Why does modifying the [math]x[/math] value of a function transform its graph? Why does [math]f(x[/math][color=#c51414][b][math]+[/math][/b][/color][math]2)[/math] make [math]f(x)[/math] shift [i]left[/i] while [math]f(x[/math][color=#c51414][b][math]-[/math][/b][/color][math]2)[/math] shifts it [i]right[/i]? Why does [math]f[/math]([color=#c51414][b][math]2[/math][/b][/color][math]x)[/math] [i]compress [/i][math]f(x)[/math] while [math]f([/math][color=#c51414][b][math]0.5[/math][/b][/color][math]x)[/math] [i]stretches [/i][math]f(x)[/math]? This seems backwards, doesn't it?

Set the large vertical slider to the bottom, "Step 1". Set the small vertical selector switch at the top of the right-hand pane to line up with either the [math]h[/math] slider or the [math]b[/math] slider, The [math]h[/math] slider will create horizontal [i]translations [/i](shifts), while the [math]b[/math] slider controls horizontal [i]dilations[/i] (stretches/compressions). [br][br]In Step 1, we see our original function in purple, called [math]f(x)[/math]. (Feel free to enter a different function in the "[color=#000000]f(x) =[/color]" box). Move the Step slider up one notch. In Step 2, we can select any value of [math]x[/math] in [math]f[/math]'s domain by dragging the red [color=#ff0000][math]x[/math][/color] point. Next, in Step 3, we modify the [math]x[/math]-value by subtracting [math]h[/math] or multiplying it by [math]b[/math], depending on how the selector switch is set. Notice that if we [i]subtract [/i]a [i]negative [/i]value of [math]h[/math], we are actually [i]adding [/i]a [i]positive[/i] number to [math]x[/math], and thus we are moving to the [i]right[/i]. Thus, we have [math]x-h[/math] or [math]b \times x[/math] as a new [math]x[/math]-value. In Step 4 we take this new [math]x[/math]-value and plug it into [math]f[/math], giving us the [math]y[/math]-value that goes with [math]x-h[/math] or [math]b \times x[/math]. But remember that we want the [i]transformation[/i] [math]y(x)=f(x-h)[/math] or [math]y(x)=f(bx)[/math]: in other words, we want the [math]y[/math]-value at [math]x[/math] to be the same as the [math]y[/math]-value at [math]x-h[/math] or [math]bx[/math]. So in Step 5, we "pull" the new [math]y[/math]-value back to the original [math]x[/math]-value. In Step 6, we see in blue what we get when we do these steps for all values of [math]x[/math]. (The original [math]f(x)[/math] is shown dashed).[br][br]Once you've understood the six steps in this app, leave the slider on Step 6 and move the other sliders to see these transformations in action. Another exercise you can do is to return the slider to Step 1, enter a new function, and then predict what will happen at each successive step, checking your prediction as you go. Above the "Step" slider, you'll see a brief description of what is happening at that step.[br][br]Special cases to think about: What happens when [math]b=0[/math] and when [math]h=0[/math]? Why?