Sketch the graph of $y = f(x) \pm g(x)$ by adding or subtracting ordinates: at a zero of one function the sum meets the other, opposite ordinates give a zero, equal ordinates double, and an oblique asymptote can emerge

What this dot point is asking

NESA wants you to start from the graphs of two functions $y = f(x)$ and $y = g(x)$ and sketch their sum $y = f(x) + g(x)$ and difference $y = f(x) - g(x)$, working from the two pictures rather than from a combined formula. The method is called adding ordinates: the word "ordinate" just means the $y$-coordinate of a point, and the whole technique is the single act of adding (or subtracting) the two heights at each value of $x$.

That one idea is the engine. Everything else on this page is a consequence of it, read off the graph at a few special places, the same way graphing the reciprocal is one idea (zero has no reciprocal) applied feature by feature. Because the heights add, the most useful $x$-values are the ones where one height is zero, where the two heights are equal, and where they are exact opposites. Pin those, join them smoothly, and the curve is fixed.

The answer

The adding-ordinates dictionary

Let $s(x) = f(x) + g(x)$ be the sum of two graphed functions. Reading off the two graphs of $f$ and $g$, here is what each special place tells you.

• At a zero of one function, the sum meets the other. If $f(a) = 0$, then $s(a) = 0 + g(a) = g(a)$, so the sum touches the graph of $g$ at $x = a$. Likewise at a zero of $g$, the sum meets $f$. These are the easiest points to plot.
• Opposite ordinates give a zero of the sum. If $f(a) = -g(a)$ (the two heights are equal in size but opposite in sign), then $s(a) = 0$. So the sum crosses the $x$-axis exactly where the graphs of $f$ and $-g$ cross.
• Equal ordinates double. If the two curves meet at $x = a$, so $f(a) = g(a)$, then $s(a) = 2f(a)$: the sum sits at twice the common height.
• The sign of the sum is usually clear once the zeroes are marked. Between consecutive zeroes the sum keeps one sign, which you read from a single test ordinate.
• Asymptotes carry through. If $f \to +\infty$ (or $g \to +\infty$) while the other stays finite or also grows, then $s \to +\infty$; if one tends to a finite limit while the other runs off, the sum follows the one that runs off. A vertical asymptote of either function (where the other is finite) is a vertical asymptote of the sum.
• An oblique asymptote can emerge. If $f \to L$ (a finite constant) as $x \to \pm\infty$ while $g$ is a line, then for large $|x|$ the sum is almost $L + g(x)$, a line: the sum has an oblique asymptote parallel to $g$.

The difference $d(x) = f(x) - g(x)$ is just the sum of $f$ and $-g$, so the same dictionary applies with the second graph reflected in the $x$-axis. In particular, where the two curves meet ($f = g$) the difference is zero, and at a zero of $f$ the difference is $-g$.

Starting clean: a parabola plus a line

The clearest first example combines a parabola and a line. Take $f(x) = x^2$ and $g(x) = 2x$. Every landmark in the dictionary appears, at a different point, so the sketch is built entirely by reading heights.

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At the zero of the parabola, $x = 0$, the line $g$ is also zero, so the sum passes through the origin. The ordinates are opposite where $f = -g$, that is $x^2 = -2x$, giving $x = 0$ and $x = -2$, so the sum has zeroes at $(0, 0)$ and $(-2, 0)$. The curves meet where $f = g$, that is $x^2 = 2x$, at $x = 0$ and $x = 2$; at $x = 2$ both are $4$, so the sum doubles to $s(2) = 8$. Joining these with the vertex $(-1, -1)$ gives the parabola $y = x^2 + 2x$, exactly $s(x) = x(x + 2)$.

The data dots make the addition concrete: at $x = 2$ the two muted heights ($4$ and $4$) stack to the accent height $8$, while at $x = -2$ the heights $4$ and $-4$ cancel to a zero. You did not need the formula at any point; the heights told the whole story, which is the skill NESA is testing.

A rational plus a line: the oblique asymptote

The richest payoff of adding ordinates is spotting an oblique asymptote, a slanted line the curve approaches. Take the hyperbola $f(x) = \dfrac{2}{x}$ and the line $g(x) = x$. Their sum is $s(x) = x + \dfrac{2}{x}$.

For large $|x|$ the term $\dfrac{2}{x}$ shrinks to nothing, so the sum is almost exactly the line $g(x) = x$: that is the oblique asymptote $y = x$. The justification is precise, not hand-waving: $s(x) - x = \dfrac{2}{x} \to 0$ as $x \to \pm\infty$, so the vertical gap between the curve and the line closes. Near $x = 0$, the line is finite but the hyperbola blows up, so the sum has a vertical asymptote at $x = 0$. The sum has no zeroes, since $s(x) = \dfrac{x^2 + 2}{x}$ and $x^2 + 2$ is never zero. The curves meet where $\dfrac{2}{x} = x$, at $x = \pm\sqrt{2}$, and the doubled ordinate there gives the two turning points.

The minimum on the right branch is $\left(\sqrt{2}, 2\sqrt{2}\right)$ and the maximum on the left branch is $\left(-\sqrt{2}, -2\sqrt{2}\right)$, both lying where the line and hyperbola cross and so where the sum doubles. Notice the curve never touches the oblique asymptote; it only homes in on it as $|x|$ grows. This is the single most "beat the textbook" idea on the page: the oblique asymptote is not guessed, it is read off the fading of the $\dfrac{2}{x}$ term.

A sum of straight pieces stays straight

Absolute-value graphs are made of straight segments, and a fact worth pinning is that the sum of two linear pieces is linear. So when you add two absolute-value graphs, the result is again made of straight-line sections, with corners only where one of the originals bends. Take $f(x) = |x + 2|$, which bends at $x = -2$, and $g(x) = |x - 1|$, which bends at $x = 1$. The sum $s(x) = |x + 2| + |x - 1|$ has corners at those two $x$-values and is straight everywhere else.

Working piece by piece: for $x < -2$ both insides are negative, so $s = -(x + 2) - (x - 1) = -2x - 1$; for $-2 \le x \le 1$ the first is positive and the second negative, so $s = (x + 2) - (x - 1) = 3$, a flat segment; for $x > 1$ both are positive, so $s = (x + 2) + (x - 1) = 2x + 1$. The flat middle is the signature of this kind of sum: $|x + 2| + |x - 1|$ is the total distance from $x$ to the points $-2$ and $1$, and that total cannot be less than the gap $3$ between them, achieved everywhere between. So the minimum value is exactly $3$, attained on the whole interval $[-2, 1]$.

Subtracting: where the curves meet, the difference is zero

The difference $d(x) = f(x) - g(x)$ is handled by adding $f$ and $-g$: reflect the second graph in the $x$-axis and add ordinates as before. The one rule worth memorising is the mirror of the sum's "doubling": where the two curves meet ($f = g$), the difference is zero, because subtracting equal heights gives nothing. Take $f(x) = x^2$ and $g(x) = x$. They meet at $x = 0$ and $x = 1$, so the difference $d(x) = x^2 - x$ has zeroes exactly there.

The difference dips to a minimum of $-\dfrac14$ at $x = \dfrac12$, the vertex of $y = x^2 - x$. Between the meeting points the parabola $f$ sits below the line $g$, so $d = f - g$ is negative there; outside, $f$ is above $g$, so the difference is positive. As always, the sign came straight from comparing the two heights, with no need to factor first.

How exam questions ask about sums and differences

The wording points to which landmark to use first.

• "By adding ordinates, sketch $y = f(x) + g(x)$." Add the two heights. Plot the easy points first: at each zero of $f$ the sum meets $g$, and at each zero of $g$ it meets $f$. Then mark the zeroes of the sum (opposite ordinates), double the height where the curves cross, fix the sign of each interval, and finish the tails (including any asymptote).
• "Sketch the difference $y = f(x) - g(x)$." Reflect $g$ in the $x$-axis and add, or subtract heights directly. The zeroes of the difference are where the original curves meet; at a zero of $f$ the difference is $-g$.
• "Find / explain the oblique asymptote." Look for a function with a finite limit added to a line. Show $s(x) - (\text{line}) \to 0$ as $x \to \pm\infty$; the line is the oblique asymptote. State it as an equation $y = mx + c$.
• "Where does $y = f(x) + g(x)$ cross the $x$-axis?" Set $f(x) + g(x) = 0$, that is $f(x) = -g(x)$: the sum is zero exactly where the graphs of $f$ and $-g$ intersect (opposite ordinates).
• "State the domain (and range) of the sum." The domain of $s$ (and of $d$) is the intersection of the domains of $f$ and $g$: a value of $x$ must be allowed for both to add their heights. Read the range from the finished sketch.
• "Is the sum even or odd?" Even $+$ even is even and odd $+$ odd is odd (both terms transform the same way), but even $+$ odd is neither. Decide each function's symmetry first, then apply the rule.

The centrepiece: a parabola plus a line, stage by stage

The cleanest case to drill is a parabola plus a line, because each landmark in the dictionary lands at a separate, tidy point. Take $f(x) = x(x - 3)$ (a parabola with zeroes at $x = 0$ and $x = 3$) and $g(x) = -(x - 3)$ (a line with a zero at $x = 3$). We build $s(x) = f(x) + g(x)$ in four stages.

Stage 1, plot the two graphs and read their zeroes. The parabola $f$ has zeroes at $x = 0$ and $x = 3$ and $y$-intercept $f(0) = 0$. The line $g$ has its zero at $x = 3$ and $y$-intercept $g(0) = 3$. These are the heights we will add.

Stage 2, mark the landmark ordinates. Run the dictionary. At the zero of $f$ at $x = 0$, the sum will meet $g$ at $(0, 3)$. The curves meet where $f = g$, that is $x(x - 3) = -(x - 3)$, so $(x - 3)(x + 1) = 0$, at $x = -1$ (and $x = 3$); at $x = -1$ both equal $4$, so the sum will double to $8$. The ordinates are opposite where $f = -g$, that is $x(x - 3) = x - 3$, so $(x - 3)(x - 1) = 0$, at $x = 1$ (and $x = 3$): the sum will have a zero at $x = 1$. At $x = 3$ both functions are zero, so the sum is zero there too.

Stage 3, draw the sum through the landmarks. Plot the sum at each landmark: $(0, 3)$ where it meets $g$, the doubled point $(-1, 8)$, and the zeroes $(1, 0)$ and $(3, 0)$. Adding the heights everywhere else, $s(x) = x(x - 3) - (x - 3) = x^2 - 4x + 3$, an upward parabola through those points, with vertex at $x = 2$ where $s(2) = -1$.

Stage 4, finish and read off the features. The sum is $s(x) = x^2 - 4x + 3 = (x - 1)(x - 3)$, an upward parabola with zeroes at $x = 1$ and $x = 3$, $y$-intercept $(0, 3)$ and minimum $(2, -1)$. It is positive outside $[1, 3]$ and negative between, and its range is $y \ge -1$. Every feature was found by adding heights, then confirmed against the factored formula.

Two edge cases the textbook rushes

Two subtleties separate a careful answer from a careless one.

Domains must overlap. You can only add two heights where both functions are defined, so the domain of $s(x) = f(x) + g(x)$ (and of the difference) is the intersection of the two domains. For example, $f(x) = \sqrt{x}$ (domain $x \ge 0$) plus $g(x) = \dfrac{1}{x}$ (domain $x \ne 0$) gives a sum defined only for $x > 0$. Never extend the sum into a region where one of the originals does not exist.

When the two functions pull in opposite directions, the tail is not automatic. The asymptote rules are easy when both functions head the same way: if $f \to +\infty$ and $g \to +\infty$ then $s \to +\infty$. But if $f \to +\infty$ while $g \to -\infty$, the sum could go either way (or settle on a finite limit), and you cannot decide from the graphs alone, you need the actual sizes. So flag such a case and, if you have the equations, combine them before claiming a tail. This is exactly why the rational-plus-line example works: there $\dfrac{2}{x} \to 0$ is finite, so the line wins cleanly.