Sketching a polynomial at sight
Tonight's idea
You can sketch a factorised polynomial without plotting a table of points. Three facts do it.
The ends. Far from zero, the highest power swamps the rest: at \(x = 1000\), \(x^3\) is a billion and \(100x^2\) is only a tenth of that. So the ends of the graph copy the leading term \(ax^n\). Even \(n\): both ends go the same way, up if \(a > 0\). Odd \(n\): opposite ways, down on the left and up on the right if \(a > 0\). A negative \(a\) flips both.
The roots. Each factor \((x - r)\) is zero at \(x = r\), so the graph meets the \(x\)-axis there (the factor theorem).
Cross or touch. Close to \(x = r\) the other factors keep their sign, so \((x - r)^m\) decides. If \(m\) is odd it changes sign as \(x\) passes \(r\), and the graph crosses. If \(m\) is even it is positive on both sides, and the graph touches the axis and turns back. The number \(m\) is the multiplicity of the root.
Last, put \(x = 0\) to get the \(y\)-intercept.
Watch one: \(y = (x + 2)(x - 1)^2\)
Multiplied out it would start with \(x^3\): odd degree, positive coefficient, so down on the left and up on the right. The root \(-2\) comes from a single factor, so the graph crosses there. The root 1 comes from a squared factor, so it touches. The intercept is \(2 \cdot (-1)^2 = 2\).
The three facts do not say exactly where the peak is. That takes a derivative, next year.
Your turn
- Say which way each end points: (a) \(y = -2x^4 + 5x\), (b) \(y = x^5 - 9x^3\), (c) \(y = (3 - x)(x + 1)^2\).
- Sketch \(y = x(x - 3)(x + 2)\). Mark the roots and show the ends. For which \(x\) is \(y > 0\)?
- Sketch \(y = -x^2(x - 4)(x + 3)\). Say where it crosses, where it touches, and where \(y > 0\).
- A polynomial of degree 4 touches the \(x\)-axis at \(-1\), crosses it at 1 and at 3, and meets the \(y\)-axis at \(-6\). Find its formula in factorised form. Which way do its ends point?
Stretch: counting crossings
Why must every polynomial of odd degree have at least one real root? Then decide: can a polynomial of degree 4 cross the \(x\)-axis at exactly three points and touch it nowhere? Think about where the two ends are.
Answers
- (a) Leading term \(-2x^4\): both ends down. (b) Leading term \(x^5\): down on the left, up on the right. (c) Multiplied out, the leading term is \(-x^3\): up on the left, down on the right.
- Degree 3, positive: down left, up right. It crosses at \(-2\), 0 and 3 (all single roots), passing through the origin. \(y > 0\) for \(-2 \lt x \lt 0\) and for \(x > 3\). Check: at \(x = -1\), \(y = (-1)(-4)(1) = 4\).
- Leading term \(-x^4\): both ends down. Crosses at \(-3\) and at 4, touches at 0 (double root) from above. \(y > 0\) for \(-3 \lt x \lt 0\) and \(0 \lt x \lt 4\), and \(y = 0\) at \(x = 0\) in between. Check: \(y = 10\) at \(x = -1\) and \(y = 12\) at \(x = 1\).
- Touching at \(-1\) needs \((x + 1)^2\); the crossings give \((x - 1)\) and \((x - 3)\). That is already degree 4, so \(y = a(x + 1)^2(x - 1)(x - 3)\). At \(x = 0\): \(a \cdot 1 \cdot (-1)(-3) = 3a = -6\), so \(a = -2\) and \(y = -2(x + 1)^2(x - 1)(x - 3)\). The leading term is \(-2x^4\): both ends point down.
Stretch. With odd degree, one end goes down and the other up. The graph has no breaks, so somewhere it must pass through \(y = 0\). For degree 4, both ends are on the same side of the axis. Each crossing changes the side, so to finish on the side it started the graph must cross an even number of times. Three crossings are impossible: with three, one of the meetings must be a touch (as in question 4) or there is a fourth crossing.