Why does y equals (x plus 3) squared move 3 to the left and not to the right?

Because the change is inside the bracket, acting on the input x. The vertex sits where the bracket is zero. Solving x plus 3 equals 0 gives x equals minus 3, so the vertex moves from the origin to (minus 3, 0): a move 3 to the left. A change inside the bracket moves the graph horizontally and the opposite way to the sign.

What is the difference between a change inside the bracket and a change outside it?

A change inside the bracket, such as the 3 in (x plus 3) squared, acts on the input and moves the graph horizontally, the opposite way to the sign: f(x plus a) moves left, f(x minus a) moves right. A change outside the bracket, such as the 2 in x squared plus 2, acts on the output and moves the graph vertically, in the same direction as the sign: plus 2 moves up.

How do you write a horizontal translation as a column vector?

Set the bracket equal to zero to find where the key feature lands. For y equals (x plus 3) squared the vertex is at x equals minus 3, a move 3 to the left, so the horizontal component is minus 3 and the column vector is (minus 3, 0). For y equals (x minus 2) cubed the point of inflection is at x equals 2, a move 2 to the right, so the vector is (2, 0). Left is negative, right is positive, and the vertical component is 0.

GCSE Maths Higher · AQA · Graph transformations

Horizontal shifts: f(x+a) moves left, not right

When students transform $y=x^2$ into $y=(x+3)^2$ , the instinct is to read the $+3$ and slide the curve 3 to the right, or to treat the change as a move up. Both are wrong. A change inside the bracket acts on the input $x$ , so it moves the graph horizontally and the opposite way to the sign. The vertex sits where the bracket is zero: $x+3=0$ gives $x=-3$ , so $y=(x+3)^2$ slides 3 to the left to vertex $(-3,0)$ .

The rule is: $f(x+a)$ moves left by $a$ , and $f(x-a)$ moves right by $a$ . A change outside the bracket, $f(x)+a$ , is a vertical move in the same direction as the sign. Mixing these up is one of the most reliable grade 7 to 9 mark-losers on AQA Higher graph-transformation questions.

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How to spot it in your own work

You moved $y=(x+3)^2$ 3 to the right instead of 3 to the left, reading the $+3$ literally rather than solving $x+3=0$ to find the vertex at $x=-3$ .
You treated the inside-bracket change in $y=(x+3)^2$ as a vertical shift (3 up) instead of a horizontal one.
You over-applied the flip and sent $y=x^2+2$ downward, when an outside change moves the curve up (same direction as the sign) to vertex $(0,2)$ .
You gave the translation vector with the wrong sign, writing $(3,0)$ for the left shift $y=(x+3)^2$ instead of $(-3,0)$ .

An exam question that triggers it

Here is the structure of a typical AQA Higher graph-transformation question:

The graph of $y=x^2$ is transformed to $y=(x+3)^2$ .

Describe the single transformation, and write it as a column vector.

The misconception writes $(3,0)$ (3 to the right). The fix: solve $x+3=0$ to get $x=-3$ , so the vertex moves to $(-3,0)$ , a translation 3 to the left, vector $(-3,0)$ .

Why students fall for this

The sign inside the bracket is the salient, visible thing. Students see $+3$ and reach for a rightward move, because positive feels like right and forward. But the bracket changes the input: $(x+3)^2$ reaches each output value 3 steps sooner in $x$ , so the whole curve slides left.

A second confusion is mixing the axis of the move. A change inside the bracket is horizontal; a change outside, $x^2+2$ , is vertical. Students who have only seen one of these often apply the wrong axis, sliding a vertical shift sideways or vice versa.

A third error appears once the flip is learnt: over-generalising it. Having learnt that $(x+3)^2$ goes the opposite way to the sign, students wrongly send $x^2+2$ downward. The flip belongs only to the inside-bracket horizontal move; the vertical move goes the same way as its sign.

The fix: Solve bracket = 0 for the new feature: f(x+a) goes left, f(x−a) goes right; outside changes go vertically, same as the sign

Step 1: locate the change. Is the number inside the bracket (acting on $x$ ) or outside (acting on the output)? For $y=(x+3)^2$ it is inside.

Step 2: solve bracket = 0. $x+3=0$ gives $x=-3$ , so the vertex lands at $(-3,0)$ .

Step 3: read the direction. The vertex moved from $(0,0)$ to $(-3,0)$ , that is 3 to the left. So $f(x+a)$ goes left, $f(x-a)$ goes right.

Step 4: write the vector. Left is negative, so the column vector is $(-3,0)$ . The vertical component is 0 because the move is purely horizontal.

Step 5: separate from vertical shifts. For $y=x^2+2$ the change is outside the bracket, so it moves 2 up to $(0,2)$ , in the same direction as the sign. No flip outside.

Worked example

The graph of $y=x^2$ is transformed to $y=(x+3)^2$ . Describe the transformation and give the column vector.

Find the new vertex. Solve $x+3=0$ to get $x=-3$ , vertex $(-3,0)$ .
Read the direction. The vertex moved from $(0,0)$ to $(-3,0)$ : 3 to the left.
Vector.
$\begin{pmatrix} -3 \\ 0 \end{pmatrix}$
Trap: $(3,0)$ would move the curve 3 to the right, the wrong way.

The graph of $y=x^3$ is translated to give $y=(x-2)^3$ . State the translation vector.

Find the new point of inflection. Solve $x-2=0$ to get $x=2$ , so the inflection moves from $(0,0)$ to $(2,0)$ .
Direction. 2 to the right, because $f(x-a)$ goes right by $a$ .
Vector.
$\begin{pmatrix} 2 \\ 0 \end{pmatrix}$
The same inside-the-bracket rule works for cubics.

Contrast: where does $y=x^2+2$ go?

Locate the change. The $+2$ is outside the bracket, acting on the output.
Direction. The curve moves 2 up, in the same direction as the sign, to vertex $(0,2)$ . The flip does not apply to vertical shifts.

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Common questions

Why does $y=(x+3)^2$ move left and not right?: Because the change is inside the bracket, acting on the input $x$ . The vertex is where the bracket is zero: $x+3=0$ gives $x=-3$ , so the vertex moves to $(-3,0)$ , a move 3 to the left. An inside-bracket change moves the graph horizontally and the opposite way to the sign.
What is the difference between $(x+3)^2$ and $x^2+3$ ?: In $(x+3)^2$ the 3 is inside the bracket, so the curve moves 3 to the left to vertex $(-3,0)$ . In $x^2+3$ the 3 is outside, so the curve moves 3 up to vertex $(0,3)$ . Inside the bracket is horizontal and flipped; outside is vertical and in the same direction as the sign.
How do I write the translation as a column vector?: Solve the bracket equals zero to find where the key feature lands. For $y=(x+3)^2$ the vertex is at $x=-3$ , a move 3 to the left, so the vector is $(-3,0)$ . For $y=(x-2)^3$ the inflection is at $x=2$ , a move 2 to the right, so the vector is $(2,0)$ . Left is negative, right is positive, and the vertical component is 0.

Related misconceptions

Reflections: −f(x) is the x-axis, f(−x) is the y-axisWriting y=(−x)² for an x-axis reflection of y=x² instead of y=−x², a companion graph-transformation misconception about which sign change maps to which axis.
Turning point x-coordinate: the sign flips from vertex formCopying the printed sign from (x−7)²+8 to get x=−7 instead of solving (x−7)=0 to get x=7, the same opposite-to-the-sign trap in a different completing-the-square context.

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