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Match Polynomial Functions to Graphs Using Roots and End Behavior
Learn to pair factored polynomial equations with candidate graphs by checking degree, roots, and end behavior, with worked examples and practice.
Free lesson
Four exercises
Worked answers
Algebra and Function Practice
Matching a factored polynomial to a graph is a process of elimination using three visible features: the degree tells you the maximum number of turning points and the end behavior, the roots tell you where the graph crosses or touches the x-axis, and the sign of the leading coefficient tells you whether the ends go up or down. This lesson shows how to read those features from an equation and compare them to labeled candidate graphs without graphing from scratch.
Practice materials: Printable downloads are not available for this lesson yet. Use the instructions and four exercises below.
Read degree and end behavior from factored form
A polynomial written as a product of factors, such as f(x) = (x - 2)(x + 1)(x - 3), has degree equal to the sum of the exponents on its linear factors. Odd degree means the two ends of the graph point in opposite directions; even degree means they point the same way. The leading coefficient is the product of the coefficients of the highest-power terms in each factor. For monic linear factors the leading coefficient is positive, so odd degree gives down on the left and up on the right, while even degree gives up on both ends.
When a factor is repeated, the degree still counts the repetition, but the graph touches the x-axis at that root instead of crossing it. For example, (x - 1)^2 contributes degree 2 and a touch at x = 1. Counting multiplicity is therefore essential before comparing to candidate graphs.
- Expand mentally only enough to count the highest power: add the exponents of all factors.
- Decide whether the degree is odd or even.
- Multiply the signs of the leading coefficients of each factor to get the end behavior direction.
- List each root and its multiplicity from the factors.
Use roots and multiplicity to eliminate candidates
Each factor (x - r) gives a root at x = r. A multiplicity of 1 means the graph crosses the axis at that point; an even multiplicity means it touches and turns back; an odd multiplicity greater than 1 means it crosses but flattens near the root. Candidate graphs that show a crossing where the equation predicts a touch can be eliminated immediately.
Work through the roots in order and mark which candidates survive. Often one root with an unusual multiplicity, such as a double root at a negative value, removes most wrong graphs before you even check end behavior.
- Write every root and its multiplicity in a small table.
- For each candidate graph, note whether the curve crosses or touches at each labeled x-intercept.
- Eliminate any graph whose crossing or touching behavior disagrees with the table.
- Keep the surviving candidates for the end-behavior check.
Confirm with end behavior and a test point
After roots have narrowed the field, compare the left and right ends of each remaining graph with the predicted end behavior. If two candidates still look alike, evaluate the equation at one convenient x-value, such as x = 0, and compare the sign of the result with the graph's height at that x-value. This final check resolves ties without needing a full sketch.
Keep arithmetic simple and consistent: choose integer test points, compute the product of factors, and compare only the sign and rough size. A positive value means the graph is above the axis at that x; a negative value means below.
- Check the far-left and far-right direction of each surviving graph.
- Compute f(0) or another easy value from the equation.
- Compare the sign of that value with the graph at the same x.
- Select the single candidate that satisfies roots, multiplicity, and end behavior.
Worked example
Match f(x) = -(x + 2)(x - 1)^2 to one of these described candidate graphs. Graph A: crosses at x = -2, touches at x = 1, rises on both ends. Graph B: crosses at x = -2, touches at x = 1, rises on the left and falls on the right. Graph C: touches at x = -2, crosses at x = 1, falls on both ends.
Worked answer
Graph B.
Why: The factors give roots at x = -2 with multiplicity 1 (crossing) and x = 1 with multiplicity 2 (touching). The degree is 1 + 2 = 3, which is odd, so the ends point in opposite directions. The leading coefficient is negative because of the minus sign, so the graph rises on the left and falls on the right. Graph A has even-degree end behavior (rises on both ends), which is wrong for this odd-degree function; Graph C has the multiplicities reversed and the wrong end behavior. Graph B matches all three features: crossing at x = -2, touching at x = 1, and rising left/falling right. As a check, f(0) = -(2)(-1)^2 = -2, which is negative, so the graph should be below the x-axis at x = 0, consistent with Graph B.
Four practice exercises
Exercise 1
Match g(x) = (x - 4)(x + 1)(x - 2) to one of these described graphs. Graph P: crosses at x = -1, 2, and 4; falls on the left and rises on the right. Graph Q: crosses at x = -1, 2, and 4; rises on both ends. Graph R: touches at x = 2 and crosses at x = -1 and 4; falls on the left and rises on the right.
Show answer and check
Graph P.
Why: The degree is 3 (odd), and the leading coefficient is positive, so the graph falls on the left and rises on the right. All three roots have multiplicity 1, so the graph crosses at x = -1, 2, and 4. Graph Q has even-degree end behavior; Graph R shows a touch at x = 2, which would require an even multiplicity.
Check: Evaluate g(0) = (0 - 4)(0 + 1)(0 - 2) = (-4)(1)(-2) = 8, which is positive. The matching graph should be above the x-axis at x = 0.
Exercise 2
Match h(x) = (x + 3)^2(x - 1) to one of these described graphs. Graph S: touches at x = -3, crosses at x = 1, falls on the left and rises on the right. Graph T: crosses at x = -3, touches at x = 1, falls on both ends. Graph U: touches at x = -3, crosses at x = 1, falls on both ends.
Show answer and check
Graph S.
Why: The degree is 2 + 1 = 3, which is odd, and the leading coefficient is positive, so the graph falls on the left and rises on the right. The factor (x + 3)^2 gives a touch at x = -3, and (x - 1) gives a crossing at x = 1. Graph T has the multiplicities reversed and the wrong end behavior; Graph U has the wrong end behavior for a positive leading coefficient.
Check: Evaluate h(0) = (3)^2(-1) = -9, which is negative, so the graph should be below the x-axis at x = 0.
Exercise 3
Match k(x) = -2(x - 1)(x + 2)(x - 3) to one of these described graphs. Graph V: crosses at x = -2, 1, and 3; rises on the left and falls on the right. Graph W: crosses at x = -2, 1, and 3; falls on the left and rises on the right. Graph X: touches at x = 1 and crosses at x = -2 and 3; rises on the left and falls on the right.
Show answer and check
Graph V.
Why: The degree is 3 (odd), and the leading coefficient is -2, which is negative, so the graph rises on the left and falls on the right. All roots have multiplicity 1, so the graph crosses at x = -2, 1, and 3. Graph W has the wrong end behavior for a negative leading coefficient; Graph X shows a touch at x = 1, which is incorrect.
Check: Evaluate k(0) = -2(-1)(2)(-3) = -2(6) = -12, which is negative, so the graph should be below the x-axis at x = 0.
Exercise 4
Match m(x) = (x - 2)^2(x + 1)^2 to one of these described graphs. Graph Y: touches at x = -1 and x = 2, rises on both ends. Graph Z: crosses at x = -1 and x = 2, falls on both ends. Graph AA: touches at x = -1 and x = 2, falls on both ends.
Show answer and check
Graph Y.
Why: The degree is 2 + 2 = 4, which is even, and the leading coefficient is positive, so the graph rises on both ends. Both factors have even multiplicity, so the graph touches at x = -1 and x = 2 without crossing. Graph Z shows crossings, which is wrong; Graph AA has the wrong end behavior for a positive leading coefficient.
Check: Evaluate m(0) = (-2)^2(1)^2 = 4, which is positive, so the graph should be above the x-axis at x = 0.
Common mistakes
- Counting factors instead of adding multiplicities when finding the degree. Add the exponents of all factors, including repeated ones, to get the true degree.
- Treating a double root as a crossing point. An even multiplicity means the graph touches the axis and turns back; check the exponent on that factor.
- Assuming all odd-degree graphs rise to the right. The leading coefficient sign matters; a negative leading coefficient makes an odd-degree graph fall to the right.
- Skipping the test-point check when two candidates look similar. Evaluate the equation at x = 0 or another easy integer and compare the sign with the graph.
Self-check rubric
- Degree correctly counted from exponents, including multiplicities.
- Each root listed with correct multiplicity and crossing or touching behavior.
- End behavior direction stated for both left and right ends.
- Candidate graphs eliminated with a stated reason, not by guesswork.
- Final match supported by at least one test-point or y-intercept check.
- Arithmetic at test points is correct and units are consistent.
Questions and answers
Do I need to expand the polynomial fully to match it?
No. Counting degree from exponents, listing roots with multiplicities, and reading the leading coefficient sign are enough for matching.
What if a candidate graph has an extra turning point?
A polynomial of degree n has at most n - 1 turning points. A graph with more turning points than the degree allows cannot be the match.
How do I handle a factor like (2x - 3)?
The root is x = 3/2, and the leading coefficient contributes a factor of 2. Include that 2 when determining end behavior.
Can two different equations have the same roots and end behavior?
Yes, if they differ only by a positive constant multiple. In matching tasks, the candidate graphs usually differ in at least one visible feature, such as multiplicity or a y-intercept sign.