= Solution
For each sample, include both its marker and its bar. A bar's lower vertical limit is $\min(0,y_i)$, including negative bars below the horizontal axis. Its horizontal half-width is $w_i/2$, rather than $w_i$. Its marker extends by $C$ in both directions. Therefore that sample's <axis-aligned bounding box> is
$$
\left[x_i-\max(C,w_i/2),x_i+\max(C,w_i/2)\right]
\times
\left[\min(0,y_i-C),\max(0,y_i+C)\right].
$$
Taking successive bounding unions includes every sample. Finally unite with $[-M,M]\times[-M,M]$ to include the stipulated axes. The printed upper-corner formula has one extra closing parenthesis; both the lower-corner formula and the picture confirm the coordinatewise interpretation used here.<Object Oriented Programming examination paper of the University of Oxford (2015)>[p. 5, Question 2(f), axes formula and Figure 2]{ref}
The following commented <Java> <object method> may be placed in a graph-view utility <software class>. It uses `Rect` from part (a) and one <iterator snapshot> from part (e).
``
static Rect barBounds(DataList data, double C, double M) {
if (!Double.isFinite(C) || !Double.isFinite(M) || C <= 0 || M <= 0)
throw new IllegalArgumentException("positive finite margins required");
List<Point> p = new ArrayList<Point>();
for (Point point : data) p.add(point);
int n = p.size();
if (n < 2) throw new IllegalArgumentException("at least two points required");
// This central square supplies the required minimum axis extents.
Rect bounds = new Rect(new Point(-M, -M), new Point(M, M));
for (int i = 0; i < n; ++i) {
double x = p.get(i).getX(), y = p.get(i).getY();
double width;
if (i == 0) {
width = p.get(1).getX() - x;
} else if (i == n - 1) {
width = x - p.get(i - 1).getX();
} else {
width = Math.min(x - p.get(i - 1).getX(),
p.get(i + 1).getX() - x);
}
// Include whichever is wider: marker or half of the bar.
double half = Math.max(C, width / 2.0);
// The bar reaches the baseline; the marker reaches y +/- C.
Rect item = new Rect(new Point(x - half, Math.min(0.0, y - C)),
new Point(x + half, Math.max(0.0, y + C)));
bounds = bounds.Union(item);
}
return bounds;
}
``
This uses the usual assumption that coordinate differences and expanded bounds are representable as finite `double` values. `Rect` rejects overflowed bounds; numerical range extension is not part of the requested geometry. As with all <floating-point arithmetic>, the calculation approximates the corresponding exact <real number> formulas.
Every side produced by the coordinatewise extrema is required by some bar, marker enclosure or axis extent. The result is therefore the smallest guaranteed <rectangle> for the specified marker enclosures and axes. If $X_l,Y_l,X_h,Y_h$ denote the graph-only bounds, the final answer is
$$
\boxed{[\min(-M,X_l),\max(M,X_h)]\times[\min(-M,Y_l),\max(M,Y_h)].}
$$
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