Curve Sketching: The Step-by-Step Method (and the Mistakes That Cost Marks)

Curve sketching is the point where fifth-year analysis has to work all together: limits, derivatives, rational inequalities, in the right order, without a mistake from the second step surviving intact all the way to the seventh. It comes up in the seconda prova, the Maturità maths paper, almost every year, either as a full problem or split across the short questions. Marking class tests and mock papers, in our courses we see that almost nobody skips a step: marks are lost inside the steps, in small errors that drag through to the graph. An $x = 1$ written instead of $x = \pm 1$ in the domain wipes out an asymptote; a limit with no sign makes the drawing impossible; a derivative with the minus in the wrong place swaps the maximum for the minimum. This guide works through the whole method on a single function, $f(x) = \dfrac{x^3}{x^2-1}$, stopping at every step on the exact point where the grade slips.

What are the steps of curve sketching, and why does the order matter?

There are seven steps: domain; symmetry (even or odd function); intercepts and sign; limits at the edges of the domain, giving the asymptotes; the first derivative, for increasing/decreasing behaviour and maxima/minima; the second derivative, for concavity and points of inflection; the graph. The order is not a textbook convention. The domain tells you where to calculate the limits. The sign tells you which side of the $x$-axis the curve approaches each asymptote from. The arrows from the first derivative have to agree with the limits: a stretch that is falling cannot end up at $+\infty$. The concavity has to agree with the maxima and minima: a minimum inside a concave-down stretch cannot exist. Every step checks the one before it — but only if you actually do it afterwards.

The function we shall work on is $f(x) = \dfrac{x^3}{x^2-1}$. It is rational, so the calculations stay manageable, but it contains almost every trap we see in class tests: two points excluded from the domain, a symmetry, an oblique asymptote, a point where the derivative vanishes without giving either a maximum or a minimum, a point of inflection with horizontal tangent, and a change of concavity that is not a point of inflection.

Step 1: how do you work out the domain, and why does it need rewriting in every table?

The domain gathers up the conditions for existence: denominators different from zero, even-index radicands greater than or equal to zero, logarithm arguments strictly positive. If there is more than one condition, the domain is their intersection. For our function there is only one:

The first mistake shows up right here, and it only looks trivial: from $x^2 = 1$ people write $x = 1$ and stop there. Half the domain disappears, and with it the asymptote $x = -1$, a limit, an interval of monotonicity. Whoever makes this mistake does not notice again until the graph, where the curve sails calmly through $x = -1$ as if the function were defined there.

The second mistake is more subtle and costs more: the domain gets calculated, written down, and then forgotten. The sign table has no double bar at $-1$ and $1$; the intervals of increase are written as if the function existed everywhere; the point of inflection gets looked for even at the excluded points. The domain is the first row of every table you write from here on, not a result to file away and forget. When there are two conditions, the classic mistake is union instead of intersection: for $\sqrt{x-1} + \ln(3-x)$ the conditions $x \geq 1$ and $x < 3$ give $[1, 3)$, not "the whole of $\mathbb{R}$" as we sometimes see written.

Step 2: is the function even or odd, and how do you use that?

A function is even if $f(-x) = f(x)$, and its graph is symmetric about the $y$-axis; it is odd if $f(-x) = -f(x)$, and its graph is symmetric about the origin. Either way, the domain has to be symmetric about $0$, otherwise the question makes no sense: $\ln x$ is neither even nor odd, because it does not exist to the left of $0$. Our domain is symmetric, so we calculate:

The function is odd. From here on we could study it only for $x \geq 0$ and get the rest by central symmetry: a minimum at $x_0$ corresponds to a maximum at $-x_0$ with the opposite ordinate, a limit of $+\infty$ at $x_0^+$ corresponds to $-\infty$ at $-x_0^-$. In a class test that is a real saving of time, provided you write explicitly "the function is odd, so..." every time you use it.

There are two mistakes here. The first is declaring the parity by eye: you see $x^2$ in the denominator and write "even", without noticing $x^3$ in the numerator; or you calculate $(-x)^3 = x^3$, losing the sign of the odd power. The second only shows up at the end, and it is the most recognisable: the right-hand half of the graph is correct and the left-hand half is flipped as in a mirror, giving two minima instead of one maximum and one minimum. An odd function is rotated by half a turn about the origin, not reflected. If you are not sure, forget the shortcut and do all the working on both sides: it costs five minutes, not marks.

Step 3: intercepts with the axes, and the sign of the function

The intercept with the $y$-axis is found by calculating $f(0)$, provided $0$ belongs to the domain: here $f(0) = 0$, so the curve passes through the origin. The intercepts with the $x$-axis are the solutions of $f(x) = 0$, that is the zeros of the numerator that lie in the domain: $x^3 = 0$, again $x = 0$. A single point for both axes.

The sign is worked out with the inequality $f(x) > 0$, and for a fraction this means studying the numerator and the denominator separately and combining the signs:

On the interval $x < -1$ the numerator is negative and the denominator positive: $f < 0$. Between $-1$ and $0$, negative over negative: $f > 0$. Between $0$ and $1$, positive over negative: $f < 0$. Beyond $1$, positive over positive: $f > 0$. In summary, the curve sits above the $x$-axis on $(-1, 0)$ and on $(1, +\infty)$, and below it on $(-\infty, -1)$ and on $(0, 1)$. This is consistent with the function being odd: the regions swap over as you rotate about the origin.

The mistake we correct most often at this step is moving the denominator across: from $\dfrac{x^3}{x^2-1} > 0$ to $x^3 > x^2 - 1$, and from there to a cubic inequality nobody knows how to solve. Multiplying by $x^2 - 1$ is only valid where $x^2 - 1$ is positive; where it is negative the direction of the inequality flips, and the table with the separate sign study does exactly that job for you. The other mistake is reading the sign off the numerator alone, "positive for $x > 0$", which is false on the interval $(0, 1)$. Keep the result of this step in mind: the limits will have to confirm it shortly, and if they do not, one of the two is wrong.

Step 4: limits at the edges of the domain, and the asymptotes

The edges of the domain are the points where a limit has to be calculated: $-\infty$, $-1^-$, $-1^+$, $1^-$, $1^+$, $+\infty$. Six limits, one per row, and the empty row is almost always the same one: $-\infty$, or the left-hand side of an excluded point. Because the function is odd, we can calculate the three on the right and get the others from them.

At $x = 1$ the numerator is $1$ and the denominator tends to $0$; the limit is infinite, and all that remains is to settle the sign with a test value. At $x = 1.1$, $x^2 - 1 = 0.21 > 0$; at $x = 0.9$ it is $-0.19 < 0$.

Vertical asymptote $x = 1$. By symmetry, $x = -1$ is a vertical asymptote too, with $-\infty$ coming in from the left and $+\infty$ from the right. The cross-check against step 3: on $(0, 1)$ the function is negative, and indeed it drops to $-\infty$; on $(1, +\infty)$ it is positive and starts from $+\infty$. If the sign had said one thing and the limit the opposite, one of the two pieces of working would need redoing before going any further.

At infinity the numerator has degree $3$ and the denominator degree $2$: the limit is infinite, there is no horizontal asymptote, and there is a candidate oblique one, because the gap in degree is exactly one. The shortest route is polynomial division: $x^3 = x \cdot (x^2 - 1) + x$, so

Oblique asymptote $y = x$ on both sides. The remainder $\dfrac{x}{x^2-1}$ also tells you which side of the line the curve runs on: positive for $x > 1$, so above the line on the right; negative for $x < -1$, so below it on the left. The full procedure with $m$ and $q$, the case where the excluded point is a hole rather than an asymptote, and the two different horizontal asymptotes of non-rational functions are all in our guide to vertical, horizontal and oblique asymptotes; if the problem lies further back, in calculating the limits, start again from our guide to limits for the Maturità.

The mistake worth flagging here, because it ruins everything that follows, is the limit written without a sign: "$\lim_{x \to 1} f(x) = \infty$", a single line for two sides. With that result the graph cannot be drawn, because you do not know whether the curve is rising or falling as it approaches the line, and you usually end up guessing.

Step 5: the first derivative — increasing and decreasing, maxima and minima

The derivative of a quotient is $\dfrac{u'v - uv'}{v^2}$, with $u = x^3$ and $v = x^2 - 1$:

The denominator is a square, positive over the whole domain, so it plays no part in the sign. In the numerator, $x^2 \geq 0$, with a zero at $x = 0$; the sign of $f'$ is decided by $x^2 - 3$, positive for $x < -\sqrt{3}$ and $x > \sqrt{3}$, negative in between. Remembering that the function does not exist at $\pm 1$:

• $f$ is increasing on $(-\infty, -\sqrt{3})$ and on $(\sqrt{3}, +\infty)$;
• $f$ is decreasing on $(-\sqrt{3}, -1)$, on $(-1, 1)$ and on $(1, \sqrt{3})$.

At $x = -\sqrt{3}$ the derivative goes from positive to negative: a relative maximum, with $f(-\sqrt{3}) = \dfrac{-3\sqrt{3}}{3-1} = -\dfrac{3\sqrt{3}}{2} \approx -2.60$. At $x = \sqrt{3}$ it goes from negative to positive: a relative minimum, $f(\sqrt{3}) = \dfrac{3\sqrt{3}}{2} \approx 2.60$. At $x = 0$ the derivative vanishes but stays negative on both sides: the tangent is horizontal, the function keeps falling, and there is neither a maximum nor a minimum. What happens at that point is for the second derivative to tell us.

Three mistakes cluster here, and they are the costliest in the whole problem.

The first is the sign in the quotient rule: anyone who writes $x^3 \cdot 2x - 3x^2(x^2-1)$ in the numerator gets $f'$ with the opposite sign, the maximum becomes a minimum and vice versa, and the final graph ends up with a maximum on a stretch where the function is climbing towards $+\infty$. It is a contradiction the limits reveal on their own, once you compare them against the table.

The second is "$f'(x_0) = 0$, so there is a maximum or a minimum". At $x = 0$ that is not true, and you can only see it from the sign of $f'$ on the left and on the right: with no change of sign there is no extremum. In class tests we regularly read "minimum at $x = 0$", and whoever writes it then draws a curve that climbs back up after the origin, against the sign and against the limit at $1^-$.

The third is a matter of form but it is penalised all the same: the maximum reported simply as "$x = -\sqrt{3}$", with no ordinate. A maximum point has two coordinates, and without $f(-\sqrt{3})$ you cannot place it on the graph. For the rules of differentiation, including the quotient rule with every step worked through, see our guide to derivatives with rules and examples.

Step 6: the second derivative — concavity and points of inflection

We differentiate $f'$ in the form $\dfrac{x^4 - 3x^2}{(x^2-1)^2}$, again with the quotient rule, and factor out $(x^2-1)$ in the numerator before expanding anything at all:

The factor $x^2 + 3$ is always positive, and the cube has the same sign as $x^2 - 1$, so the sign of $f''$ is the product of the sign of $x$ and that of $x^2 - 1$. Concave up ($f'' > 0$) on $(-1, 0)$ and on $(1, +\infty)$; concave down on $(-\infty, -1)$ and on $(0, 1)$. At $x = 0$ the second derivative vanishes and changes sign: a point of inflection at the origin, with horizontal tangent because $f'(0) = 0$. The curve crosses the origin flattening out, like $y = x^3$. The checks agree: the minimum at $\sqrt{3}$ sits on a stretch that is concave up, the maximum at $-\sqrt{3}$ on a stretch that is concave down.

The first way to lose marks here is arithmetic: differentiating $f'$ without having simplified it first, or expanding $(x^2-1)^4$ in the denominator. Factoring out $(x^2-1)$ brings the denominator down to a cube and reduces the numerator to two terms; anyone who expands everything usually does not make it to the bottom of the page.

The second is conceptual, and we see it in almost every curve-sketching problem with a vertical asymptote: "the second derivative changes sign at $x = 1$, so there is a point of inflection at $x = 1$". No. A point of inflection is a point of the graph, and at $x = 1$ the graph does not exist. Concavity can change across a vertical asymptote, and here it does so at both of them, without there being any point of inflection at all. The same rule holds in reverse: $f''(x_0) = 0$ with no change of sign is not a point of inflection — $y = x^4$ has $f''(0) = 0$ and a minimum at the origin.

Step 7: how do you draw the graph, and what to check before handing it in?

You start from the skeleton: the axes, the three dashed lines $x = -1$, $x = 1$, $y = x$, and the three points you have found — the origin, the maximum $\left(-\sqrt{3}, -\dfrac{3\sqrt{3}}{2}\right)$ and the minimum $\left(\sqrt{3}, \dfrac{3\sqrt{3}}{2}\right)$ — with a short horizontal dash at each to mark the tangent. Then the three branches, one for each piece of the domain.

For $x < -1$ the curve comes in from $-\infty$ below the line $y = x$, climbs to the maximum at $-\sqrt{3}$ and falls back to $-\infty$ against the asymptote $x = -1$, staying below the $x$-axis throughout and concave down. Between $-1$ and $1$ it falls from $+\infty$, crosses the origin with a horizontal tangent while changing concavity, and plunges to $-\infty$ against $x = 1$: an elongated S, positive to the left of the origin and negative to the right. For $x > 1$ it starts from $+\infty$, falls to the minimum at $\sqrt{3}$ and climbs back up, hugging the line $y = x$ from above, always above the $x$-axis, concave up.

Before handing it in, the graph has to be checked against every table that produced it, not just looked at:

1. The curve never touches or crosses a vertical asymptote. If two branches meet at $x = \pm 1$, the domain has been forgotten.
2. On every interval the curve sits on the side of the $x$-axis that the sign table says it should.
3. At the maxima and minima the tangent is horizontal, and so it is at the point of inflection with horizontal tangent: the origin is not crossed on the diagonal.
4. A minimum sits on a stretch that is concave up, a maximum on a stretch that is concave down. If that does not check out, either the second derivative or the sign of the first one is wrong.
5. If the function is odd, rotating the page by half a turn should bring the graph back onto itself.

The most frequent drawing mistake is not conceptual but a matter of scale: the maximum at $-2.6$ ends up plotted at $-1$ and the minimum at $+2.6$ at $+1$, and the graph, though "correct", is no longer readable or easy to check against the working. Before you plot anything, choose a scale so that $\sqrt{3} \approx 1.73$ and $2.6$ both sit comfortably on the page.

Where does curve sketching come back after the Maturità?

In the maths section of the TOLC-I — 20 questions in 50 minutes, a little over two minutes each — you cannot do a complete curve-sketching problem, but the individual steps come up on their own, picked out one at a time: the domain of a function with a root and a logarithm, the sign of a fraction, the interval where a function is increasing, the correct graph among four options. Anyone who has the method in hand recognises from the shape of the function which step is needed and does just that; anyone who has learned it as a sequence to recite starts from the domain and runs out of time. The same goes for the other engineering and economics admission tests.

At university, in your first Analysis exam, curve sketching comes back as an exercise in its own right, usually on functions with an absolute value, roots or exponentials, with one extra step that is barely touched on at school: studying differentiability at points where $f'$ does not exist, with corners and cusps. The method in this guide stays exactly the same; it is the algebra that gets harder. If you are already looking beyond the Maturità, our admission test preparation paths work on exactly this shift, from the long problem to the two-minute question.

And if curve sketching still feels like a list of steps to learn by heart, the problem is usually further back — in rational inequalities, in limits, in the sign of a product — and needs rebuilding from there. Tell us about your situation and book your first lesson: we start from the exact point where the method broke down, not from the beginning of the syllabus.

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