Horizontal Dilation

Changing the Width of a Function Graph

Horizontal dilation changes how wide a function graph looks along the $x$-axis without changing any of its heights. Stretching the graph makes it wider, and compressing it makes the graph narrower, while every point keeps its original $y$ value.

If you have a function $f(x)$, then horizontal dilation produces a new function $g(x) = f(kx)$. Here $k \ne 0$ is the number that multiplies the input, and the horizontal scale of the new graph is $s = \frac{1}{|k|}$.

How the Input Factor Changes the Width

For any function $f(x)$ and a nonzero factor $k$ that multiplies the input, horizontal dilation is defined as:

The factor $k$ decides what happens to the width:

• If $k > 1$, the graph is compressed to $\frac{1}{k}$ of its original width
• If $0 < k < 1$, the graph is stretched to $\frac{1}{k}$ times its original width
• If $k = 1$, the graph does not change
• If $k < 0$, the graph is also reflected across the $y$-axis

The value $k = 0$ is excluded: $f(0x) = f(0)$ would be constant wherever it is defined. The graph would collapse to one horizontal line.

The reciprocal matters here. A larger factor that multiplies the input makes the graph narrower, because the horizontal scale is $\frac{1}{|k|}$.

Visualization of Horizontal Dilation

For the quadratic function $f(x) = x^2$, several input multipliers show that horizontal scaling uses the reciprocal factor.

The graph shows:

• The original function $f(x) = x^2$ as reference
• Function $g(x) = (2x)^2$ uses input multiplier $2$, so its horizontal scale is $\frac{1}{2}$
• Function $h(x) = (0.5x)^2$ uses input multiplier $0.5$, so its horizontal scale is $2$
• All graphs have the same vertex at $(0, 0)$

Horizontal Dilation on Linear Functions

The same rule applies to the linear function $f(x) = x + 2$.

The three lines show:

• The original function $f(x) = x + 2$ has slope $1$
• Function $g(x) = f(2x) = 2x + 2$ has slope $2$ (horizontally compressed)
• Function $h(x) = f(0.5x) = 0.5x + 2$ has slope $0.5$ (horizontally stretched)
• All lines intersect the $y$-axis at the same point $(0, 2)$

Which Values the Horizontal Scale Moves

Once you know whether the graph stretches or compresses, you still need to know which values move. Horizontal dilation divides every horizontal coordinate by the factor $k$ that multiplies the input, so positions along the $x$-axis change while heights stay the same.

Take a point $(a, b)$ on the graph of $f(x)$. The matching point on the graph of $f(kx)$ is $(\frac{a}{k}, b)$. The $y$ value stays at $b$ because horizontal dilation never changes a height.

The domain shrinks or grows together with the graph, because it lists the allowed values of $x$. The range keeps its original values, because the height of the graph at each point never changes. A horizontal translation shifts those coordinates without changing their size.

Changing the Growth Rate of an Exponential Function

For the exponential function $f(x) = 2^x$, the input multiplier changes how quickly the graph rises.

For exponential functions:

• The horizontal asymptote remains at $y = 0$ for both functions
• The $y$-intercept remains the same at $(0, 1)$
• The growth rate of the function changes according to the input multiplier

Horizontal Dilation with a Negative Input Multiplier

A negative input multiplier combines horizontal scaling with a reflection.

When the input multiplier is negative:

• The graph undergoes reflection across the $y$-axis
• Its horizontal scale magnitude is $\frac{1}{|k|}$
• The graph shape remains the same because the quadratic function is symmetric

Exercises

The three problems below use the width rule you have just learned. Each one multiplies the input by a stated factor, so watch whether that factor is greater than one, between zero and one, or negative. Try them without the answer key first, then compare your steps with the worked answers.

1. Given the function $f(x) = x^2 - 2x + 1$. Determine the equation produced by the input multiplier $3$.

2. The graph of the function $g(x) = \sqrt{x}$ undergoes horizontal dilation with input factor $\frac{1}{2}$. Determine the equation of the resulting function, then determine the domain of that function.

3. Function $h(x) = |x - 2|$ undergoes horizontal dilation with input multiplier $-1$. Determine the vertex of the resulting function.

Answer Key

Every worked answer starts by replacing $x$ with the product of the factor and $x$, because that is where the width rule lives. The equation is then simplified and the new graph is checked to confirm whether it stretched or compressed.

1. Horizontal dilation with input multiplier $3$: $f'(x) = f(3x) = (3x)^2 - 2(3x) + 1 = 9x^2 - 6x + 1$

   Function $f(x) = x^2 - 2x + 1$ and Its Dilation Result

   Input multiplier $3$ gives horizontal scale $\frac{1}{3}$, producing a narrower parabola.

   GridAutomatic rotation

2. Equation of the resulting dilated function:

   • Horizontal dilation: $g'(x) = g(\frac{1}{2}x) = \sqrt{\frac{1}{2}x} = \sqrt{\frac{x}{2}}$
   • Domain after dilation: $[0, \infty)$ remains $[0, \infty)$ because the input multiplier is positive

   Visualization:

   Function $g(x) = \sqrt{x}$ and Its Dilation Result

   Input multiplier $0.5$ gives horizontal scale $2$, making the square root curve twice as wide.

   GridAutomatic rotation

3. The original function $h(x) = |x - 2|$ has its vertex at $(2, 0)$. After horizontal dilation with input multiplier $-1$: $h'(x) = |-x - 2| = |-(x + 2)| = |x + 2|$, the vertex becomes $(-2, 0)$.

   Function $h(x) = |x - 2|$ and Its Dilation Result

   Input multiplier $-1$ reflects the absolute value function across the $y$-axis without changing its horizontal scale.

   GridAutomatic rotation