How do I display only certain numbers from a plot?

This is for a begineer Matlab class assignment so bear with me here. The assignment is to plot a graph of a first order equation, then based on the persison set by the user, it needs to tell you which x values y is equal to zero at. The problem I am having is that I get an error with my if statement and I dont know how to fix it. All I need it to do is check if there are y values between the set persison and then display the x value at that point.

clc 
clear
a=input('What is the value of a? ');
b=input('What is the value of b? ');
r=input('What is the range? ');
s=input('What is the step-size for range of x ');
p=input('What is the persison of zero? ');
x=-r:s:r;
y=(a*x)+b;
plot(x,y,'r*')

if -p<y && y<p
    fprintf('Y=0 at x=%f',y)
    
else
    disp('Error. Enter a new range or persison')
    
end

[Image Analyst]

What does "persison" mean?

[Benjamin]

Sorry *precision

I typed this project up quick and I'm not great at spelling sorry

[Cris LaPierre]

What error message are you getting? Please share the entire message (all the red text).

[Benjamin]

This is what happens when I run my code.

What is the value of a? 1

What is the value of b? 0

What is the range? 10

What is the step-size for range of x 0.1

What is the persison of zero? 0.5

Operands to the logical AND (&&) and OR (||) operators must be

convertible to logical scalar values. Use the ANY or ALL

functions to reduce operands to logical scalar values.

Error in Project2_Practice (line 11)

if -p<y && y<p

[Cris LaPierre]

a=1;
b=0;
r=10;
s=0.1;
p=0.5;
x=-r:s:r;
y=(a*x)+b;
plot(x,y,'r*')

if -p<y && y<p
    fprintf('Y=0 at x=%f',y)  
else
    disp('Error. Enter a new range or persison') 
end

Operands to the logical AND (&&) and OR (||) operators must be convertible to logical scalar values. Use the ANY or ALL functions to reduce operands to logical scalar values.

Answer 1

[Dyuman Joshi]

"&&" can only be used when then result of each logical expression is a scalar (as the error states as well), in all other cases you need to use a single AND "&" operator.

It's not clear if you want to compare each values of y seperately or all together? In case you want to compare all values, you need to use all() (again stated in the error). Here you can use &&, as the output of these expressions will be a scalar

if all(-p<y) && all(y<p)

Answer 2

[Walter Roberson]

Your y is a vector. You want to find locations in that vector that are between -p and +p .

mask = -p < y & y < p;

Now you have a few different possibilities to consider:

• all entries in mask might be false -- ~any(mask) . In this case either there were no roots or the step size is too large or the precision is too small
• exactly one entry in mask might be true -- nnz(mask) == 1. In this case you have successfully found one root at x(mask)
• there might be one grouping in mask where several entries in a row are true. This can happen if the precision is too large and the function is not changing faster than the precision would suggest
• there might be several distinct places in mask where exactly one entry is true. This would typically correspond to multiple roots of the equation
• you might have false roots. For example x.^2 + 1e-7 has no true roots over reals, but if your precision were 1e-6 then you would think that you found a root,