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Curve sketching

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Curve Sketching in Calculus

Curve sketching uses a function's first and second derivatives to map out its shape before drawing it: domain, intercepts, asymptotes, increasing and decreasing intervals, local extrema, concavity, and inflection points. This topic walks through the full process with a worked polynomial and rational function example.

What Is Curve Sketching in Calculus?

Curve sketching is the process of using algebra and calculus together to draw an accurate graph of a function without relying on a graphing tool. Instead of plotting dozens of points, you use the function's derivatives to figure out exactly where it rises, where it falls, where it curves upward, where it curves downward, and where any breaks or asymptotes occur. Once you know this information, only a handful of key points need to be plotted, and the rest of the curve follows naturally.

Why Use Derivatives to Sketch a Curve?

The first derivative, \(f'(x)\), tells you the slope of the tangent line at every point on the graph. When \(f'(x) > 0\), the function is increasing; when \(f'(x) < 0\), it is decreasing. The points where \(f'(x) = 0\) or is undefined are the critical numbers of a function, and they are exactly where local maximums and minimums can occur.

The second derivative, \(f''(x)\), describes concavity. When \(f''(x) > 0\), the graph curves upward like a cup (concave up); when \(f''(x) < 0\), it curves downward (concave down). A point where concavity switches sign is called an inflection point.

Steps for Curve Sketching

  1. Find the domain of \(f(x)\) and note any values that are excluded.
  2. Find the \(y\)-intercept by evaluating \(f(0)\), and find \(x\)-intercepts by solving \(f(x) = 0\).
  3. Check for symmetry: if \(f(-x) = f(x)\) the graph is even, and if \(f(-x) = -f(x)\) it is odd.
  4. Determine vertical asymptotes (values that make the function undefined) and horizontal or end behavior asymptotes.
  5. Compute \(f'(x)\), solve for critical numbers, and build a sign chart to identify intervals of increase and decrease along with local maximums and minimums.
  6. Compute \(f''(x)\), find possible inflection points, and build a sign chart to identify concave up and concave down intervals.
  7. Plot the intercepts, critical points, and inflection points, then connect them following the increasing, decreasing, and concavity information you found.

Worked Example: Sketching a Polynomial Function

Sketch \(f(x) = x^3 - 3x^2 + 2\).

Domain and intercept: the domain is all real numbers, and \(f(0) = 2\), so the \(y\)-intercept is \((0, 2)\). There are no vertical or horizontal asymptotes since this is a polynomial.

First derivative: \(f'(x) = 3x^2 - 6x = 3x(x - 2)\), so the critical numbers are \(x = 0\) and \(x = 2\). Testing values in each interval shows \(f'(x) > 0\) for \(x < 0\), \(f'(x) < 0\) for \(0 < x < 2\), and \(f'(x) > 0\) for \(x > 2\). This means the graph increases, then decreases, then increases again, giving a local maximum at \((0, 2)\) and a local minimum at \((2, -2)\).

Second derivative: \(f''(x) = 6x - 6\), which equals zero at \(x = 1\). Since \(f''(x) < 0\) for \(x < 1\) and \(f''(x) > 0\) for \(x > 1\), the graph is concave down then concave up, with an inflection point at \((1, 0)\).

Graph of f of x equals x cubed minus 3x squared plus 2, showing a local maximum at x equals 0, a local minimum at x equals 2, and an inflection point at x equals 1 Plot of y = x**3 - 3*x**2 + 2 for x in [-2, 4] -2 -1 0 1 2 3 4 -20 -10 0 10 20 x f(x) Local maximum Inflection point Local minimum
Graph of \(f(x) = x^3 - 3x^2 + 2\) showing the local maximum, local minimum, and inflection point.

Putting it all together: the curve rises from the lower left, reaches a peak at \((0, 2)\), dips down through the inflection point at \((1, 0)\) to a valley at \((2, -2)\), then rises again to the upper right.

Worked Example: Sketching a Rational Function with an Asymptote

Sketch \(g(x) = \dfrac{2x}{x - 1}\).

Domain and asymptotes: \(g(x)\) is undefined at \(x = 1\), giving a vertical asymptote there. As \(x \to \pm\infty\), \(g(x) \to 2\), so \(y = 2\) is a horizontal asymptote.

Intercepts: \(g(0) = 0\), so the graph passes through the origin.

First derivative: using the quotient rule, \(g'(x) = \dfrac{-2}{(x - 1)^2}\), which is negative everywhere it is defined. There are no critical numbers, so the function is decreasing on both \((-\infty, 1)\) and \((1, \infty)\), with no local extrema.

Second derivative: \(g''(x) = \dfrac{4}{(x - 1)^3}\), which is negative for \(x < 1\) (concave down) and positive for \(x > 1\) (concave up). Concavity changes across the asymptote, but since \(g(x)\) is undefined at \(x = 1\), this is not counted as a true inflection point.

Graph of g of x equals 2x over x minus 1, showing a vertical asymptote at x equals 1 and a horizontal asymptote at y equals 2 Plot of y = (2*x)/(x-1) for x in [-6, 8] -6 -4 -2 0 2 4 6 8 -20 -10 0 10 20 30 x g(x) Intercept at origin
Graph of \(g(x) = \dfrac{2x}{x - 1}\) showing the vertical asymptote at \(x = 1\) and horizontal asymptote at \(y = 2\).

Where Curve Sketching Shows Up Later

These same techniques reappear whenever you need to describe the shape of a function rather than just its formula. For example, finding a minimum average cost in an applied problem uses the exact same first and second derivative sign charts you practiced here: locate the critical number, confirm it is a minimum, and describe the behavior around it before drawing conclusions about the cost curve.

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