Far from the origin, a polynomial's graph is controlled by just one term. You know the parts of a polynomial from Polynomials: Degree and Standard Form and the shapes of $x^2$ and $x^3$ from the Parent Functions.
Key idea
The leading term (highest power) decides the end behavior. Even degree: both ends point the same way (up if the leading coefficient is positive, down if negative). Odd degree: the ends point opposite ways (falls left and rises right if positive, the reverse if negative).
Why it works
For large $|x|$, higher powers win by a lot. When $x = 100$, $x^4$ is a hundred million while $50x^3$ is fifty million and $1000x$ is only a hundred thousand. As $|x|$ keeps growing, the ratio of the leading term to all the others put together goes to 1, so the polynomial behaves like its leading term. And $x^{\text{even}}$ is positive for big positive or negative $x$, while $x^{\text{odd}}$ keeps the sign of $x$.
| Leading term | Left end | Right end |
|---|---|---|
| even degree, positive (like $x^2$) | rises | rises |
| even degree, negative (like $-x^2$) | falls | falls |
| odd degree, positive (like $x^3$) | falls | rises |
| odd degree, negative (like $-x^3$) | rises | falls |
Worked examples
Worked example: The leading term is not written first
Describe the end behavior of $f(x) = 7 + 5x - 2x^4 + x^3$.
- The highest power is $x^4$, with coefficient $-2$. The leading term is $-2x^4$.
- Even degree, negative coefficient: both ends fall.
Answer: As $x \to \pm\infty$, $f(x) \to -\infty$.
Worked example: A polynomial in factored form
Describe the end behavior of $g(x) = -3(x - 1)^2(x + 4)$.
- Multiplying out, the leading term comes from the $x$ in each factor: $-3 \cdot x^2 \cdot x = -3x^3$.
- Odd degree, negative coefficient.
Answer: Rises on the left, falls on the right.
Common mistake
Using the first term written instead of the term of highest degree. In $7 + 5x - 2x^4$ the first term is a constant, which tells you nothing about the ends. Rewrite in standard form (highest power first) before you decide. Check with a big number: $f(100)$ is hugely negative here, so the right end falls.