End Behavior of Polynomials
Far to the left and right, a polynomial behaves like its leading term — degree and leading sign decide the direction.
Reading the concept isn't enough. The score comes from practicing the real Bluebook-style format and finding the exact mistakes you keep making.
What the SAT tests here
The leading term controls the ends. Even degree: both ends go the same way; odd degree: opposite ways. A negative leading coefficient flips the whole picture.
How to solve one, step by step
Example: describe the ends of $p(x) = -2x^3 + \dots$
- Degree $3$ is odd, so the ends go opposite directions.
- The leading coefficient is negative, so as $x \to \infty$, $p \to -\infty$, and as $x \to -\infty$, $p \to +\infty$.
The mistakes that cost points
- Ignoring degree parity. Odd and even degrees behave differently at the ends — the leading coefficient alone isn't enough.
- Confusing ends with the middle. End behavior is about the far left and right, not the wiggles near the zeros.
Practice questions
Try these the way you would on test day, then open the solution to check your method.
Easy
Let $f$ be a polynomial. As $x$ becomes very large and positive, $f(x)$ approaches $\infty$. As $x$ becomes very large and negative, $f(x)$ approaches $\infty$. Which of the following could be $f(x)$?
- A$f(x) = -x^{7} - 2x^{5} + 2x^{4} + x^{3}$
- B$f(x) = x^{6} + 2x^{5} - 2x^{4} + 2$
- C$f(x) = -x^{6} + 2x^{5} - 2x^{4} + 2$
- D$f(x) = x^{7} - 2x^{5} + 2x^{4} + x^{3}$
Show solution
Answer: B, $f(x) = x^{6} + 2x^{5} - 2x^{4} + 2$. The leading term determines the end behavior. The leading term here is the highest-degree term. In standard form (descending powers), it's the first term. From its sign and degree-parity: negative leading coefficient with even degree gives the stated end behavior.
Medium
Suppose $f$ is a polynomial whose values approach $\infty$ as $x$ grows without bound, and approach $-\infty$ as $x$ decreases without bound. Which of the following could be $f(x)$?
- A$f(x) = -2x^{5} + 2x^{2} - x^{6} + x^{7}$
- B$f(x) = x^{5} - 3x^{3} + 2x^{7} - x^{8}$
- C$f(x) = -2x^{5} + 2x^{2} - x^{7} - x^{6}$
- D$f(x) = 3x^{3} + 2x^{5} + x^{8} - 3$
Show solution
Answer: A, $f(x) = -2x^{5} + 2x^{2} - x^{6} + x^{7}$. The leading term determines the end behavior — but the leading term is the term with the HIGHEST DEGREE, not necessarily the first term written. Scan all four terms to find the highest power of $x$, then check its sign and degree parity. Don't fall for the first-term trap: a polynomial like $3x^4 - 3x^7 + x^2 - 2x^6$ has leading term $-3x^7$, even though $3x^4$ is written first.
Hard
A polynomial $p(x)$ of degree 5 has zeros at $x = -3, -1, 0, 1, 3$, and its graph passes through the point $(4, -100)$. Which of the following describes the end behavior of $p$?
- AAs $x \to +\infty$, $p(x) \to -\infty$, and as $x \to -\infty$, $p(x) \to +\infty$
- BAs $x \to +\infty$, $p(x) \to +\infty$, and as $x \to -\infty$, $p(x) \to +\infty$, because all degree-5 polynomials eventually dominate and grow
- CAs $x \to +\infty$, $p(x) \to -\infty$, matching the behavior near $x = 3$ where the graph is also decreasing
- DAs $x \to +\infty$, $p(x) \to -\infty$, and as $x \to -\infty$, $p(x) \to -\infty$
Show solution
Answer: A, As $x \to +\infty$, $p(x) \to -\infty$, and as $x \to -\infty$, $p(x) \to +\infty$. Since $p(x)$ has degree 5 (odd degree), the two ends go in opposite directions. The point $(4, -100)$ lies to the right of all zeros, and $p(4) = -100 < 0$, indicating the graph is negative for large positive $x$. Therefore the leading coefficient is negative, confirming that $p(x) \to -\infty$ as $x \to +\infty$ and $p(x) \to +\infty$ as $x \to -\infty$.
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