From a Graph to a Formula
A polynomial graph can reveal several important features of its formula. Its intercepts show where the polynomial is zero; its behavior at each intercept can suggest whether the zero has odd or even multiplicity; and its far-left and far-right behavior gives information about the degree and leading coefficient.
These clues can be assembled into a factored expression. A further point on the graph can then determine the remaining numerical factor. The key is to distinguish clues that determine part of the formula from clues that establish the entire formula.
Zeros and Multiplicities in Factored Form
A real number \(r\) is a zero of a polynomial \(f\) if \(f(r)=0\). Its multiplicity is the largest positive integer \(m\) for which \((x-r)^m\) divides \(f(x)\). Thus, a zero of multiplicity \(m\) contributes a factor \((x-r)^m\) to the polynomial.
The full-degree condition matters. A graph may show all the real zeros while the polynomial also has nonreal factors. For example, the polynomial \(2(x+1)(x-1)(x^2+1)\) has just two real zeros, \(-1\) and \(1\), each of multiplicity one. Those two real zeros do not account for its degree, which is four.
In practice, the graph clues may explicitly give the degree, or state that the listed zeros and multiplicities account for it. Without such information, the product of the listed real-root factors may be only part of the polynomial.
Reading Multiplicity and End Behavior
Near a zero \(r\), a polynomial with multiplicity \(m\) has a factor \((x-r)^m\). If \(m\) is odd, this factor changes sign as \(x\) passes through \(r\), so the graph crosses the horizontal axis. If \(m\) is even, the factor has the same sign on both sides of \(r\), so the graph touches the axis and turns back rather than crossing it.
A graph’s crossing or touching behavior indicates whether the multiplicity is odd or even. It does not, by itself, always determine the exact multiplicity: crossing is consistent with multiplicity one, three, five, and so on; touching is consistent with multiplicity two, four, six, and so on. Use an exact multiplicity only when the clue supplies it or other information establishes it.
The end behavior provides a separate consistency check. For a polynomial of degree \(n\) with leading coefficient \(c\), its behavior for large positive and negative \(x\) matches that of \(cx^n\). Even degree gives ends in the same direction; odd degree gives ends in opposite directions. If \(c>0\), the right end goes up, while if \(c<0\), the right end goes down. The left end is then determined by the degree parity.
| Degree and leading coefficient | Left end | Right end |
|---|---|---|
| Even degree, positive leading coefficient | Up | Up |
| Even degree, negative leading coefficient | Down | Down |
| Odd degree, positive leading coefficient | Down | Up |
| Odd degree, negative leading coefficient | Up | Down |
End behavior usually determines the degree’s parity and the sign of the leading coefficient, not the exact degree or coefficient value. The roots and multiplicities can supply the degree when they account for all factors.
Why the Product Form Works
The factor theorem says that if \(f(r)=0\), then \((x-r)\) divides \(f(x)\). If the zero has multiplicity \(m\), the factor appears \(m\) times. Applying this to each distinct zero gives the product of all the corresponding factors as a divisor of \(f\).
Here is why accounting for the degree completes the reconstruction. Let \(Q(x)=\prod_{i=1}^k(x-r_i)^{m_i}\), and suppose the listed multiplicities sum to the degree \(n\) of \(f\). The factor theorem, applied with the stated multiplicities, gives \(f(x)=Q(x)S(x)\) for some polynomial \(S\). The degree of \(Q\) is \(n\), so \(\deg f=\deg Q+\deg S\) implies \(n=n+\deg S\). Therefore \(\deg S=0\), which means \(S\) is a nonzero constant \(c\). Hence \(f(x)=cQ(x)\). If the listed multiplicities sum to less than \(n\), this argument does not show that \(S\) is constant; additional factors may remain.
Finding the Leading Coefficient from a Point
Once the root factors are known, write \(f(x)=cQ(x)\). If a known point \((a,b)\) lies on the graph, then \(f(a)=b\), so \(b=cQ(a)\). Provided \(Q(a)\ne 0\), solve for \(c\) by dividing by \(Q(a)\).
Worked Example: One Simple Zero and One Double Zero
A degree-three polynomial has a double zero at \(-2\), a simple zero at \(1\), and passes through \((0,8)\). Its right end goes down. Reconstruct the polynomial.
The multiplicities sum to \(2+1=3\), the stated degree, so the factors account for the full degree. The factored form is \(f(x)=c(x+2)^2(x-1)\).
Substitute the known point’s \(x\)-coordinate, \(x=0\), into the product: \(Q(0)=(0+2)^2(0-1)=2^2(-1)=4(-1)=-4\). Since \(f(0)=8\), we have \(8=c(-4)\), so \(c=-2\). Therefore \(f(x)=-2(x+2)^2(x-1)\).
Check the point: \(f(0)=-2(2)^2(-1)=-2(4)(-1)=8\). The factors give a double zero at \(-2\) and a simple zero at \(1\). The degree is three and the leading coefficient is negative, so the right end goes down, as stated.
Check the Whole Set of Clues
A reconstructed formula should agree with each clue, not just the point used to find \(c\). Verify its zeros and multiplicities from the factors, its degree from the sum of their exponents when they account for the full degree, its end behavior from degree and leading-coefficient sign, and its value at the known point.
Worked Example: Two Double Zeros and Downward Ends
A degree-four polynomial has double zeros at \(-1\) and \(2\), has both ends down, and passes through \((0,-8)\). Find its formula.
The multiplicities total \(2+2=4\), so they account for the degree. Thus \(f(x)=c(x+1)^2(x-2)^2\). Evaluate the product at \(x=0\): \(Q(0)=(0+1)^2(0-2)^2=1^2(-2)^2=1(4)=4\). The point gives \(-8=4c\), and therefore \(c=-2\). The formula is \(f(x)=-2(x+1)^2(x-2)^2\).
Check: \(f(0)=-2(1)(4)=-8\). Each squared factor gives the stated double zero. The degree is four, and the leading coefficient is \(-2\). An even degree with a negative leading coefficient has both ends down, as required.
Worked Example: A Fourth-Degree Polynomial with Three Distinct Zeros
A degree-four polynomial has simple zeros at \(-3\) and \(2\), a double zero at \(0\), and passes through \((3,27)\). Both ends go up. Reconstruct it.
The multiplicities sum to \(1+2+1=4\), so all factors are accounted for. Write \(f(x)=c(x+3)x^2(x-2)\). At \(x=3\), the product without \(c\) is \(Q(3)=(3+3)(3^2)(3-2)=6(9)(1)=54\). The point gives \(27=54c\), so \(c=\frac{27}{54}=\frac12\). Thus \(f(x)=\frac12(x+3)x^2(x-2)\).
Check: \(f(3)=\frac12(6)(9)(1)=27\). The factors show simple zeros at \(-3\) and \(2\), and a double zero at \(0\). The degree is four, and the leading coefficient is positive, so both ends go up.
A Common Mistake: Treating Real Zeros as All Factors
Suppose the only real zeros shown are \(-1\) and \(1\), each simple. It would be unjustified to conclude that the polynomial must be \(c(x+1)(x-1)\) unless the degree is two or the clues otherwise establish that those factors account for the full degree.
For a concrete counterexample, let \(g(x)=2(x+1)(x-1)(x^2+1)\). For real \(x\), the factor \(x^2+1\) is never zero, so the only real zeros are indeed \(-1\) and \(1\), both simple. But \(g\) has degree four, not two. Its real-root factors alone would give \(c(x+1)(x-1)\), which has degree two and cannot equal \(g\). The correction is to check the degree before treating the product of the listed real-root factors as the whole polynomial.
There is a related limitation: when multiplicities are not given exactly, a graph that crosses at a zero does not justify choosing multiplicity one rather than another odd multiplicity. Such a choice can produce a plausible formula, but it is not uniquely forced by crossing behavior alone.
When the Clues Determine a Unique Formula
If the zeros, their exact multiplicities, and the fact that those multiplicities account for the degree are known, then the only remaining choice in the factored form is \(c\). A graph point away from the zeros determines \(c\), and the end behavior checks its sign and the degree’s parity. If the point is itself a zero, or if a factor is missing because the listed multiplicities do not account for the degree, the available clues may not determine a unique polynomial.
Use exact multiplicities when supplied; crossing or touching alone gives parity information, not always the exact value.
If they do, form \(Q(x)=\prod_i(x-r_i)^{m_i}\). If not, more information is needed to identify any remaining factors.
Substitute its coordinates into \(f(x)=cQ(x)\), ensuring the point is not at a zero.
Check the zeros, multiplicities, known point, degree, leading-coefficient sign, and end behavior.
Check Your Understanding
Use the definitions and checks in this tutorial to answer the following questions.
- What condition lets you conclude that a product of listed zero factors differs from the polynomial only by a constant?
- A graph crosses the axis at a zero. What does that indicate about the multiplicity, and what does it not determine by itself?
- Why can a known point at a zero fail to determine the leading coefficient?
- A degree-four polynomial has positive leading coefficient. What direction do its two ends take?
- What additional possibility must be considered if the listed real-zero multiplicities sum to less than the polynomial’s degree?
