What is the minimum point of y equals x squared plus 2?

The minimum is (0, 2). The plus 2 is outside the function, added after squaring, so it adds 2 to every output and lifts the whole curve 2 up; the lowest point (0, 0) of y equals x squared rises straight up to (0, 2). The x-coordinate is unchanged. The trap (minus 2, 0) is not even on the curve: at x equals minus 2, y equals (minus 2) squared plus 2 equals 6, not 0.

What is the difference between f of x plus a and f of x plus a inside the bracket?

A change outside the function, f of x plus a, acts on the output and moves the graph up or down: a vertical translation. A change inside the bracket, f of (x plus a), acts on the input and moves the graph left or right: a horizontal translation. So y equals x squared plus 2 has minimum (0, 2), while y equals (x plus 2) squared has vertex (minus 2, 0).

How do you write the equation of y equals x squared lifted so its lowest point is (0, 3)?

Add 3 outside the function: y equals x squared plus 3. An outside change moves the curve up, so the lowest point rises to (0, 3). Writing y equals (x plus 3) squared would be wrong: that puts the 3 inside the bracket, moving the curve left to (minus 3, 0), not up.

GCSE Maths Higher · AQA · Graph transformations

Translations: outside the function is up/down, inside is left/right

When students translate $y=x^2$ to $y=x^2+2$ , the instinct is to read the $+2$ as a sideways move and give the minimum as $(-2,0)$ or $(2,0)$ . But the $+2$ is outside the function: it is added after squaring, so it lifts every output by 2 and moves the whole curve 2 up. The minimum is $(0,2)$ .

The rule is: a change outside the function, $f(x)+a$ , moves the graph up or down, so the x-coordinate stays put. A change inside the bracket, $f(x+a)$ , moves it left or right. Reading an outside change as a sideways move 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 wrote $(-2,0)$ or $(2,0)$ for the minimum of $y=x^2+2$ , reading the $+2$ as a horizontal shift. The minimum is $(0,2)$ .
You moved the curve sideways for a change added after the function, when an outside change moves it up or down.
You wrote $y=(x+3)^2$ for an upward parabola with lowest point $(0,3)$ , when the answer is $y=x^2+3$ ; $y=(x+3)^2$ moves left to $(-3,0)$ .
You gave $(-4,0)$ for the inflection of $y=x^3-4$ instead of $(0,-4)$ , treating the outside $-4$ as a sideways move.

An exam question that triggers it

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

The curve $y=x^2+2$ is a translation of $y=x^2$ .

Write down the coordinates of its minimum point.

The misconception writes $(-2,0)$ . The fix: the $+2$ is outside the function, so the curve moves 2 up and the minimum is $(0,2)$ . At $x=-2$ , $y=(-2)^2+2=6$ , not 0, so $(-2,0)$ is not even on the curve.

Why students fall for this

The number is the salient, visible thing, and students attach it to a sideways move because so many transformations they meet first are horizontal. But the direction of a translation is decided by where the number sits, not by its value. A number added outside the function changes the output, so the whole curve rises or falls.

The trap is sharpest when the same number can sit in two places. In $y=x^2+2$ the 2 is outside and the curve moves up to $(0,2)$ ; in $y=(x+2)^2$ the 2 is inside the bracket and the curve moves left to $(-2,0)$ . The same value, a different move.

A cubic such as $y=x^3-4$ confirms the rule on a different curve: the outside $-4$ moves the graph 4 down, so the point of inflection goes from $(0,0)$ to $(0,-4)$ , not to $(-4,0)$ .

The fix: Outside the function moves up/down, f(x)+a; inside the bracket moves left/right, f(x+a); check the moved key point

Step 1: find where the number sits. Is it added after the function, $f(x)+a$ , or inside the bracket acting on the input, $f(x+a)$ ?

Step 2: choose the direction. Outside means up or down (a vertical move). Inside means left or right (a horizontal move).

Step 3: apply it. For $y=x^2+2$ the outside $+2$ lifts the minimum to $(0,2)$ . For $y=(x+2)^2$ the inside $+2$ slides the vertex to $(-2,0)$ .

Step 4: check the moved point. An outside change keeps the x-coordinate and changes the y; an inside change keeps the y-coordinate and changes the x. For $y=x^2+2$ the x stays 0, confirming $(0,2)$ .

Step 5: substitute to reject the trap. If you wrote $(-2,0)$ , test it: at $x=-2$ , $y=(-2)^2+2=6$ , not 0, so the point is not on the curve.

Worked example

The curve $y=x^2+2$ is a translation of $y=x^2$ . Write the coordinates of its minimum point.

Find where the number sits. The $+2$ is outside the function, so the move is vertical.
Apply it. The curve moves 2 up, so the minimum $(0,0)$ rises to $(0,2)$ .
Check. $x^2\ge 0$ so $y\ge 2$ , with equality at $x=0$ . Trap: at $x=-2$ , $y=(-2)^2+2=6$ , so $(-2,0)$ is not on the curve.

A parabola is $y=x^2$ lifted so its lowest point is $(0,3)$ . Write its equation.

Choose the direction. The lowest point moved 3 up, a vertical move, so add 3 outside the function.
Apply it. $y=x^2+3$ .
Reject the inside form. $y=(x+3)^2$ sets $x+3=0$ , giving vertex $(-3,0)$ , a left shift, not the upward parabola asked for.

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Common questions

What is the minimum of $y=x^2+2$ ?: $(0,2)$ . The $+2$ is outside the function, so it lifts every output by 2 and moves the curve 2 up; the minimum $(0,0)$ of $y=x^2$ rises to $(0,2)$ . The x-coordinate stays 0. The trap $(-2,0)$ is not on the curve: at $x=-2$ , $y=6$ .
What is the difference between $f(x)+a$ and $f(x+a)$ ?: $f(x)+a$ changes the output and moves the graph up or down: a vertical translation. $f(x+a)$ changes the input and moves the graph left or right: a horizontal translation. For $y=x^2$ these are $y=x^2+2$ (minimum $(0,2)$ ) and $y=(x+2)^2$ (vertex $(-2,0)$ ).
How do I write $y=x^2$ lifted so its lowest point is $(0,3)$ ?: Add 3 outside the function: $y=x^2+3$ . An outside change moves the curve up, so the lowest point rises to $(0,3)$ . Writing $y=(x+3)^2$ would put the 3 inside the bracket, moving the curve left to $(-3,0)$ .

Related misconceptions

Horizontal shifts: f(x+a) moves left, not rightSliding y=(x+3)² 3 to the right instead of 3 to the left, the companion inside-the-bracket misconception about which way a horizontal translation goes.
Reflections: −f(x) is the x-axis, f(−x) is the y-axisConfusing which axis a reflection uses, the same outside-versus-inside reading applied to a minus sign instead of an added constant.

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