Skip to main content
New YorkMathsSyllabus dot point

How do the zeros, their multiplicities, and the leading term shape the graph of a polynomial?

Find the zeros of a polynomial from its factored form; use multiplicity to decide whether the graph crosses or touches the x-axis; and use the degree and leading coefficient to determine end behavior, then sketch the graph.

A NY Regents Algebra II answer on polynomial graphs: finding zeros from factored form, how multiplicity makes a graph cross or touch the x-axis, and how degree and leading coefficient set the end behavior.

Generated by Claude Opus 4.89 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this topic is asking
  2. Zeros from factored form
  3. Multiplicity: cross or touch
  4. End behavior
  5. Reading it for credit
  6. Try this

What this topic is asking

The Regents Algebra II exam (the Arithmetic with Polynomials, A-APR, and Interpreting Functions, F-IF, clusters) wants you to find a polynomial's zeros from factored form, use each zero's multiplicity to decide whether the graph crosses or touches the x-axis there, and use the degree and leading coefficient to determine end behavior, then sketch the curve. This combines algebra with graph reading.

Zeros from factored form

The zeros of a polynomial are the inputs where its value is zero. In factored form they are read off instantly: f(x)=(x−2)(x+3)(x−5)f(x) = (x - 2)(x + 3)(x - 5) has zeros x=2,−3,5x = 2, -3, 5. These are the x-intercepts of the graph. Each zero comes from a factor, and the highest power tells you the polynomial's degree.

Multiplicity: cross or touch

The multiplicity of a zero is the exponent on its factor, and it decides the graph's behavior at that x-intercept.

For f(x)=(x−1)3(x+2)2f(x) = (x - 1)^3(x + 2)^2, the zero x=1x = 1 has multiplicity 3 (odd, crosses, with a flattened S-shape) and x=−2x = -2 has multiplicity 2 (even, touches and turns). Recognizing this from the exponents lets you sketch the curve near each intercept without plotting points.

End behavior

The end behavior describes the graph's direction as xx goes to positive and negative infinity, and it is fixed by the leading term (the highest-degree term).

So a positive even-degree polynomial rises on both sides (like a parabola), and a positive odd-degree polynomial falls to the left and rises to the right (like y=x3y = x^3).

Reading it for credit

A clarifying point worth stressing is that multiplicity and end behavior are independent tools: multiplicity tells you the local behavior at each x-intercept (cross versus bounce), while the leading term tells you the global behavior at the far ends. A complete description or sketch uses both, and the Regents commonly awards a credit for explicitly tying the end behavior to the odd or even degree and the sign of the leading coefficient. Stating only the directions, without that justification, leaves a credit on the table.

Try this

Q1. State the zeros and their multiplicities for f(x)=(x−4)2(x+1)f(x) = (x - 4)^2(x + 1). [2 credits]

  • Cue. x=4x = 4 multiplicity 2 (touches), x=−1x = -1 multiplicity 1 (crosses).

Q2. Describe the end behavior of f(x)=x4−3x2+1f(x) = x^4 - 3x^2 + 1. [2 credits]

  • Cue. Even degree, positive leading coefficient: both ends go up.

Exam-style practice questions

Practice questions written in the style of NYSED exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Regents (style)2 marksPart I (multiple choice). The polynomial f(x)=(x−2)2(x+5)f(x) = (x - 2)^2(x + 5) has which behavior at x=2x = 2? (1) crosses the x-axis (2) touches the x-axis and turns around (3) has a vertical asymptote (4) is undefined
Show worked answer →

The correct answer is (2).

The factor (x−2)2(x - 2)^2 has even multiplicity 2, so the graph touches the x-axis at x=2x = 2 and turns around (a bounce) rather than crossing. The factor (x+5)(x + 5) has odd multiplicity 1, so the graph crosses at x=−5x = -5. Even multiplicity means touch-and-turn; odd multiplicity means cross.

Regents (style)2 marksPart II (constructed response). Describe the end behavior of g(x)=−2x3+4x−1g(x) = -2x^3 + 4x - 1 as x→∞x \to \infty and as x→−∞x \to -\infty, and explain how the leading term determines it.
Show worked answer →

A 2-credit question: 1 credit for each end, tied to the leading term.

End behavior is set by the leading term −2x3-2x^3. The degree 3 is odd and the leading coefficient −2-2 is negative, so the ends go in opposite directions with a downward right end: as x→∞x \to \infty, g(x)→−∞g(x) \to -\infty, and as x→−∞x \to -\infty, g(x)→+∞g(x) \to +\infty. A response that gives the directions without referencing the odd degree and negative leading coefficient earns partial credit.

Related dot points

Sources & how we know this