Why is y equals (minus x) squared not the reflection of y equals x squared in the x-axis?

Because (minus x) squared equals x squared: squaring removes the minus sign, so y equals (minus x) squared is the original curve unchanged. Reflecting in the x-axis negates the output, y equals minus f of x, so y equals x squared becomes y equals minus x squared, a downward parabola through (2, minus 4) and (minus 2, minus 4).

What is the difference between minus f of x and f of minus x?

Minus f of x negates the output, flipping the curve top to bottom: a reflection in the x-axis. F of minus x negates the input, flipping the curve left to right: a reflection in the y-axis. For y equals x squared the x-axis reflection is y equals minus x squared, and the y-axis reflection is y equals (minus x) squared, which equals x squared and so looks unchanged.

How do you reflect the line y equals 5x plus 4 in the y-axis?

Replace x with minus x: y equals 5 times minus x plus 4, which is y equals minus 5x plus 4. The gradient flips sign but the y-intercept (0, 4) stays fixed, because a reflection in the y-axis leaves every point on the y-axis where it is. The x-axis reflection would be y equals minus (5x plus 4), which is y equals minus 5x minus 4, moving the intercept to (0, minus 4).

GCSE Maths Higher · AQA · Graph transformations

Reflections: −f(x) is the x-axis, f(−x) is the y-axis

When students reflect $y=x^2$ in the x-axis, the instinct is to write $y=(-x)^2$ , because the minus sign feels like the reflection. But $(-x)^2=x^2$ : squaring removes the minus, so $y=(-x)^2$ is the original curve unchanged. It is in fact the y-axis reflection, not the x-axis one.

The rule is: reflecting in the x-axis negates the output, $y=-f(x)$ , so $y=x^2$ becomes $y=-x^2$ , a downward parabola through $(2,-4)$ and $(-2,-4)$ . Reflecting in the y-axis negates the input, $y=f(-x)$ . Picking the wrong axis 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 $y=(-x)^2$ for the x-axis reflection of $y=x^2$ , not noticing that $(-x)^2=x^2$ is no change at all. The x-axis reflection is $y=-x^2$ .
You put the minus inside on the $x$ when the question asked for an x-axis reflection, which negates the output, not the input.
You reflected the line $y=5x+4$ in the y-axis but wrote $y=-5x-4$ (the x-axis reflection) instead of $y=-5x+4$ , moving the intercept off $(0,4)$ .
You swapped the axes, naming the y-axis for a $y=-f(x)$ change or the x-axis for a $y=f(-x)$ change.

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 reflected in the x-axis.

Write the equation of the reflected curve.

The misconception writes $y=(-x)^2$ . The fix: reflecting in the x-axis negates the output, so $y=x^2$ becomes $y=-x^2$ . Squaring out shows $(-x)^2=x^2$ , which is no change, the y-axis reflection.

Why students fall for this

The minus sign is the salient, visible thing, and students attach it to the variable they can see: the $x$ . But the axis of a reflection is decided by whether the output or the input is negated. Reflecting in the x-axis turns each $y$ into $-y$ , which means negating the whole function, $y=-f(x)$ .

The trap is sharpest on $y=x^2$ , because the parabola is symmetric about the y-axis. Its y-axis reflection, $y=(-x)^2=x^2$ , lands exactly on the original, so it looks like nothing happened. Students then misread $y=(-x)^2$ as a reflection that should have flipped the curve, and offer it for the x-axis instead.

A line such as $y=5x+4$ exposes the difference: its x-axis reflection $y=-5x-4$ and its y-axis reflection $y=-5x+4$ are genuinely different equations, and only the y-axis reflection keeps the intercept at $(0,4)$ .

The fix: Negate the output for the x-axis, y=−f(x); negate the input for the y-axis, y=f(−x); check the invariant point

Step 1: read the named axis. The x-axis flips top to bottom; the y-axis flips left to right.

Step 2: choose what to negate. X-axis means negate the whole output, $y=-f(x)$ . Y-axis means replace $x$ with $-x$ , $y=f(-x)$ .

Step 3: apply it. For $y=x^2$ in the x-axis, $y=-x^2$ . For $y=5x+4$ in the y-axis, $y=5(-x)+4=-5x+4$ .

Step 4: check the invariant point. An x-axis reflection fixes the x-intercepts; a y-axis reflection fixes the y-intercept. For $y=-5x+4$ the intercept $(0,4)$ is unchanged, confirming the y-axis.

Step 5: beware the even curve. $(-x)^2=x^2$ , so the y-axis reflection of $y=x^2$ shows no change. That does not make $y=(-x)^2$ the x-axis reflection; the x-axis reflection is $y=-x^2$ .

Worked example

The graph of $y=x^2$ is reflected in the x-axis. Write the equation of the reflected curve.

Choose what to negate. The x-axis flips the output, so negate the whole function: $y=-f(x)$ .
Apply it. $y=-x^2$ , a downward parabola.
Check. $(2,4)$ maps to $(2,-4)$ , and $-(2^2)=-4$ . Trap: $y=(-x)^2=x^2$ is no change, the y-axis reflection.

The line $y=5x+4$ is reflected in the y-axis. Write the equation of the reflected line.

Choose what to negate. The y-axis flips the input, so replace $x$ with $-x$ : $y=f(-x)$ .
Apply it. $y=5(-x)+4=-5x+4$ .
Check the fixed point.
$(0,4)\ \text{stays fixed: } -5(0)+4=4$
Trap: $y=-5x-4$ negates the whole output, moving the intercept to $(0,-4)$ , so it is the x-axis reflection.

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Common questions

Why is $y=(-x)^2$ not the x-axis reflection of $y=x^2$ ?: Because $(-x)^2=x^2$ : squaring removes the minus, so $y=(-x)^2$ is the original curve unchanged. Reflecting in the x-axis negates the output, $y=-f(x)$ , so the x-axis reflection of $y=x^2$ is $y=-x^2$ , a downward parabola.
What is the difference between $-f(x)$ and $f(-x)$ ?: $-f(x)$ negates the output and flips the curve top to bottom: a reflection in the x-axis. $f(-x)$ negates the input and flips the curve left to right: a reflection in the y-axis. For $y=x^2$ these are $y=-x^2$ and $y=(-x)^2=x^2$ .
How do I reflect $y=5x+4$ in the y-axis?: Replace $x$ with $-x$ : $y=5(-x)+4=-5x+4$ . The gradient flips sign but the y-intercept $(0,4)$ stays fixed, because a reflection in the y-axis leaves points on the y-axis where they are. The x-axis reflection is $y=-(5x+4)=-5x-4$ , with intercept $(0,-4)$ .

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, a companion graph-transformation misconception about which sign change moves the curve which way.
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 read-the-sign-literally trap in a different context.

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