Function transformations on the ACT

Changes made outside the function affect the graph vertically and do exactly what they look like, while changes made inside the parentheses affect it horizontally and do the opposite of what they look like. That single sentence covers every transformation the ACT asks about. The rest of this guide is just examples that make it stick.

The one idea behind all of them

Start with some graph y = f(x). A transformation rewrites the formula slightly and asks how the picture moves. Anything done to the output, outside the f, changes the y-values. Anything done to the input, inside the parentheses, changes which x-value produces a given point, so it slides the graph sideways in the direction that feels backwards at first.

Keep that split in mind and you will not need a memorized table. It does help to be comfortable with function notation first, since every rule below is written in it.

Vertical shifts: f(x) + k

Adding a number after the function moves the whole graph up. Subtracting moves it down. This is the intuitive one.

Take f(x) = x^2, whose vertex sits at (0, 0). Then f(x) + 3 = x^2 + 3, and every output is three units larger, so the vertex moves to (0, 3). Likewise f(x) - 2 drops the vertex to (0, -2). Check a point: at x = 1, the original gives 1 and the shifted version gives 4.

Horizontal shifts: the counterintuitive one

Here is the rule students get backwards more than any other: f(x + h) moves the graph LEFT by h units, and f(x - h) moves it RIGHT by h units.

Do not fight it, just verify it once. With f(x) = x^2, the graph of f(x + 4) is y = (x + 4)^2, and the squared part is smallest when x + 4 = 0, which happens at x = -4. The vertex sat at 0 and now sits at -4, four units to the left, even though the formula says plus four. In reverse, f(x - 3) has its vertex where x - 3 = 0, so at x = 3, three units to the right.

The reason is that adding 4 inside lets a smaller x produce the old result, so the picture slides toward smaller x.

Reflections: -f(x) and f(-x)

Putting a negative outside the function flips it over the x-axis, because every output changes sign. Putting the negative inside flips it over the y-axis, because every input changes sign.

• If f(x) = x^2 - 1, the point (2, 3) is on the graph. On -f(x), that becomes (2, -3). The x stays put and the y flips.
• If f(x) is the square root of x, its graph lives to the right of the origin. Then f(-x) is the square root of -x, which only accepts negative inputs, so the graph lives to the left instead. The y-values are unchanged.

Vertical stretches: a times f(x)

Multiplying the whole function by a number a stretches the graph away from the x-axis when a is bigger than 1, and squashes it toward the x-axis when a is between 0 and 1.

With f(x) = x^2, the point (2, 4) is on the graph. On 3f(x) = 3x^2, that point becomes (2, 12), since only the output tripled. On one half times f(x), it becomes (2, 2). Outputs of 0 stay 0, so points on the x-axis never move during a vertical stretch.

Putting several together

The ACT sometimes stacks transformations. Take f(x) = x^2 and consider g(x) = 2f(x - 1) + 5, which is g(x) = 2(x - 1)^2 + 5. Read it piece by piece: inside, x - 1 shifts right 1; outside, the 2 stretches vertically and the plus 5 shifts up 5. The vertex started at (0, 0), so it lands at (1, 5). Confirm by plugging in: g(1) = 2(0) + 5 = 5, and g(2) = 2(1) + 5 = 7.

A common question type gives you a point instead of a formula. If (3, 7) is on y = f(x), what point must be on y = f(x + 2) - 4? The input has to satisfy x + 2 = 3, so x = 1, and the output drops by 4 to give 3. The point is (1, 3).

Common traps and a quick drill

• Reversing the horizontal shift. Plus moves left, minus moves right. When in doubt, ask which x makes the inside equal zero.
• Mixing inside and outside. f(x) + 2 and f(x + 2) look nearly identical and move the graph in completely different directions.
• Applying a vertical stretch to the x-values. In 3f(x), only the outputs triple.
• Forgetting that reflections keep one coordinate. A flip over the x-axis leaves x alone, and a flip over the y-axis leaves y alone.

For the drill, sketch y = x^2, then on the same axes sketch (x - 2)^2, (x - 2)^2 + 3, and -(x - 2)^2 + 3, saying out loud what each change did. Then take the point (4, 1) and find its image under f(x) - 6, f(x - 6), and -2f(x). The answers are (4, -5), (10, 1), and (4, -2). Mixed function sets on thirty-six keep these sharp. To zoom out, the ACT Math functions hub links the rest of the topic, and coordinate geometry covers the graph-reading skills these questions lean on.

This article offers general ACT prep guidance. The ACT can change from year to year, including its format, scoring, policies, test dates, and fees, so always confirm the latest details on the official ACT website at act.org before you make decisions. ACT® is a registered trademark of ACT, Inc. thirty-six is not affiliated with, endorsed by, or sponsored by ACT.