Introduction
A polynomial can have many zeros and several turning points, but its behavior far to the left and far to the right is controlled by a simple feature: its leading term. Knowing this gives a reliable global check on a curve sketch. The zeros and turning points describe important features at particular finite locations; they do not change the direction in which the graph eventually goes at either end.
For example, a graph may rise, fall, and rise again between several zeros. That local complexity is compatible with its far-right end rising, falling, or following a different pattern on the far-left end. To predict the ends, first identify the polynomial’s degree and leading coefficient.
Degree, Leading Coefficient, and Leading Term
Write a nonzero polynomial in the form \(P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\), where \(a_n\ne 0\). The degree is \(n\), the leading coefficient is \(a_n\), and the leading term is \(a_nx^n\). The degree is the highest exponent with a nonzero coefficient, so terms that cancel must be combined before identifying it.
The degree determines whether the ends go in the same or opposite vertical directions. The leading coefficient determines which vertical direction applies on the right. Together, these give four possibilities.
| Degree | Leading coefficient | As \(x\to-\infty\) | As \(x\to+\infty\) |
|---|---|---|---|
| Even | Positive | \(P(x)\to+\infty\) | \(P(x)\to+\infty\) |
| Even | Negative | \(P(x)\to-\infty\) | \(P(x)\to-\infty\) |
| Odd | Positive | \(P(x)\to-\infty\) | \(P(x)\to+\infty\) |
| Odd | Negative | \(P(x)\to+\infty\) | \(P(x)\to-\infty\) |
The constant polynomial case is degree zero: if \(P(x)=c\ne0\), its graph stays at the fixed height \(c\) at both ends. The table concerns positive degrees.
Why the Leading Term Determines the Ends
The leading term dominates because its power of \(x\) is higher than every other power in the polynomial. Here is a direct justification. For a degree-\(n\) polynomial with \(n\ge1\), separate its leading term from the rest:
Let \(C=\sum_{k=0}^{n-1}|a_k|\). When \(|x|\ge1\), each \(|x|^k\) in the lower-degree terms is at most \(|x|^{n-1}\). Consequently,
Choose \(|x|\) large enough that \(|x|>2C/|a_n|\), as well as \(|x|\ge1\). Then
Thus the total contribution from all lower-degree terms is less than half the magnitude of the leading term. If \(a_nx^n\) is positive, adding \(E(x)\) cannot make the result negative; if \(a_nx^n\) is negative, adding \(E(x)\) cannot make the result positive. So sufficiently far out, \(P(x)\) has the same sign as \(a_nx^n\). Its magnitude also grows without bound, since it is at least half the magnitude of that leading term. The parity of \(n\) and the sign of \(a_n\) therefore give the end-behavior table.
Worked Examples: Compare Ends, Zeros, and Turning Points
Worked Example: A Cubic with a Positive Leading Coefficient
Consider \(P(x)=x^3-3x\). Its degree is \(3\), which is odd, and its leading coefficient is \(1\), which is positive. Therefore the left end goes down and the right end goes up:
To locate the zeros, factor the polynomial:
The zeros are \(x=-\sqrt3,\ 0,\ \sqrt3\). Substitution checks them: \(P(0)=0\); \(P(\sqrt3)=(\sqrt3)^3-3\sqrt3=3\sqrt3-3\sqrt3=0\); and \(P(-\sqrt3)=(-\sqrt3)^3-3(-\sqrt3)=-3\sqrt3+3\sqrt3=0\).
For turning points, differentiate and find where the derivative is zero:
The derivative is positive for \(x<-1\), negative for \(-1<x<1\), and positive for \(x>1\). Thus the graph changes from rising to falling at \(x=-1\), and from falling to rising at \(x=1\). The turning-point values are \(P(-1)=(-1)^3-3(-1)=-1+3=2\) and \(P(1)=1-3=-2\). Directly checking the derivative gives \(P'(-1)=3-3=0\) and \(P'(1)=3-3=0\).
The graph’s zeros and turning points occur at finite \(x\)-values. They fit the prediction of a left end that falls and a right end that rises; neither turning point changes those eventual directions.
Worked Example: An Even-Degree Polynomial with Both Ends Up
Consider \(Q(x)=x^4-5x^2+4\). The degree is \(4\), which is even, and the leading coefficient is \(1>0\). Both ends therefore rise.
To find its zeros, factor using \(u=x^2\):
The zeros are \(-2,-1,1,2\). Substituting them into the factored form gives \(Q(-2)=Q(-1)=Q(1)=Q(2)=0\), because in each case one of the factors is zero.
For turning points, differentiate:
Solving \(Q'(x)=0\) gives \(x=0\) or \(2x^2-5=0\), so \(x=\pm\sqrt{5/2}\). The derivative is negative for \(x<-\sqrt{5/2}\), positive for \(-\sqrt{5/2}<x<0\), negative for \(0<x<\sqrt{5/2}\), and positive for \(x>\sqrt{5/2}\). Therefore there are turning points at all three locations.
The values are \(Q(0)=4\), and, for either \(x=\pm\sqrt{5/2}\),
Checking the derivative gives \(Q'(0)=0\). At either \(x=\pm\sqrt{5/2}\), the factor \(2x^2-5\) equals \(2(5/2)-5=0\), so \(Q'(x)=0\) there as well. The local turns and four zeros are compatible with both ends eventually rising. In particular, the zeros do not imply that the graph must continue downward beyond the last zero.
Worked Example: An Even-Degree Polynomial with Both Ends Down
Consider \(R(x)=-x^4+4x^2\). Its degree is \(4\), and its leading coefficient is \(-1\). The degree is even and the leading coefficient is negative, so both ends fall.
Factor to find the zeros:
The zeros are \(-2,0,2\). Substitution confirms them: \(R(-2)=-16+16=0\), \(R(0)=0\), and \(R(2)=-16+16=0\).
Differentiate to locate turning points:
The derivative is zero at \(x=-\sqrt2,0,\sqrt2\). For \(x<-\sqrt2\), \(R'(x)>0\); on \((-\sqrt2,0)\), \(R'(x)<0\); on \((0,\sqrt2)\), \(R'(x)>0\); and for \(x>\sqrt2\), \(R'(x)<0\). The graph therefore turns at each of those three values. The corresponding heights are
Checking the derivative, \(R'(0)=0\); at \(x=\pm\sqrt2\), \(x^2-2=0\), so \(R'(x)=0\). The graph can rise to two turning points and cross the axis at finite locations while still falling at both far ends. Its leading term requires that eventual behavior.
Worked Example: An Odd-Degree Polynomial with a Negative Leading Coefficient
Consider \(S(x)=-x^3+3x\). Its degree is \(3\), which is odd, and its leading coefficient is \(-1\). The left end rises and the right end falls:
Factoring gives \(S(x)=-x(x^2-3)=-x(x-\sqrt3)(x+\sqrt3)\), so the zeros are \(-\sqrt3,0,\sqrt3\). Substitution yields \(S(0)=0\), \(S(\sqrt3)=-3\sqrt3+3\sqrt3=0\), and \(S(-\sqrt3)=3\sqrt3-3\sqrt3=0\).
The derivative is \(S'(x)=-3x^2+3=3(1-x^2)\), which is zero at \(x=-1,1\). It is negative for \(x<-1\), positive for \(-1<x<1\), and negative for \(x>1\), so both are turning points. Their heights are \(S(-1)=-(-1)^3+3(-1)=1-3=-2\) and \(S(1)=-1+3=2\). The derivative checks give \(S'(-1)=-3+3=0\) and \(S'(1)=-3+3=0\). The turns and zeros fit the predicted left-up, right-down behavior.
Reading the Whole Picture
End behavior and local features answer different questions. The leading term predicts what the graph does sufficiently far to either side. A zero records where the graph meets the horizontal axis. A turning point records where the graph changes from rising to falling or vice versa. Zeros and turning points help describe the finite part of a graph, but they cannot overturn the leading-term prediction.
When checking a sketch, examine the two ends separately. A positive even-degree polynomial must eventually be above the axis on both sides; a negative even-degree polynomial must eventually be below it on both sides. Odd-degree polynomials must have opposite end directions, with the right end determined by the sign of the leading coefficient. A sketch that contradicts any of these statements cannot represent that polynomial, regardless of whether its plotted zeros appear correct.
Common Pitfall: Using the Wrong Term
A common error is to use a term that looks large near the origin, or to choose the term with the largest coefficient, instead of the term with the highest power. For \(T(x)=-2x^5+100x^2-1\), one might incorrectly focus on \(100x^2\) because \(100\) is the largest coefficient. That would suggest that both ends rise. But the degree is \(5\) and the leading coefficient is \(-2\), so the correct prediction is left end up and right end down. The fifth-degree term eventually dominates the quadratic term.
Another error is to treat a zero as an end-behavior rule. Knowing several zeros does not determine what happens at both ends; a polynomial with those zeros could have different degrees or leading-coefficient signs. First determine the degree and leading coefficient. Then use zeros and turning points to check the rest of the graph against that global prediction.
Check Your Understanding
For each polynomial, identify its degree and leading coefficient before answering.
- What is the end behavior of \(4x^6-3x^2+1\)?
- What is the end behavior of \(-5x^3+x-7\)?
- Why can the lower-degree terms change the zeros and turning points without changing the eventual end directions?
- A sketch of a degree-five polynomial with positive leading coefficient has both ends pointing down. What is inconsistent?
- For \(x^4-4x^2\), factor to find its zeros and use the leading term to predict both ends. Do those zeros contradict the prediction?
