.01, this is expressed as:
0: 0.01 : 1

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