Curve Sketching

Here is a function: f(x) = \dfrac{x^2}{x^2 - 1}. Draw its graph. No calculator, no plotting table, no guessing.

If that seems impossible, here is the thing: the graph is completely determined by six pieces of information, every one of which can be computed by hand in a minute. The shape of any reasonable curve lives in the intersection of these six constraints. Once you know where the function is defined, where it crosses the axes, whether it is symmetric, what it does near infinity and near its trouble spots, where it turns around, and how it bends, the only curve matching all six is the actual graph.

The graph looks like this:

By the end of this article, you will be able to reproduce that picture — including the asymptotes, the local maximum at the origin, and the fact that the two outer branches hug y = 1 as x runs off to infinity — without ever plugging a value into the function except at a handful of key points.

The six questions

Sketching a curve is not one skill but six, applied in a fixed order so that each answer constrains the next. The order matters: if you do concavity before you know the domain, you waste effort on places the function doesn't live.

The order is not arbitrary. Each step uses information from the previous one. Asymptotes depend on the domain. Extrema live inside the domain. Inflections are constrained by both domain and extrema. Once you have all six answers, the graph is the unique curve that satisfies them.

Let's go through each question in turn, then sketch two full examples.

Question 1: Domain

The domain of f is every x for which the formula makes sense. The common obstructions are:

• Division by zero. \frac{1}{x - 2} is undefined at x = 2.
• Even roots of negative numbers. \sqrt{x - 3} is undefined for x < 3.
• Logarithms of non-positives. \ln(x^2 - 4) is undefined where x^2 - 4 \leq 0, i.e. for -2 \leq x \leq 2.
• Trig obstructions. \tan x is undefined at x = \frac{\pi}{2} + k\pi.

Write the domain down before anything else. You now know the set of x-values your graph can occupy. Points outside this set are holes in the graph — no ink there.

Question 2: Intercepts

• y-intercept: plug in x = 0, if 0 is in the domain. The y-intercept is the single point (0, f(0)).
• x-intercepts: solve f(x) = 0. These are the points where the curve crosses or touches the x-axis.

x-intercepts can be hard to find — solving f(x) = 0 is sometimes impossible in closed form. If so, leave this step vague ("one zero somewhere between 1 and 2") and move on; the other steps will recover the rough location.

Question 3: Symmetry

Three quick symmetry checks, each a one-line computation.

Even function. If f(-x) = f(x) for every x in the domain, f is even. The graph is a mirror image across the y-axis. You can draw the right half, then flip.

Odd function. If f(-x) = -f(x) for every x, f is odd. The graph has 180° rotational symmetry about the origin. You can draw the right half, then rotate.

Periodic. If f(x + T) = f(x) for some T > 0, f is periodic with period T. You only need to draw one period; the rest is copies.

For f(x) = x^2/(x^2 - 1), check: f(-x) = \frac{(-x)^2}{(-x)^2 - 1} = \frac{x^2}{x^2 - 1} = f(x). Even. You can sketch the right half of the graph and mirror it.

Polynomials are even if they contain only even powers, odd if they contain only odd powers, and neither otherwise. Rational functions built from such polynomials inherit the parity.

Question 4: Asymptotes

An asymptote is a line (or curve) that the graph approaches but does not meet, in the limit as x runs to some boundary. There are three kinds.

Vertical asymptote at x = a: f(x) \to \pm\infty as x \to a. These happen at the boundaries of the domain — places where the function blows up because of a division-by-zero or similar. To find them, look at every excluded point from the domain step and check the one-sided limits. If \lim_{x \to a^-} f(x) or \lim_{x \to a^+} f(x) is \pm \infty, there is a vertical asymptote at x = a.

Horizontal asymptote at y = L: f(x) \to L as x \to \infty or x \to -\infty. Compute \lim_{x \to \infty} f(x) and \lim_{x \to -\infty} f(x). If either limit is a finite number, that's a horizontal asymptote.

Oblique asymptote y = mx + b: the graph approaches a non-horizontal line at infinity. For a rational function \frac{p(x)}{q(x)}, this happens when the degree of p is exactly one more than the degree of q. Do polynomial long division: \frac{p(x)}{q(x)} = (mx + b) + \frac{\text{remainder}}{q(x)}, and since the remainder fraction vanishes as x \to \infty, the graph approaches the line y = mx + b.

For f(x) = x^2/(x^2 - 1): the denominator vanishes at x = \pm 1, so check those. Near x = 1, the numerator approaches 1, the denominator approaches 0, so f \to \pm \infty — vertical asymptote at x = 1. Similarly at x = -1. For x \to \pm \infty, divide numerator and denominator by x^2: f(x) = \frac{1}{1 - 1/x^2} \to 1. Horizontal asymptote at y = 1, in both directions. No oblique asymptote.

Question 5: Monotonicity and extrema

Compute f'(x), find its zeroes and the points where it is undefined, and build a sign chart. From the first derivative test, you get the intervals of increase and decrease, and every local extremum.

For f(x) = x^2/(x^2 - 1):

Zeroes: -2x = 0 gives x = 0. Undefined: x = \pm 1 (but those are already outside the domain). Sign: the denominator (x^2 - 1)^2 is always positive where defined, so the sign of f' is the sign of -2x. Positive for x < 0, negative for x > 0.

So f is increasing on (-\infty, -1) and (-1, 0), decreasing on (0, 1) and (1, \infty). At x = 0, f' changes from positive to negative — local maximum. f(0) = 0, so the local max value is 0.

Question 6: Concavity and inflection

Compute f''(x), find where it equals zero or is undefined, and build a sign chart. From the second derivative, you get concavity: f'' > 0 means concave up (smile), f'' < 0 means concave down (frown). Points where concavity actually changes are inflection points.

For f(x) = x^2/(x^2 - 1), differentiate f'(x) = \frac{-2x}{(x^2 - 1)^2} using the quotient rule:

Numerator: 6x^2 + 2, always positive. So f'' has the sign of (x^2 - 1)^3, which is the same as the sign of x^2 - 1. Positive when |x| > 1, negative when |x| < 1 (excluding \pm 1, which are outside the domain).

So: concave up on (-\infty, -1) and (1, \infty); concave down on (-1, 1). No inflection point inside the domain (the sign change happens at x = \pm 1, which are holes, not inflection points).

Putting it all together

Six answers, one graph. For f(x) = x^2/(x^2 - 1):

Read this table back as a drawing. The middle branch lives between -1 and 1, passes through the origin (where it has a local max), and is concave down — so it looks like the top of a frown with peak at the origin. As x \to \pm 1, the function dives to -\infty (because near x = 1, the numerator approaches 1 and the denominator approaches 0 from the negative side for x slightly less than 1, so f \to -\infty). The outer branches live for |x| > 1, are concave up, decreasing on the right and increasing on the left, approach y = 1 as x \to \pm\infty, and come down from +\infty at the asymptote. That is exactly the graph at the top of this article.

Two worked examples

Notice that without drawing a single extra point, the sketch is determined by six numbers and a few signs.

The second example shows two things worth remembering. First, oblique asymptotes happen whenever the function is eventually linear plus something small. Rewriting f(x) so that the linear part is visible (as here, where x is already sitting next to a vanishing \frac{1}{x}) saves you from polynomial long division. Second, concavity changes across a gap in the domain do not produce inflection points — the definition of an inflection point requires the change to happen at a point on the curve.

Common confusions

• Thinking the checklist must be done in exact order, with no looping back. The order is a suggestion, not a law. If step 5 reveals a critical point you missed in step 2 (because it happens to also be an intercept), go back and add it. Sketching is about building up a consistent picture; you can revise.

• Assuming every zero of f'' is an inflection. An inflection requires an actual sign change of f'', not just f'' = 0. f(x) = x^4 at x = 0 is the standard counterexample.

• Missing oblique asymptotes. For rational functions where \deg(p) = \deg(q) + 1, there is always an oblique asymptote. Polynomial long division will find it. Students often stop at horizontal asymptotes and miss this case.

• Drawing the curve crossing a vertical asymptote. Vertical asymptotes are hard barriers — the function explodes to \pm\infty on approach. The graph does not cross them. Different branches on either side of a vertical asymptote are disconnected.

• Drawing a horizontal asymptote as a hard barrier. Horizontal asymptotes are the opposite — the graph can cross them, repeatedly. The asymptote says only what happens in the limit x \to \pm\infty. Near the origin, the function is free to weave above and below.

• Confusing "concave down" with "decreasing." A function can be concave down and still increasing (think of the right half of a parabola opening downward — concave down, but increasing on (-\infty, 0) if its vertex is at 0). Concavity and monotonicity are independent.

Going deeper

Oblique asymptotes beyond rational functions

Oblique asymptotes can appear for functions that are not rational. The general test: a function f has an oblique asymptote y = mx + b as x \to \infty if

and both limits exist and are finite. If only m exists and b does not, there is no asymptote even though the function grows approximately linearly.

Example: f(x) = \sqrt{x^2 + 1}. Compute m = \lim_{x \to \infty} \frac{\sqrt{x^2+1}}{x} = 1. Then b = \lim_{x \to \infty}(\sqrt{x^2+1} - x). Rationalise: \sqrt{x^2+1} - x = \frac{1}{\sqrt{x^2+1} + x} \to 0. So b = 0, and the oblique asymptote is y = x. For x \to -\infty, by similar calculation, the asymptote is y = -x. These two asymptotes together show that the graph of \sqrt{x^2+1} looks like a hyperbola — which it is, with equation y^2 - x^2 = 1.

Curve sketching for parametric and implicit curves

The six-step checklist is for functions y = f(x). For a parametric curve (x(t), y(t)) or an implicit curve F(x, y) = 0, the picture requires a different toolkit, but the spirit is the same: identify the features that constrain the shape, and sketch each feature separately before combining them. The main change is that monotonicity and concavity now have to be computed parametrically — for a curve given parametrically, \frac{dy}{dx} = \frac{dy/dt}{dx/dt}, and the sign of this ratio (plus the direction of motion along the curve) tells you about monotonicity.

Relation to Taylor polynomials

Every local feature in the checklist — intercept, extremum, inflection — is captured by a local Taylor expansion. Near a point c, the function looks like

The first term says where the curve is; the second term gives the tangent line; the third term gives the bending. So the first-, second-, and third-order behaviour of the curve near c is exactly the information the sketching checklist collects. In a deeper sense, sketching a curve and understanding its local Taylor polynomial at every point are the same activity.

Where this leads next

Curve sketching is the synthesis article for differential calculus — almost everything you have learned about derivatives shows up here. The next ideas apply these tools to real problems.

• Optimization Problems — using the same extrema tools to solve "largest / smallest" problems from geometry and physics, without having to draw the full curve.
• Concavity and Points of Inflection — a deeper look at the second-derivative piece of the checklist.
• Mean Value Theorems — the theoretical engine that connects sign of the derivative to shape of the curve.
• Integration - Introduction — what happens when you start accumulating the area under the curve you just sketched.
• Tangent and Normal — finding the tangent line at a specific point, the local version of "how does the curve go."