Analyse a cubic graph or determine its rule

Two distinct real roots r and s are supplied: r is a double root and s is a simple root. Which representation is most direct?

Choosing a cubic representation

Different forms expose different features. Factorised form displays roots and multiplicities. The special form a(x−h)³+k displays a stationary inflection; general coefficient form also covers nonstationary inflections and four point conditions with distinct x-coordinates. Every real cubic has a point of inflection and symmetry centre (h,k). In general coefficient form, a is nonzero, h=−b/(3a) and k=P(h). Expanding about h removes the quadratic term, but a linear term may remain. Therefore a general inflection need not be stationary

1. 1Match form to information Use the special form a(x−h)³+k, with a nonzero, when the inflection is stationary. Not every cubic has this form \(\displaystyle P(x)=a\left(x-h\right)^3+k\)
2. 2Use factors for roots Use factorised form when roots or their multiplicities are supplied \(\displaystyle P(x)=2\left(x+1\right)^2\left(x-3\right)\)
3. 3Keep multiplicity The exponent on a factor records how many times its root occurs \(\displaystyle x=-1\text{ has multiplicity }2\)
4. 4Use the leading coefficient For a cubic, the sign of the leading coefficient determines the end behaviour \(\displaystyle a<0\quad\Rightarrow\quad y\to-\infty\text{ as }x\to\infty\)
5. 5Find the general cubic centre Every real cubic has a point of inflection and symmetry centre (h,k). In general coefficient form, a is nonzero, h=−b/(3a) and k=P(h). Expanding about h removes the quadratic term, but a linear term may remain. Therefore a general inflection need not be stationary \(\displaystyle \begin{gathered}P(x)=ax^3+bx^2+cx+d\\a\ne0,\quad h=-\frac{b}{3a}\\k=P(h)\end{gathered}\)
6. 6Distinguish stationary from nonstationary inflection Put m=c−b²/(3a). The centred form is at³+mt+k, so values at h+t and h−t add to 2k. For nonzero t, the secant slope from the centre is m+at². As |t| becomes small this approaches m; the tangent is horizontal exactly when m=0. For x³−3x the centre is (0,0) but m=−3, so it is not stationary. For (x−2)³+1 the centre is (2,1) and m=0; its x-intercept (1,0) is simple. The later drawing task also covers (x−2)³, whose centre is a triple-root x-intercept \(\displaystyle P(h+t)=at^3+\left(c-\frac{b^2}{3a}\right)t+k\)

Using the factorised rule P(x)=2(x+1)^2(x-3), calculate the vertical intercept

Expand enough of the factorised rule to find the coefficient of the squared term

Which expression retains the double root at negative one?

Turn features into a complete drawing

Work on paper or a blank drawing surface. Find the features, construct the whole labelled curve, and then review it. The complete drawings and explanations are ungraded; the checked entry assesses one feature

1. 1Prepare the whole drawing State the domain and range. Label axes and scale; find intercepts, endpoints, asymptotes and exact key points before joining anything
2. 2Show the correct connections Draw every allowed branch with the correct direction and shape. Use open/closed endpoints, leave gaps at excluded inputs, and show end behaviour. A handful of correct dots is not a complete sketch
3. 3Check the mathematics without calculus Substitute labelled points, use factors and signs between roots, compare symmetry and both ends, and use tables or simple inequalities to check the shape. Distinguish a local extremum from a bound on the whole domain. A plotted model is feedback, not your submitted drawing

A complete cubic from factors and algebra

For f(x)=x³−3x=x(x²−3), the roots are −√3, 0 and √3; each is simple and the curve crosses at each root. The signs from left to right are negative, positive, negative, positive. The identities f(x)−2=(x+1)²(x−2) and f(x)+2=(x−1)²(x+2) show a local maximum (−1,2) and local minimum (1,−2): near −1, f(x)≤2; near 1, f(x)≥−2. Also f(−x)=−f(x), so the graph has origin symmetry. Draw one smooth cubic through all these features, with the left tail falling and the right tail rising. Domain and range are R. No differentiation is needed for these algebraic checks

Rule

\(\displaystyle f(x)=x^3-3x\)

Substitute x=−1 into the worked cubic. What is its value?

Your complete hand-sketch task

On separate axes, draw the complete graphs g(x)=(x−1)³−3(x−1) and h(x)=(x−2)³+1. Label every intercept, its local extrema, symmetry centre, domain/range and both ends. For h, label the centre, both intercepts and both ends; explain why the flattened crossing through its centre is not a maximum or minimum. Verify an extra exact point on each graph. Work by hand from algebra and key points, not differentiation

Review both complete drawings

For g, the roots are 1−√3, 1 and 1+√3, all crossings. Its y-intercept/local maximum is (0,2), local minimum (2,−2), and symmetry centre (1,0). An extra point is (3,2). The left tail falls and the right tail rises; domain/range R. For h, the centre is (2,1), x-intercept (1,0), y-intercept (0,−7), and extra point (3,2). It is increasing through its centre: inputs just below 2 give outputs below 1, and inputs just above 2 give outputs above 1. Thus the centre is a stationary inflection, not an extremum. Domain/range R, with the left tail falling and right tail rising. Join smooth curves, not straight segments between the key points

1. 1Review the whole curve Compare every labelled feature, branch and end direction with your own drawing. Annotate corrections and discuss discrepancies with a teacher or peer

A flattened crossing on the x-axis

Translate the previous h(x)=(x−2)³+1 down one unit to obtain r(x)=(x−2)³. Draw its complete graph on R, with both intercepts, the symmetry centre, a short horizontal tangent at the centre, two further exact points and both end directions. State the domain and range. Use the triple factor and values on both sides to explain why (2,0) is a stationary point of inflection on the x-axis, not a touch point or a turning point. Work without differentiation

Review the stationary x-intercept

The only x-intercept is the triple root (2,0); the y-intercept is (0,−8). Other exact points are (1,−1), (3,1) and (4,8). The centre is (2,0), the domain and range are R, the left end falls and the right end rises. The curve remains increasing through the flattened crossing, so there is no local maximum or minimum. For a small nonzero t, the secant from the centre to (2+t,t³) has slope t², which becomes small as |t| becomes small; draw the horizontal tangent at the centre. Inputs on opposite sides give opposite output signs, unlike the touching double roots of (x²−4)². Draw one smooth curve through every labelled feature