Stretching a Graph Horizontally

Start from $y=x^2-4$y = x² − 4

Take $f\left(x\right)=x^2-4$f(x) = x² − 4. Its graph is a parabola with its vertex at (0, −4). It meets the x-axis where $x^2-4=0$x² − 4 = 0, which is where $x^2=4$x² = 4, at x = −2 and x = 2. It also passes through (3, 5), because f(3) = 9 − 4 = 5.

$y=x^2-4$y = x² − 4 meets the x-axis at (−2, 0) and (2, 0), and its vertex is (0, −4).

Replace x by 2x

Now put 2x in place of x: $y=f\left(2x\right)={\left(2x\right)}^2-4=4x^2-4$y = f(2x) = (2x)² − 4 = 4x² − 4. The 2 is inside the bracket, so it acts on the input: x is doubled before f does anything to it.

Where does the new graph meet the x-axis? f gives 0 when its input is 2 or −2, so f(2x) = 0 when 2x = 2 or 2x = −2, which is at x = 1 and x = −1. And f gives 5 when its input is 3, so f(2x) = 5 when 2x = 3, at x = 1.5. Each time, the new graph reaches the height at half the x that the old graph needed.

Every height, at half the distance

The reason is the same as for a translation inside the bracket. In f(x + a), the bracket is always a more than x, so the graph reaches every height a units earlier and moves left. In f(2x), the bracket is always twice x, so it reaches any value when x is only half that value. So the new graph reaches every height at half the distance from the y-axis.

Put as a rule: if (p, q) is on y = f(x), then $\left(\frac{p}{2},q\right)$(p/2, q) is on y = f(2x), because the bracket then holds $2\times \frac{p}{2}=p$2 × p/2 = p, and f(p) = q. Every point keeps its height and moves to half its distance from the y-axis. The graph is squashed toward the y-axis: every width is divided by 2.

This is called a horizontal stretch, or a stretch parallel to the x-axis, with scale factor ½. In general, y = f(ax) moves each point (p, q) of y = f(x) to $\left(\frac{p}{a},q\right)$(p/a, q), a stretch with scale factor $\frac{1}{a}$1/a.

The white curve is $y=x^2-4$y = x² − 4 and the gold curve is $y={\left(2x\right)}^2-4$y = (2x)² − 4. The crossings have moved from −2 and 2 in to (−1, 0) and (1, 0), and at the height 5 the point (3, 5) has moved to (1.5, 5).

The y-axis stays where it is

A point on the y-axis has x = 0, and 0 ÷ 2 = 0, so it does not move. The vertex (0, −4) is on both graphs: f(2 × 0) = f(0) = −4.

No height changes either. The lowest value of f(2x) is −4, the same as the lowest value of f(x). A horizontal stretch changes where each height is reached, never the height itself.

A factor between 0 and 1

Now halve x instead: $y=f\left(\frac{x}{2}\right)={\left(\frac{x}{2}\right)}^2-4$y = f(x/2) = (x/2)² − 4. f gives 0 when its input is 2 or −2, so this is 0 when $\frac{x}{2}=2$x/2 = 2 or $\frac{x}{2}=-2$x/2 = −2, which is at x = 4 and x = −4. The bracket $\frac{x}{2}$x/2 is always half of x, so x has to be twice as big for the bracket to reach each value.

Here a = ½, and dividing each x by ½ doubles it: (2, 0) moves to (4, 0) and (3, 5) moves to (6, 5). The graph is stretched away from the y-axis, a horizontal stretch with scale factor 2.

The white curve is $y=x^2-4$y = x² − 4 and the gold curve is $y={\left(\frac{x}{2}\right)}^2-4$y = (x/2)² − 4, which crosses the x-axis further out, at (−4, 0) and (4, 0).

Outside multiplies, inside divides

Follow the point (3, 5) of $y=x^2-4$y = x² − 4 through both stretches with the factor 2. On y = 2f(x), the 2 is outside the bracket, so it multiplies the height: (3, 5) moves to (3, 10). On y = f(2x), the 2 is inside, so it divides the x: (3, 5) moves to (1.5, 5). Check both: 2f(3) = 2 × 5 = 10, and f(2 × 1.5) = f(3) = 5.

So a number outside the bracket acts as it reads, on the heights, and leaves the points on the x-axis fixed. A number inside the bracket acts in reverse, on the widths, dividing where it seems to multiply, and leaves the points on the y-axis fixed. It is the same reversal as with a translation, where a plus sign inside the bracket moves the graph left.

A negative factor inside

With a = −1, y = f(−x) moves each point (p, q) to (−p, q): the same height, on the other side of the y-axis. So f(−x) is the reflection of y = f(x) in the y-axis.

For $f\left(x\right)={\left(x-1\right)}^2$f(x) = (x − 1)², with its vertex at (1, 0), $f\left(-x\right)={\left(-x-1\right)}^2={\left(x+1\right)}^2$f(−x) = (−x − 1)² = (x + 1)², since a number and its negative have the same square. Its vertex is at (−1, 0), the mirror image of (1, 0) in the y-axis.

The usual mistakes

Multiplying the x instead of dividing. If (6, 5) is on y = f(x), then on y = f(2x) it moves to (3, 5), because the bracket holds 2 × 3 = 6. (12, 5) would be a stretch outward.

Changing the height. A number inside the bracket never touches the heights: (6, 5) keeps its height 5 on y = f(2x). Doubling the height to 10 is what y = 2f(x) does.

Leaving the crossings where they were. Only a vertical stretch keeps the points on the x-axis. If y = f(x) crosses at x = 6, y = f(2x) crosses at x = 3.