A number added on the outside of \(f\) changes the height of every point and leaves its \(x\)-coordinate alone: the whole curve slides \(3\) in the \(y\) direction.

So \(P(3, 1)\) moves to

\((3, 1 + 3) = (3, 4)\)

A number added on the inside of \(f\) moves the graph sideways, and it goes the opposite way to the sign: \(y = f(x + 3)\) is the curve slid \(3\) in the negative \(x\) direction.

The height is unchanged, so \(P(3, 1)\) moves to

\((3 - (3), 1) = (0, 1)\)

The two parts are the whole idea in one place. Outside the bracket does what it says; inside the bracket does the opposite.

The minus sign is outside \(f\), so it acts on the height. Every \(y\)-coordinate changes sign and every \(x\)-coordinate stays where it is, which is a reflection in the \(x\)-axis.

\(P(5, 7)\) moves to \((5, -7)\)

This time the minus sign is inside \(f\), so it acts on the input. Every \(x\)-coordinate changes sign and the height stays as it was, which is a reflection in the \(y\)-axis.

\(P(5, 7)\) moves to \((-5, 7)\)

Same rule as the translations: outside the bracket works on \(y\), inside the bracket works on \(x\).

The whole curve slides \(-4\) in the \(y\) direction, so the turning point goes with it and its \(x\)-coordinate does not change.

\((-1, -1 - 4) = (-1, -5)\)

Inside the bracket moves the curve sideways, the opposite way to the sign, so \(y = f(x - 3)\) slides it \(-3\) in the negative \(x\) direction.

\((-1 - (-3), -1) = (2, -1)\)

Reflecting in the \(x\)-axis changes the sign of every height and leaves every \(x\)-coordinate alone.

\((-1, 1)\)

The curve now opens the other way, so what was a minimum is a maximum — but it is still the turning point.

A number added outside the brackets changes every height by that amount and leaves every \(x\)-coordinate alone.

Add \(-1\) to each \(y\)-coordinate:

\((-1, -2) \to (-1, -3)\)

\((0, 1) \to (0, 0)\)

\((1, 2) \to (1, 1)\)

\((2, 3) \to (2, 2)\)

\((3, 1) \to (3, 0)\)

Plot these five points and join them with straight lines.

The minus sign is inside the brackets, so it changes every \(x\)-coordinate and leaves every height alone. This reflects the graph in the \(y\)-axis. This graph starts again from \(y = f(x)\): nothing from part (a) is carried over.

Change the sign of each \(x\)-coordinate:

\((-1, -2) \to (1, -2)\)

\((0, 1) \to (0, 1)\)

\((1, 2) \to (-1, 2)\)

\((2, 3) \to (-2, 3)\)

\((3, 1) \to (-3, 1)\)

The point on the \(y\)-axis stays where it is, and is still a point of the new graph: plot it too. Plot the five points and join them with straight lines.

Replacing \(x\) by \(x - 1\) inside the bracket moves the curve sideways. The direction is the opposite of what the sign suggests: \(f(x - 1)\) moves the curve right by \(1\).

So \(1\) is added to each \(x\)-coordinate, and each \(y\)-coordinate stays the same:

\((-2, -14) \to (-1, -14)\)

\((-1, -6) \to (0, -6)\)

\((0, -4) \to (1, -4)\)

\((1, -2) \to (2, -2)\)

\((2, 6) \to (3, 6)\)

Draw a smooth curve of the same shape through these five points.

The dashed graph crosses the \(y\)-axis at \((0, 3)\) and has the \(x\)-axis, \(y = 0\), as its asymptote.

Adding to the function moves the whole graph up 2 units. The point \((0, 3)\) goes to \((0, 5)\).

The asymptote moves with the graph. Asymptote: \(y = 2\).

The dashed graph crosses the \(y\)-axis at \((0, 3)\) and has the \(x\)-axis, \(y = 0\), as its asymptote.

The minus sign reflects the graph in the x-axis, so every height changes sign. The point \((0, 3)\) goes to \((0, -3)\).

The curve is now below the axis and still gets closer to it without reaching it, so the asymptote is unchanged. Asymptote: \(y = 0\).

The two curves are the same shape and the same way up, and one sits directly above the other, so this is a translation.

Compare the turning points. The solid curve turns at \((2, -3)\) and the dashed curve turns at \((2, 0)\): the same \(x\), and a height that has changed by

\(0 - (-3) = 3\)

Translation by the vector \(\begin{pmatrix} 0 \\ 3 \end{pmatrix}\)

"Describe fully" wants the word translation and the vector. A direction in words alone scores less, because it does not say how far.

A move up or down adds a number on the outside of \(f\), and the number is the distance the curve has moved.

\(y = f(x) + 3\)

The two curves are the same shape and at the same height, so this is a translation along the \(x\)-axis.

Compare the turning points. The solid curve turns at \((1, -4)\) and the dashed curve turns at \((3, -4)\), so the curve has moved

\(3 - (1) = 2\) in the \(x\) direction

Translation by the vector \(\begin{pmatrix} 2 \\ 0 \end{pmatrix}\)

A sideways move goes inside the bracket, and the sign there is the opposite of the direction travelled: moving \(2\) in the \(x\) direction is written \(f(x - 2)\).

\(y = f(x - 2)\)

Check it on the turning point. Putting \(x = 3\) into \(f(x - 2)\) gives \(f(1)\), which is where the solid curve turns — so the dashed curve turns at \(x = 3\), as drawn.

One curve opens upwards and the other opens downwards, and the two meet each other on the \(x\)-axis. That is a mirror image rather than a slide.

Check it on the turning points, \((1, -4)\) and \((1, 4)\). The \(x\)-coordinate is unchanged and the height has changed sign, which is exactly what reflecting in the \(x\)-axis does.

Reflection in the \(x\)-axis

A reflection is only described fully once the mirror line is named, so "reflection" on its own is not the whole answer.

Every height has changed sign and no \(x\)-coordinate has moved, so the minus sign goes outside \(f\).

\(y = -f(x)\)

The solid curve turns at \((1, -4)\). Reflecting in the \(x\)-axis keeps the \(x\)-coordinate and changes the sign of the height.

\((1, 4)\)

Moving the graph 5 units to the right replaces \(x\) by \((x - 5)\):

\(f(x) = (x - 5)^3 + 9\)

Expand the cube with \((x + k)^3 = x^3 + 3kx^2 + 3k^2x + k^3\), where \(k = -5\):

\((x - 5)^3 = x^3 + 3(-5)x^2 + 3(-5)^2x + (-5)^3\)

\(= x^3 - 15x^2 + 75x - 125\)

Now add the \(9\) from the original equation to the constant, \(-125 + 9 = -116\):

\(f(x) = x^3 - 15x^2 + 75x - 116\)