d3 simple line graph, showing 'step-after' interpolation

This example shows a problem with the 'step-after' interpolation. It does not draw the 'last' step -- i.e., the step for the last data point. starting from bostock's example. then using different interpolation. and without an url for data.

by Abhishek Hingu

CSS

body {
  font: 10px sans-serif;
}

.axis path,
.axis line {
  fill: none;
  stroke: #000;
  shape-rendering: crispEdges;
}

.x.axis path {
  display: none;
}

.line {
  fill: none;
  stroke: steelblue;
  stroke-width: 1.5px;
}

JavaScript

var myData = "date	avg	min	max\n\
01-Jul-17	344251.9118	10000.0000	1847283.0000\n\
01-Oct-17	244474.6831	10000.0000	1443736.0000\n\
01-Jan-18	334264.7699	33052.0000	1443736.0000\n\
01-Apr-18	348314.0586	15380.0000	2410046.0000\n\
01-Jul-18	202586.6371	14005.0000	2322936.0000\n\
01-Oct-18	140596.8278	14005.0000	2322936.0000\n\
";
var quarter = function (date, i) {
                    if (i >= 0) {
                        var date2 = new Date();
                        date2.setMonth(date.getMonth() - 1);
                        q = Math.ceil((date2.getMonth()) / 3);
                        return "Q" + q;
                    }
                };
var margin = {top: 20, right: 20, bottom: 30, left: 60},
    width = 400 - margin.left - margin.right,
    height = 400 - margin.top - margin.bottom;

var parseDate = d3.time.format("%d-%b-%y").parse;

var x = d3.time.scale()
    .rangeRound([0, width],0.8);
    //.padding(8.8);

var x2 = d3.time.scale()
    .rangeRound([0, width],0.8);
    //.padding(0.8);

var y = d3.scale.linear()
    .range([height, 0]);

var xAxis = d3.svg.axis()
    .scale(x)
    .ticks(d3.time.months, 3)
    .tickSize(5, 0)
    .tickFormat(quarter)
    .orient("bottom");

var xAxis2 = d3.svg.axis()
    .scale(x)
    .ticks(d3.time.year,1)
    .tickFormat(d3.time.format("%Y"))
    .orient("top");

var yAxis = d3.svg.axis()
    .scale(y)
    .orient("left");

var line = d3.svg.line()
    .x(function(d) { return x(d.date); })
    .y(function(d) { return y(d.avg); });

// try this to see interpolation issues.
// Note that the data is "backwards"-going in time.
// Note that with step-after, the last data point (i.e. earliest) shows no "step". Why?
line.interpolate('spline');
// what if we use the defined method of line?
//line.defined(function (d) { return d.close; });
//line.interpolate('step-before');

var svg = d3.select("body").append("svg")
    .attr("width", width + margin.left + margin.right)
    .attr("height", height + margin.top +...