Multiplicity and Shape

Near $x = 2$, the value of $(x-2)$ is a small number. Work out what happens to that small number when you square it, and when you cube it, for a value just above $2$ and a value just below. Which of the three keeps its sign, and which does not?

The exponent on a factor is its multiplicity, and it decides the shape of the graph at that zero. Odd multiplicity crosses the axis; even multiplicity touches and turns back without crossing.

When a polynomial is fully factorised, a factor may appear more than once:

$$P(x) = (x+1)(x-3)^2(x-5)^3$$

The exponent on each factor is the multiplicity of that zero. Here $-1$ has multiplicity $1$, $3$ has multiplicity $2$, and $5$ has multiplicity $3$.

Multiplicities and the degree are linked. Adding them gives $1 + 2 + 3 = 6$, which is the degree of $P$, because degrees add when polynomials are multiplied. If a factorisation also contains an irreducible quadratic, that contributes $2$ to the degree while contributing no real zero at all.

Checking that the multiplicities add up to the degree is a fast and genuine check on a factorisation. If they do not, something has been dropped.

Look closely at a zero. Near $x = a$, write $P(x) = (x-a)^m S(x)$, where $S(a) \neq 0$. Very close to $a$ the factor $S(x)$ barely changes and keeps the sign of $S(a)$, so the behaviour of $P$ is decided by $(x-a)^m$ alone.

Now watch what $(x-a)^m$ does as $x$ passes through $a$. The quantity $(x-a)$ is negative just below $a$ and positive just above.

$m$ odd. An odd power keeps the sign, so $(x-a)^m$ is negative below and positive above. $P$ changes sign, and the curve crosses the axis.

$m$ even. An even power destroys the sign: $(x-a)^m$ is positive on both sides. $P$ does not change sign, so the curve comes down to the axis, meets it, and returns to the side it came from. It touches.

$$\text{odd } m \ \Rightarrow \ \text{cross} \qquad \text{even } m \ \Rightarrow \ \text{touch}$$

This is why the sign-testing method of the last lesson needed the warning. Signs alternate at every zero only when every multiplicity is odd. A double zero leaves the sign unchanged, and two adjacent intervals then have the same sign.

Multiplicity controls more than crossing or touching. It also controls how flat the curve is where it meets the axis.

Compare $(x-2)$, $(x-2)^2$ and $(x-2)^3$ at $x = 2.1$, where $(x-2) = 0.1$:

$$0.1, \qquad 0.01, \qquad 0.001$$

Each extra power multiplies by another factor of $0.1$, so the higher the multiplicity, the closer to zero the polynomial stays as $x$ moves away from the zero. The curve hugs the axis for longer.

In pictures:

a simple zero cuts through cleanly, like a straight line;
a double zero touches like the vertex of a parabola;
a triple zero crosses, but flattens against the axis on both sides first, like the graph of $y = x^3$ at the origin.

That flattening is a point of inflection: the curve's bending reverses there. Recognising it by eye is how you tell a triple zero from a simple one on an unlabelled sketch, since both cross. A crossing that arrives flat is odd multiplicity of at least three.

The full sketching procedure, now that multiplicity is available:

1. Factorise completely and list each zero with its multiplicity.
2. Find the $y$-intercept, $P(0)$.
3. Fix the two ends from the leading term: degree parity and sign of the leading coefficient.
4. Start from one end and work across, crossing at odd zeroes and touching at even ones.

Sketch $y = -(x+2)(x-1)^2$.

Zeroes: $-2$ with multiplicity $1$, and $1$ with multiplicity $2$. The degree is $1 + 2 = 3$, and the leading term is $-(x)(x^2) = -x^3$: odd degree, negative coefficient, so up on the left and down on the right.

Start at the upper left. Come down and cross at $-2$, since that multiplicity is odd. Now the curve is below the axis. Come up to touch at $1$ without crossing, since that multiplicity is even, and turn back down. Continue down to the right, which matches the predicted end.

The $y$-intercept confirms it: $-(2)(1)^2 = -2$, which is below the axis, and $x = 0$ lies between $-2$ and $1$ where the sketch has the curve below. Two predictions agreeing is the check worth doing.

The same rules run in reverse. Given a sketch, you can write down a possible equation.

Suppose a quartic graph touches the axis at $-1$, crosses at $2$, crosses at $3$, and has both ends pointing upward.

Turn each behaviour into a multiplicity. A touch is even, so at least $2$. A crossing is odd, so at least $1$. That gives $2$, $1$ and $1$.

Check against the degree. $2 + 1 + 1 = 4$, exactly the degree of a quartic. There is no spare degree, so no multiplicity can be raised, and the values are forced.

$$y = k(x+1)^2(x-2)(x-3)$$

Fix the sign of $k$ from the ends. The degree is even, so both ends agree; they point upward, so the leading coefficient is positive and $k > 0$. Any positive $k$ gives a graph matching the description, and one further point would pin it down exactly.

The degree check in the middle step is the one that earns its keep. If the multiplicities do not sum to the degree, something is missing: either an unseen zero off the edge of the sketch, or an irreducible quadratic factor contributing degree $2$ without any real zero at all.

Five items on multiplicity. Work each one, then reveal the answer.

1. 1 Does $y = (x-4)^2(x+1)$ cross or touch at $x = 4$?

   Show answer
   Multiplicity $2$ is even.Touches

2. 2 Does $y = x^3(x-2)$ cross or touch at $x = 0$?

   Show answer
   Multiplicity $3$ is odd.Crosses, flattening

3. 3 State the degree of $(x-1)^2(x+3)^3(x-5)$.

   Show answer
   Add the multiplicities.$6$

4. 4 A cubic touches the axis at $2$ and crosses at $-1$. State the multiplicities.

   Show answer
   They must sum to $3$, with the touch even.$2$ at $x=2$, $1$ at $x=-1$

5. 5 Can a quadratic touch the axis at two different points?

   Show answer
   Each touch needs multiplicity at least $2$, so two touches need degree at least $4$.No