var arr = [3, 6, 2, 56, 32, 5, 89, 32];
var largest = arr[0];

for (var i = 0; i < arr.length; i++) {
  if (arr[i] > largest ) {
    largest = arr[i];
  }
}
console.log(largest);

• You need to define i or else it become a global variable.
• Don't redefine largest in the loop.
• Since you're looping through the array, use i < array.length instead of i <= largest.
• Since you're comparing each of the items in the array to largest, use if(largest < array[i]) instead of if(array > largest)
• You should set largest equal to the first element in the array because what if all the numbers are negative?
• array is a bad variable name because it's too similar to Array (the array constructor). Try arr instead.

One liner:

var largest = Math.max.apply(0, array);

More info here: Javascript max() function for 3 numbers

Some ES6 magic for you, using the spread syntax:

function biggestNumberInArray(arr) {
  const max = Math.max(...arr);
  return max;
}

Actually, a few people have answered this question in a more detailed fashion than I do, but I would like you to read this if you are curious about the performance between the various ways of getting the largest number in an array.

Resig to the rescue:

Array.max = function( array ){
    return Math.max.apply( Math, array );
};

Warning: since the maximum number of arguments is as low as 65535 on some VMs, use a for loop if you're not certain the array is that small.