Cube first, then deal with the other two terms.

\((-2)^3 = -8\), so \(y = -8 - 3(-2) + 4 = 2\)

Cubing keeps the sign: a negative number cubed is still negative, unlike a negative number squared.

\((1)^3 = 1\), so \(y = 1 - 3(1) + 4 = 2\)

\((3)^3 = 27\), so \(y = 27 - 3(3) + 4 = 22\)

Plotted, these seven points give the double bend of a cubic — down, up, and away — rather than the single turn of a parabola.

An equation is solved by a graph wherever \(y\) takes the value on the right — here \(0\), which is the \(x\)-axis itself.

So read off every place the curve cuts the \(x\)-axis. There are three:

\(x = -1\), \(x = 1\) and \(x = 4\)

Three crossings is the most a cubic can have, and it is what the double bend in the middle of the curve makes room for.

Find \(2\) on the \(x\)-axis and go up or down to the branch of the graph on the right, then read across to the \(y\)-axis.

\(y = 4\)

The arithmetic agrees: \(8 \div (2) = 4\).

This time start at \(1\) on the \(y\)-axis and go across to the curve, then read down or up to the \(x\)-axis.

\(x = 8\)

Again it checks out: \(8 \div (8) = 1\).

\(y = 6 \div 2000 = 0.003\)

On the grid above, a point that high would sit right on top of the \(x\)-axis — too close to draw apart from it. It is still not \(0\), though, and that is the whole point.

A point is on the \(x\)-axis when \(y = 0\).

A fraction is zero only when its top is zero, and the top here is always \(6\). So \(y\) can never be exactly \(0\), no matter how big \(x\) is.

Part (a) shows what happens instead: \(y\) gets smaller and smaller — \(0.003\) at \(x = 2000\), smaller still further out — but it keeps going without ever arriving.

A point is on the \(y\)-axis when \(x = 0\).

Putting \(x = 0\) into \(y = \frac{ 6 }{ x }\) asks for \(6 \div 0\), which has no value. There is no point on the graph at \(x = 0\) at all.

That is why the graph is drawn as two separate pieces rather than one: there is nothing to join them through.

\(y = 38 \div (-4) = -9.5\)

Dividing a positive number by a negative one gives a negative answer, which is why the whole left-hand half of the table is below zero.

\(y = 38 \div 16 = 2.375\)

\(y = 38 \div (4) = 9.5\)

Every pair of columns the same distance from the middle gives the same answer with opposite signs, so once one half of the table is done the other half is a mirror image of it.

The point is on the curve, so its coordinates fit the equation. Put \(x = 2\) and \(y = 6\) into \(y = \frac{ a }{ x }\):

\(6 = \frac{ a }{ 2 }\)

Multiply both sides by \(2\):

\(a = 6 \times 2 = 12\)

For a reciprocal graph, \(x\) times \(y\) is the same number at every point on the curve, and that number is \(a\).

The equation is \(y = \frac{ 12 }{ x }\), so

\(y = 12 \div 6 = 2\)

Put \(y = -3\) into \(y = \frac{ 12 }{ x }\):

\(-3 = \frac{ 12 }{ x }\)

Multiplying both sides by \(x\) and then dividing by \(-3\) gives

\(x = 12 \div (-3) = -4\)

A negative \(y\) puts the point on the other branch, where \(x\) is negative too — which is the check that the sign is right.

Average speed = distance \(\div\) time, so for this journey \(\text{speed} = \frac{ 24 }{ \text{time} }\).

In 1 hour the whole 24 miles is covered, so the speed is 24 mph. At 2 hours: \(24 \div 2 = 12\). At 3 hours: \(24 \div 3 = 8\).

At 4 hours: \(24 \div 4 = 6\). At 6 hours: \(24 \div 6 = 4\).

Plot \((1, 24)\), \((2, 12)\), \((3, 8)\), \((4, 6)\) and \((6, 4)\), with time across and speed up.

Join them with one smooth curve. It falls steeply at first and flattens out as the time grows, and it does not touch either axis.

Put each missing \(x\) value into \(y = 1 - x^3\), cubing first.

\(x = -2\): \((-2)^3 = -8\), so \(y = 1 - (-8) = 9\)

\(x = 0\): \((0)^3 = 0\), so \(y = 1 - (0) = 1\)

\(x = 2\): \((2)^3 = 8\), so \(y = 1 - (8) = -7\)

A negative number cubed stays negative.

Each column of the completed table is one point: \(x\) across, \(y\) up.

Plot \((-2, 9)\), \((-1, 2)\), \((0, 1)\), \((1, 0)\) and \((2, -7)\).

Join them with one smooth curve, not with straight lines. A cubic bends through the middle, so the curve passes through \((0, 1)\) and keeps going in the same direction at both ends.

The four options share their constant, so the shape of the graph has to decide.

The curve is a single U shape, symmetrical about the y-axis, with a lowest point on the y-axis. That is a parabola, an \(x^2\) graph.

Option D has the right constant, but a cubic keeps climbing on the right where this U shape turns round.

The equation is option C: \(y = x^2 + 2\).

The number on the top of the fraction is \(8\), which is positive.

A positive number divided by a negative number is negative, so only a positive value of x can give a positive y. Dividing by a positive number closer to zero makes y larger, so the greatest y comes from the positive value of x nearest to zero.

Here that is \(x = 0.25\), which gives \(y = \frac{ 8 }{ 0.25 } = 32\).

The answer is \(0.25\).