DWG NO. 2.2 — Lesson 2 of 6

Graphing Polynomial Functions

Unit 2: Polynomial and Rational Functions · ~25 min

Objective Sketch a polynomial function using its zeros, the multiplicity of each zero, and the maximum number of turning points.

Zeros and multiplicity

When a polynomial is written in factored form, f(x) = (x−c\()^{m}\)·…, each factor gives a zero at x = c, and the exponent m on that factor is its multiplicity — how many times that same root repeats. Multiplicity controls exactly how the graph behaves at that zero:

Odd multiplicity (1, 3, 5, …): the graph crosses the x-axis at that zero, passing from one side to the other.
Even multiplicity (2, 4, 6, …): the graph touches the x-axis and turns back, never crossing — like a parabola's vertex sitting on the axis.
Multiplicity higher than 1: the graph flattens out near that zero — it looks locally straight or even briefly horizontal — before continuing.

Turning points

A turning point is a local maximum or minimum — a spot where the graph switches from increasing to decreasing or back. A polynomial of degree n has at most n−1 turning points. It can have fewer, but never more — this is a hard ceiling set by the degree alone.

Putting it together

To sketch a polynomial by hand: find the zeros and their multiplicities to see where and how the graph meets the x-axis, use the leading coefficient test from 2.1 for the two end behaviors, and use the turning-point limit to know roughly how much wiggling to expect in between. Plotting one or two extra points (like the y-intercept) ties the picture together.

f(x) = (x+2)(x−1)² — crosses at x = −2 (multiplicity 1), touches and turns at x = 1 (multiplicity 2)

Worked Example 1 · Identifying zeros and multiplicity

ProblemList the zeros of f(x) = (x+1)(x−3)² and describe the graph's behavior at each.

1x = −1 comes from the factor (x+1)¹, so its multiplicity is 1 (odd) — the graph crosses here.

2x = 3 comes from (x−3)², so its multiplicity is 2 (even) — the graph touches and turns back here.

x = −1: crosses · x = 3: touches (does not cross)

Worked Example 2 · End behavior + turning points, from the diagram above

Problemf(x) = (x+2)(x−1)² has degree 3, leading coefficient +1 (odd, positive) → falls left, rises right.

1Degree 3 allows at most 3−1 = 2 turning points.

2The graph does show exactly 2: a local max near x = −1 and a local min at the double root x = 1 — using the full allowance.

End behavior: down left, up right · Turning points: 2 (the maximum possible)

Worked Example 3 · Higher (odd) multiplicity flattening

ProblemDescribe the graph of f(x) = (x−1)³(x+2) at each of its zeros.

1x = 1 has multiplicity 3 (odd) — the graph crosses, but flattens out near x = 1 before continuing, instead of crossing at a steady slope.

2x = −2 has multiplicity 1 (odd) — the graph crosses normally, at a steady slope, with no flattening.

x = 1: crosses with flattening · x = −2: crosses normally

Guided practice

1.List the zeros and multiplicities of f(x) = (x−4)(x+1)², and describe the graph's behavior at each.
Show answer

x = 4, multiplicity 1 (crosses). x = −1, multiplicity 2 (touches, turns back).
2.A polynomial has degree 6. What is the maximum number of turning points its graph can have?
Show answer

Maximum turning points = degree − 1 = 6 − 1. 5 turning points.
3.Describe the behavior of f(x) = x²(x−3)³ at x = 0 and at x = 3.
Show answer

x = 0 has multiplicity 2 (even): touches and turns back. x = 3 has multiplicity 3 (odd, greater than 1): crosses, but flattens near the axis first.
4.A polynomial's graph crosses the x-axis at x = −3, touches (without crossing) at x = 1, and crosses again at x = 5. What is the smallest possible degree for this polynomial?
Show answer

Crossing needs at least multiplicity 1; touching needs at least multiplicity 2. Minimum multiplicities: 1 (at −3) + 2 (at 1) + 1 (at 5) = 4. Smallest possible degree: 4.
5.List the zeros and multiplicities of f(x) = (x+2)²(x−1)(x−4), and state the maximum number of turning points its graph could have.
Show answer

Zeros: x = −2 (multiplicity 2, touches), x = 1 (multiplicity 1, crosses), x = 4 (multiplicity 1, crosses). Degree = 2+1+1 = 4, so at most 3 turning points.

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