CA7 - Sketching curves using stationary points

A sketch records a curve's important features without plotting a large table of points. CA7 uses stationary points, intercepts and the direction of travel to make the shape precise. You will also use a finished sketch to explain how many times a horizontal line meets a curve.

Stationary points fix the shape

How can the equation $x^3-3x=k$ have one, two or three distinct real solutions just by changing $k$ ? The curve stays the same; a horizontal line moves up and down. Its stationary points will tell us where the number of intersections changes.

You already know

Polynomial roots locate horizontal-axis intersections, and $x=0$ gives the vertical intercept. A positive leading cubic rises overall from bottom left to top right. Stationary points have zero gradient; maxima and minima have different changes of direction.

A list of points does not yet make a sketch. You must connect them with the correct behaviour. For $y=x^3-3x$ , the derivative is $3(x-1)(x+1)$ , so the only stationary coordinates are $-1$ and $1$ . The corresponding points are $(-1, 2)$ , a maximum, and $(1, -2)$ , a minimum. The derivative is positive to the left of $-1$ , negative between $-1$ and $1$ , and positive to the right of $1$ . There is no extra turn hidden between these coordinates.

Sketch

A sketch is a drawing that shows the main mathematical features and their relative positions, rather than an accurately plotted graph. Label requested intercepts and stationary points with coordinates; show horizontal tangents at stationary points and the correct directions at the ends.

Make a skeleton before drawing the smooth curve. Place the maximum above the minimum, arrange all points in horizontal-coordinate order, then connect them by rising or falling stretches as appropriate. A stationary point is smooth and flat; drawing a sharp corner there gives it the wrong tangent behaviour.

Range of $x$ for $y=x^3-3x$	Gradient sign	Curve's behaviour
$x<-1$	Positive	Rises towards the maximum
$-1<x<1$	Negative	Falls towards the minimum
$x>1$	Positive	Rises away from the minimum

A known stationary point need not be a turn. For $y=x^3$ , the origin is stationary but the gradient $3x^2$ is positive on both sides. Draw the curve flattening at the origin and continuing to rise through it, rather than turning back. Use the known nature of each stationary point, not just its zero gradient.

The derivative sign rules and the use of stationary points to sketch a curve are not on the formulae sheet: learn them. If stationary points are supplied, you can use them directly; do not spend time finding them again unless asked to verify them.

Key idea

An accurate sketch preserves positions, crossings, contacts, flat points and directions, even when the drawing is not to scale.

Use the shape skeleton to choose a valid connection.

Combining intercepts and turning points

Use roots and intercepts as fixed positions, and stationary points as instructions for how the curve joins those positions.

A repeated root can be both an axis contact and a turning point. If the same point appears in two lists, plot it once and label both of its roles. Read the far-left and far-right directions from the leading term, then check that every connection agrees with the stationary-point data.

Worked example 1: a cubic that touches the axis

Question. Sketch $y=x^3-3x+2$ , labelling its axis intersections and stationary points. You are given a maximum at $(-1, 4)$ and a minimum at $(1, 0)$ .

Plan. Factorise to find roots, find the vertical intercept and combine these with the given turns and leading-term directions.

\begin{aligned} x^3-3x+2&=(x-1)^2(x+2) && \text{factorise}\\ y=0&\quad\Rightarrow\quad x=-2\text{ or }1 && \text{horizontal intercepts}\\ y(0)&=2 && \text{vertical intercept}\\ \frac{\mathrm{d}y}{\mathrm{d}x}&=3(x-1)(x+1) && \text{check the direction intervals} \end{aligned}

From the bottom left, rise through $(-2, 0)$ to the maximum $(-1, 4)$ . Fall through $(0, 2)$ to the minimum $(1, 0)$ , touching the axis there, then rise towards the top right.

Why does the graph touch rather than cross at $x=1$ ? The factor $(x-1)^2$ does not change sign there; the given minimum confirms the curve turns back upwards.

Answer. A smooth positive cubic crossing at $(-2, 0)$ , passing through $(0, 2)$ , and touching at $(1, 0)$ , with maximum $(-1, 4)$ .

Check. Both stated stationary points give zero derivative. The graph is above the axis on both sides close to $x=1$ , consistent with its minimum height $0$ .

Change only the sign of the entire function. The roots stay put, every other height changes sign, and each turn reverses.

Worked example 2: reflecting the complete sketch

Question. Sketch $y=-x^3+3x-2$ , labelling its axis intersections and stationary points. You are given a minimum at $(-1, -4)$ and a maximum at $(1, 0)$ .

Plan. Reflect the preceding features in the horizontal axis and check them algebraically.

\begin{aligned} y&=-(x-1)^2(x+2) && \text{same roots}\\ y=0&\quad\Rightarrow\quad x=-2\text{ or }1 && \text{horizontal intercepts}\\ y(0)&=-2 && \text{vertical intercept}\\ \frac{\mathrm{d}y}{\mathrm{d}x}&=-3(x-1)(x+1) && \text{reversed gradient signs} \end{aligned}

From the top left, fall through $(-2, 0)$ to $(-1, -4)$ . Rise through $(0, -2)$ to $(1, 0)$ , touching the axis from below, then fall towards the bottom right.

Why do the roots not move? Negating a value changes its sign, but zero remains zero.

Answer. The reflection of Worked example 1: a negative cubic, minimum $(-1, -4)$ , maximum $(1, 0)$ , vertical intercept $(0, -2)$ .

Check. The derivative signs are negative, positive, negative in the successive intervals, matching fall, rise, fall.

The two complete sketches show which features reflection preserves and which it reverses.

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CA7 - Sketching curves using stationary points

Stationary points fix the shape

Combining intercepts and turning points

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