Thread: NEED HELP NOW, DUE TODAY

It helps to post the full stack trace - which tells what line the exception is being thrown, and should point you in the right direction. I'd suggest reading this link: Arrays (The Java™ Tutorials > Learning the Java Language > Language Basics)
add some System.out.println statements to debug and print out the values of the indexes you are trying to access, and look at those loops carefully.

Also, suggested reading: Ease Up at JavaRanch

Did you read read my last post? None of what you just said suggests otherwise, and stating 'its not working' is fairly meaningless - what's not working? Whats the behavior? Stack trace? Output (expected and received)? Suggested reading: How To Ask Questions The Smart Way

I suggest reading - Debugging with System.out.println - Java Programming Forums

Your inner for loop should probably not be reseting "i", you should probably choose a different variable name such as "j".

double lastY = 0;
 
for(int i=0;i<numPoints;i++){
   System.out.print(i+"\t");
   x[0]=minX;		
   x[numPoints-1]= maxX;
 
 
   x[i]=x[0]+ i*((maxX-minX)/(numPoints-1));
   y[i]=(((x[i])*(x[i])*(x[i]))-(2*((x[i])*(x[i])))-(5*x[i])+4);
 
   System.out.print(i);
   System.out.print("\t");
   System.out.print(x[i]);
   System.out.print("\t");
   System.out.println(y[i]);
 
   //i gets reset here
   for(i=0;i<numPoints;i++){
      if (y[i]>=0 && y[i+1]<0){
                   ...
 
   }//for
}//for

Also "y[i+1]", will be null since it hasn't yet been populated with a value. The OutOfBounds exception is occuring when you reach the next to last interation in "y[i+1]". If numPoints = 5, then plugging the values in during the next to last iteration we see:

for (i=0; i<=5;i++) { 
    if (y[5] >=0 && y[5+1]<0) ...

y[5+1]" also referred to as "y[6]" doesn't exist. Only y[0], y[1], y[2], y[3], y[4], y[5] exist. You could probably eliminate the inner for loop and just keep track of the last y[i] value and compare it to the current y[i] value. Try 1 of the following two code examples:

Since the function appears linear (and a linear function will only cross 0 once) you could do the following:

Scanner input= new Scanner(System.in);
 
double minX =0.0;
double maxX=0.0;
int numPoints=0;
 
double lastX=0.0;
double lastY=0.0;
 
System.out.print("Enter number of points: ");
numPoints=input.nextInt();
System.out.print("Enter min x: ");
minX=input.nextDouble();
System.out.print("Enter max x: ");
maxX=input.nextDouble();
 
System.out.println("Numpoints: " + numPoints);
double [] x= new double[numPoints];
double [] y= new double[numPoints];
int change=0;
System.out.println("i"+"\t"+"x[i]"+"\t"+"y[i]");
 
for(int i=0;i<numPoints;i++){
   System.out.print(i+"\t");
   x[0]=minX;		
   x[numPoints-1]= maxX;
 
 
   x[i]=x[0]+ i*((maxX-minX)/(numPoints-1));
   y[i]=(((x[i])*(x[i])*(x[i]))-(2*((x[i])*(x[i])))-(5*x[i])+4);
 
   System.out.print(i);
   System.out.print("\t");
   System.out.print(x[i]);
   System.out.print("\t");
   System.out.println(y[i]);
 
  //go to else if for first iteration
   if (i>0){
       if(lastY >=0 && y[i]<=0){
          //System.out.println("Inside inner if");
          change=i;
       }//if
      else if(lastY<= 0 && y[i]>=0){
         //System.out.println("Inside inner else if");
         change=i;
     }//else if
  }//if
 
  //in case the first number starts on 0
  else if(y[i]==0){                           
    //System.out.println("Inside inner if");
   change=i;
  }//else if
 
  //this will print out the value of change during every iteration,
  //probably not what you want
  //if you only want to print it out once, move it to the outside of the "for" loop
  //since the function appears linear it will only cross the "y" axis 1 time,
  //then you don't need to use an array, you could just use  "double change;"
  //and eliminate "numChange"
  System.out.print("\t"+change);
  System.out.println();
 
  //set lastY=y[i]
   lastY = y[i];
 
}//for

Otherwise, you can do the following for both linear and non-linear functions (which may cross the y axis more than once):

Scanner input= new Scanner(System.in);
 
double minX =0.0;
double maxX=0.0;
int numPoints=0;
int numChange=0;
 
double lastX=0.0;
double lastY=0.0;
 
System.out.print("Enter number of points: ");
numPoints=input.nextInt();
System.out.print("Enter min x: ");
minX=input.nextDouble();
System.out.print("Enter max x: ");
maxX=input.nextDouble();
 
System.out.println("Numpoints: " + numPoints);
double [] x= new double[numPoints];
double [] y= new double[numPoints];
int [] change= new int[numPoints];
 
System.out.println("i"+"\t"+"x[i]"+"\t"+"y[i]");
 
 
System.out.println("i"+"\t"+"x[i]"+"\t"+"y[i]");
 
for(int i=0;i<numPoints;i++){
   System.out.print(i+"\t");
   x[0]=minX;		
   x[numPoints-1]= maxX;
 
 
   x[i]=x[0]+ i*((maxX-minX)/(numPoints-1));
   y[i]=(((x[i])*(x[i])*(x[i]))-(2*((x[i])*(x[i])))-(5*x[i])+4);
 
   System.out.print(i);
   System.out.print("\t");
   System.out.print(x[i]);
   System.out.print("\t");
   System.out.println(y[i]);
 
  //go to else if for first iteration
   if (i>0){
       if(lastY >=0 && y[i]<=0){
          //System.out.println("Inside inner if");
          change[numChange]=i;
          numChange++;
       }//if
      else if(lastY<= 0 && y[i]>=0){
         //System.out.println("Inside inner else if");
         change[numChange]=i;
 
        //added numChange++
        numChange++;
     }//else if
  }//if
 
  //in case the first number starts on 0
  else if(y[i]==0){                           
    //System.out.println("Inside inner if");
   change[numChange]=i;
   numChange++;
  }//else if
 
  //we will print out the crossing points outside of this for loop
  // System.out.print("\t"+change[0]);
  //System.out.print("Outside inner for \t"+change[numChange]);
  System.out.println();
 
  //set lastY=y[i]
   lastY = y[i];
 
}//for
 
System.out.print("The function crosses the y-axis at the following i value(s):");
System.out.print("\t");
for (int j=0; j<numChange;j++){
 
   if (j>0){
      System.out.print(", " + change[j]);
   }//if
   else{
      System.out.print(change[j]);
   }//else
}//for
 
System.out.println();                
System.out.println();