4.9 - Sketching curves with known turning points

4.9 - Sketching curves with known turning points

Build a reliable curve sketch from its known maximum and minimum points, axis intercepts and far-end behaviour. You will distinguish crossings from touches and use the finished shape to interpret gradients and count solutions without inventing coordinates.

Building a cubic sketch from its anchors

Can a cubic have two turning points but only two different xx-intercepts? A familiar S-shaped curve often has three crossings, but one of its turns can change that count. We will reach that case after assembling a complete ordinary cubic sketch.

You already know

Set y=0y=0 for xx-intercepts and x=0x=0 for the yy-intercept. A maximum has a horizontal tangent with rising then falling behaviour; a minimum has falling then rising behaviour. A sketch shows correct features and their order rather than a table of accurately plotted points.

Use the supplied turning points as anchors. If their coordinates and natures are given, you can use them directly; there is no need to repeat the calculus unless the question asks you to establish them. Place them and the intercepts before joining anything.

The leading term controls a polynomial's far ends, where its highest power dominates the lower powers. For example, a positive x3x^3 term eventually gives negative outputs far to the left and positive outputs far to the right.

Polynomial far-end behaviour

For a nonconstant polynomial with leading term axnax^n, where a≠0a\ne0 and nn is a positive integer,

a>0a<0n oddleft down, right upleft up, right downn evenboth ends upboth ends down\begin{array}{c|c|c} &a>0&a<0\\\hline n\text{ odd}&\text{left down, right up}&\text{left up, right down}\\ n\text{ even}&\text{both ends up}&\text{both ends down} \end{array}

“Far ends” describes behaviour for very large positive or negative inputs, not the direction at every point.
Not on the formulae sheet: learn it.

Connect the anchors with a smooth curve that respects those ends and the given turns. Do not insert another turn merely to make the sketch look tidy.

Worked example 1: three distinct crossings

Question. The curve y=x3−3xy=x^3-3x has a maximum at (−1,2)(-1,2) and a minimum at (1,−2)(1,-2). Sketch it, labelling the intercepts and turning points.

Plan. Factorise for the intercepts, place the known turns and use the positive cubic term for the ends.

y=x(x2−3)factorisey=0⟹x=−3, 0, 3solve each factorx=0⟹y=0vertical interceptknown turns: (−1,2) maximum, (1,−2) minimumplace the supplied anchorsx→−∞⟹y→−∞positive odd leading termx→∞⟹y→∞right end rises\begin{aligned} y&=x(x^2-3) &&\text{factorise}\\ y=0&\Longrightarrow x=-\sqrt3,\ 0,\ \sqrt3 &&\text{solve each factor}\\ x=0&\Longrightarrow y=0 &&\text{vertical intercept}\\ \text{known turns}&:\ (-1,2)\text{ maximum},\ (1,-2)\text{ minimum} &&\text{place the supplied anchors}\\ x\to-\infty&\Longrightarrow y\to-\infty &&\text{positive odd leading term}\\ x\to\infty&\Longrightarrow y\to\infty &&\text{right end rises} \end{aligned}

Why this step? The intercepts fix where the curve meets the axis; the known turns fix the rise-fall-rise route between its far ends.

Answer. Draw a smooth cubic rising from the lower left through (−3,0)(-\sqrt3,0) to the maximum, falling through (0,0)(0,0) to the minimum, then rising through (3,0)(\sqrt3,0) towards the upper right. The left panel below shows the sketch.

Check. −3<−1<0<1<3-\sqrt3<-1<0<1<\sqrt3, so all labelled features occur in the stated order. The simple factors change sign at each root, so all three are crossings.

Worked example 2 multiplies the entire rule by −1-1. The roots stay fixed, while heights and end directions reverse.

Worked example 2: a negative-leading cubic

Question. The curve y=−x3+3xy=-x^3+3x has a minimum at (−1,−2)(-1,-2) and a maximum at (1,2)(1,2). Sketch it, labelling the intercepts and turning points.

Plan. Use the unchanged roots and reflected turning points, then reverse the far-end directions.

y=−x(x2−3)factorisey=0⟹x=−3, 0, 3same zero factorsknown turns: (−1,−2) minimum, (1,2) maximumreflected anchorsx→−∞⟹y→∞negative cubic termx→∞⟹y→−∞right end falls\begin{aligned} y&=-x(x^2-3) &&\text{factorise}\\ y=0&\Longrightarrow x=-\sqrt3,\ 0,\ \sqrt3 &&\text{same zero factors}\\ \text{known turns}&:\ (-1,-2)\text{ minimum},\ (1,2)\text{ maximum} &&\text{reflected anchors}\\ x\to-\infty&\Longrightarrow y\to\infty &&\text{negative cubic term}\\ x\to\infty&\Longrightarrow y\to-\infty &&\text{right end falls} \end{aligned}

Why this step? Multiplication by −1-1 reflects the whole graph in the xx-axis. It cannot move a point that already has height zero.

Answer. Draw a smooth cubic falling from the upper left through (−3,0)(-\sqrt3,0) to the minimum, rising through (0,0)(0,0) to the maximum, then falling through (3,0)(\sqrt3,0). See the right panel below.

Check. Every point (x,y)(x,y) from the first sketch becomes (x,−y)(x,-y), including both turning points and all three unchanged intercepts.

The comparison makes the leading coefficient's role visible. Both curves have the same roots and the same turning-point inputs, but travel through them in opposite directions.

[DIAGRAM: asset_name: Reflecting a complete cubic sketch; asset_slug: l2fm-cubic-sketch-reflection; caption: Multiplying by minus one preserves the roots while reversing the turns and both far ends; description: Two coordinate graphs show y equals x cubed minus 3x and y equals minus x cubed plus 3x. Both label roots at minus square root 3, zero and square root 3. The first labels maximum minus 1 comma 2 and minimum 1 comma minus 2; the second labels minimum minus 1 comma minus 2 and maximum 1 comma 2. Each curve continues in the correct far-end directions]
Multiplying by minus one preserves the roots while reversing the turns and both far ends

A sketch communicates coordinates through labels, so an accurately placed but unlabelled intercept can still leave the reader uncertain.

Watch out

Wrong: connect the anchors with straight segments or draw a sharp corner at a turning point. A smooth polynomial has a horizontal tangent there. Draw the curve smoothly through each turn.

Prepare the anchors for a new cubic without needing to draw in the answer box.

Recognising touches and repeated roots

A root does not always make the curve cross the axis: a repeated factor can make it turn back on the same side.

For y=(x−1)2(x+2)y=(x-1)^2(x+2), the squared factor is positive on both sides of x=1x=1. The other factor is also positive near 11, so the curve stays above the axis as it reaches and leaves that root. At x=−2x=-2, the factor x+2x+2 changes sign while the square stays positive, so the curve crosses.

Root multiplicity and the axis

If a polynomial can be written near a root rr as

f(x)=(x−r)mg(x),g(r)≠0\mathrm{f}(x)=(x-r)^m\mathrm{g}(x),\qquad \mathrm{g}(r)\ne0

where mm is a positive integer, then an even mm gives a touch with no sign change, while an odd mm gives a crossing. The nonzero remaining factor keeps the same sign sufficiently close to rr.
Not on the formulae sheet: learn it.

This includes an important boundary: y=x3y=x^3 has a repeated root of odd multiplicity and crosses with a horizontal tangent. “Repeated” alone does not decide between a crossing and a touch.

Worked example 3: a minimum on the axis

Question. Sketch y=(x−1)2(x+2)y=(x-1)^2(x+2). Its maximum is (−1,4)(-1,4) and its minimum is (1,0)(1,0). Label the axis intercepts and turning points.

Plan. Read the roots and their multiplicities, calculate the vertical intercept and combine them with the supplied turns.

y=0⟹x=−2 or 1zero factorsx=−2:simple root, crossingthe linear factor changes signx=1:double root, touchingthe square stays non-negativey(0)=(−1)2(2)=2vertical interceptleading term=x3left down, right up\begin{aligned} y=0&\Longrightarrow x=-2\text{ or }1 &&\text{zero factors}\\ x=-2&:\quad\text{simple root, crossing} &&\text{the linear factor changes sign}\\ x=1&:\quad\text{double root, touching} &&\text{the square stays non-negative}\\ y(0)&=(-1)^2(2)=2 &&\text{vertical intercept}\\ \text{leading term}&=x^3 &&\text{left down, right up} \end{aligned}

Why this step? The known minimum (1,0)(1,0) lies on the axis. The curve must touch it smoothly and rise again, rather than pass below it.

Answer. Draw a cubic crossing at (−2,0)(-2,0), rising to (−1,4)(-1,4), falling through (0,2)(0,2) to touch at (1,0)(1,0), then rising. It has two distinct xx-intercepts, answering the opening question.

Check. Expanding gives y=x3−3x+2y=x^3-3x+2. Its derivative is 3(x−1)(x+1)3(x-1)(x+1), confirming the two supplied stationary inputs.

Worked example 4 changes only the constant in the unsquared factor, from 22 to 55. The double root stays at 11, while the other crossing and maximum move.

Worked example 4: moving the other root

Question. Sketch y=(x−1)2(x+5)y=(x-1)^2(x+5). Its maximum is (−3,32)(-3,32) and its minimum is (1,0)(1,0). Label the axis intercepts and turning points.

Plan. Read the new simple root, retain the double root and calculate the new vertical intercept.

y=0⟹x=−5 or 1zero factorsx=−5:crossingsimple rootx=1:touchingdouble rooty(0)=(−1)2(5)=5vertical interceptknown turns: (−3,32) maximum, (1,0) minimumplace the anchors\begin{aligned} y=0&\Longrightarrow x=-5\text{ or }1 &&\text{zero factors}\\ x=-5&:\quad\text{crossing} &&\text{simple root}\\ x=1&:\quad\text{touching} &&\text{double root}\\ y(0)&=(-1)^2(5)=5 &&\text{vertical intercept}\\ \text{known turns}&:\ (-3,32)\text{ maximum},\ (1,0)\text{ minimum} &&\text{place the anchors} \end{aligned}

Why this step? Changing the other factor does not change the root of (x−1)2(x-1)^2, but it changes heights away from that root.

Answer. Draw the left end down, cross at (−5,0)(-5,0), rise to (−3,32)(-3,32), fall through (0,5)(0,5), touch at (1,0)(1,0) and rise to the right.

Check. Expanding gives x3+3x2−9x+5x^3+3x^2-9x+5, whose derivative is 3(x−1)(x+3)3(x-1)(x+3). At −3-3, the original product is 16×2=3216\times2=32.

The panels show the two sketches on suitable vertical scales. Neither curve needs a third crossing just because it is cubic.

[DIAGRAM: asset_name: Two cubic sketches with a double root; asset_slug: l2fm-cubic-double-root; caption: Both cubics cross at their simple root and touch at their shared double root; description: Two coordinate graphs show y equals open bracket x minus 1 close bracket squared times open bracket x plus 2 close bracket and the same squared factor times open bracket x plus 5 close bracket. The first labels crossing minus 2 comma 0, maximum minus 1 comma 4, vertical intercept zero comma 2 and touching minimum 1 comma 0. The second labels crossing minus 5 comma 0, maximum minus 3 comma 32, vertical intercept zero comma 5 and touching minimum 1 comma 0. Both have left end down and right end up]
Both cubics cross at their simple root and touch at their shared double root

The first curve hides a different kind of symmetry. If f(x)=x3−3x+2\mathrm{f}(x)=x^3-3x+2, then f(x)+f(−x)=4\mathrm{f}(x)+\mathrm{f}(-x)=4. Rotating the graph through 180∘180^\circ about (0,2)(0,2) maps it onto itself. For example, (2,4)(2,4) pairs with (−2,0)(-2,0), and its maximum pairs with its minimum. Cubic symmetry can be a half-turn rather than a mirror line.

Watch out

Wrong: draw every root as a crossing, or every repeated root as a touch. Check the complete factor's multiplicity: even powers preserve its sign across the root; odd powers reverse it.

The next cubic touches the axis at a maximum rather than a minimum. Use its factors to decide what happens at each root.

Continue this lesson.

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