How to Use CSS to Make a Background Image "Fade" or Gradient the Bottom Portion to Transparent So That a Background Color Shows

Is it possible to use css to make a background image fade or gradient the bottom portion to transparent so that a background color shows?

It is possible - in CSS3 you can set multiple values for background

body {    background: #837960 url("https://i.stack.imgur.com/MUsp6.jpg") 0 0 no-repeat;
    background: -moz-linear-gradient(top, rgba(255,255,255,0) 0%, rgba(130,91,0,1) 100%);   /* FF3.6+ */    background: -webkit-gradient(linear, left top, left bottom, color-stop(0%,rgba(255,255,255,0)), color-stop(100%,rgba(130,91,0,1))); /* Chrome,Safari4+ */    background: -webkit-linear-gradient(top, rgba(255,255,255,0) 0%,rgba(130,91,0,1) 100%); /* Chrome10+,Safari5.1+ */    background: -o-linear-gradient(top, rgba(255,255,255,0) 0%,rgba(130,91,0,1) 100%); /* Opera 11.10+ */    background: -ms-linear-gradient(top, rgba(255,255,255,0) 0%,rgba(130,91,0,1) 100%); /* IE10+ */    background: linear-gradient(to bottom, rgba(255,255,255,0) 0%,rgba(130,91,0,1) 100%); /* W3C */    filter: progid:DXImageTransform.Microsoft.gradient( startColorstr='#00ffffff', endColorstr='#825b00',GradientType=0 ); /* IE6-9 */}

How to fade a background image to transparent, in a gradient fashion?

This is not possible with CSS as it stands as background images cannot be affected by opacity.

However, you could overlay the bg-image background with a background gradient with opacity but it would have to end in a definite color, in your case white.

body {  min-height: 100vh;  background-image: linear-gradient(transparent, white 75%), url(http://www.fillmurray.com/460/300);}

Transparent gradient for top and bottom of background image css

You can have multiple stops in a gradient, so if you wanted the top 10% to fade to transparent and the bottom 10% to fade back, you could do something like this:

background-image: linear-gradient(
    to bottom,
    rgba(64, 64, 64, 1) 0%,
    rgba(64, 64, 64, 0) 10%,
    rgba(64, 64, 64, 0) 90%,
    rgba(64, 64, 64, 1) 100%
);

Demo with <img> tag: http://jsfiddle.net/sh6Hh/ or without the extra <div>: http://jsfiddle.net/sh6Hh/262/

Demo with css background picture: http://jsfiddle.net/sh6Hh/1/

Fade image to transparent like a gradient

If you want this:

Sample Image

You can do this:

<html>  <style type='text/css'>    div, img { position:absolute; top:0; left:0; width:250px; height:250px; }    img {      -webkit-mask-image:-webkit-gradient(linear, left top, left bottom, from(rgba(0,0,0,1)), to(rgba(0,0,0,0)));      mask-image: linear-gradient(to bottom, rgba(0,0,0,1), rgba(0,0,0,0));    }  </style>  <body>    <div>Lorem ipsum dolor sit amet, consectetur adipiscing elit. Morbi sit amet porttitor massa. Morbi eget tortor congue, aliquet odio a, viverra metus. Ut cursus enim eu felis sollicitudin, vitae eleifend urna lobortis. Mauris elementum erat non facilisis cursus. Fusce sit amet lacus dictum, porta libero sed, euismod tellus. Aenean erat augue, sodales sed gravida ac, imperdiet ac augue. Ut condimentum dictum mauris. Donec tincidunt enim a massa molestie, vel volutpat massa dictum. Donec semper odio vitae adipiscing lacinia.</div>    <img src='https://i.imgur.com/sLa5gg2.jpg' />  </body></html>

Transparent Background Image with a Gradient

Keep in mind that a CSS gradient is actually an image value, not a color value as some might expect. Therefore, it corresponds to background-image specifically, and not background-color, or the entire background shorthand.

Essentially, what you're really trying to do is layering two background images: a bitmap image over a gradient. To do this, you specify both of them in the same declaration, separating them using a comma. Specify the image first, followed by the gradient. If you specify a background color, that color will always be painted underneath the bottom-most image, which means a gradient will cover it just fine, and it will work even in the case of a fallback.

Because you're including vendor prefixes, you will need to do this once for every prefix, once for prefixless, and once for fallback (without the gradient). To avoid having to repeat the other values, use the longhand properties¹ instead of the background shorthand:

#mydiv .isawesome { 
    background-color: #B1B8BD;
    background-position: 0 0;
    background-repeat: no-repeat;

    /* Fallback */
    background-image: url('../images/sidebar_angle.png');

    /* CSS gradients */
    background-image: url('../images/sidebar_angle.png'), 
                      -moz-linear-gradient(top, #ADB2B6 0%, #ABAEB3 100%);
    background-image: url('../images/sidebar_angle.png'), 
                      -webkit-gradient(linear, left top, left bottom, color-stop(0%, #ADB2B6), color-stop(100%, #ABAEB3));
    background-image: url('../images/sidebar_angle.png'), 
                      linear-gradient(to bottom, #ADB2B6, #ABAEB3);

    /* IE */
    filter: progid:DXImageTransform.Microsoft.gradient(startColorstr='#ADB2B6', endColorstr='#ABAEB3', GradientType=0);
}

Unfortunately this doesn't work correctly in IE as it uses filter for the gradient, which it always paints over the background.

To work around IE's issue you can place the filter and the background image in separate elements. That would obviate the power of CSS3 multiple backgrounds, though, since you can just do layering for all browsers, but that's a trade-off you'll have to make. If you don't need to support versions of IE that don't implement standardized CSS gradients, you have nothing to worry about.

¹ _{Technically, the background-position and background-repeat declarations apply to both layers here because the gaps are filled in by repeating the values instead of clamped, but since background-position is its initial value and background-repeat doesn't matter for a gradient covering the entire element, it doesn't matter too much. The details of how layered background declarations are handled can be found here.}

How to make background SLIGHTLY fade from top and bottom?

You can create two CSS pseudo elements and configure the gradients independently with the element's height and colour gradient stops.

Example: https://codepen.io/giumagnani/pen/ZZpmje

div {
  margin: 100px 0;
  background: #2222ff;
  height: 600px;
  position: relative;
}

div::after {
  position: absolute;
  content: "";
  height: 30%;
  width: 100%;
  top: 0;
  left: 0;
  background: linear-gradient(white 0%, transparent 100%);
}

div::before {
  position: absolute;
  content: "";
  height: 30%;
  width: 100%;
  bottom: 0;
  left: 0;
  background: linear-gradient(transparent 0%, white 100%);
}

CSS - Transparency Gradient on an Image

My solution to my problem is to simply state that this is not possible with the current technology. An alternative option would be to use a simple transparency gradient. Until A better solution arrises this is what I will end up doing.

background: linear-gradient(to bottom, rgba(0,0,0,1) 0%,rgba(0,0,0,0) 100%);