Sketching a curve from its derivative

Factor the function to locate exact horizontal intercepts and substitute zero for the vertical intercept. Solve f′(x) = 0 to locate stationary inputs and use the original function to find their heights. A sign table for the derivative gives the direction of travel between them. You can now sketch the main shape of the curve.

Check that your sketch agrees with the leading term for large positive and negative inputs. A curve can cross an axis at a horizontal tangent, as y = x³ does at zero. Label the important coordinates accurately; a smooth sketch does not need a dense table of values. If there is a restricted domain, show only the permitted part and mark its endpoints.

Worked example

Sketch y = x³ − 3x², labelling intercepts and stationary points.

1. Factor y = x²(x − 3). The horizontal intercepts are (0,0) and (3,0); zero is a repeated root.
2. Differentiate: y′ = 3x(x − 2). The stationary points are (0,0) and (2,−4).
3. The derivative signs are positive, negative, positive across x = 0 and x = 2. The cubic leading term gives a lower left tail and upper right tail.

Answer: A local maximum at (0,0), a local minimum at (2,−4), touching the axis at 0 and crossing it at 3.

Revise first: Increasing and decreasing intervals, The factor theorem.