From algebraic features to a graph
A polynomial graph can be sketched without calculating dozens of points. Its factored form already reveals where it meets the horizontal axis, whether it crosses or touches that axis at each zero, and the sign of the polynomial between consecutive zeros. Its degree and leading coefficient determine the direction of its two ends.
These features do not specify every detail of the curve. A qualitative sketch shows the overall behavior supported by the algebra, rather than claiming that the graph has a particular precise shape between the features. The goal is to combine the information accurately and make the graph consistent with it.
Read the zeros and multiplicities
Suppose a polynomial is written in factored form as a nonzero constant times linear factors, possibly with a remaining factor that has no real zeros. A real zero \(a\) with multiplicity \(m\) means that the factor \((x-a)^m\) occurs, but \((x-a)^{m+1}\) does not. The graph meets the horizontal axis at \((a,0)\), since substituting \(x=a\) makes the polynomial zero.
The parity of the multiplicity predicts what happens at that point. An odd multiplicity gives a crossing: the polynomial changes sign as \(x\) passes through the zero. An even multiplicity gives a touch: the polynomial has the same sign on both sides, so it returns to the same side of the axis. A higher multiplicity can make the graph look flatter near the zero, but it does not change this crossing-versus-touch rule.
Why multiplicity determines crossing or touch
Near a real zero \(a\) of multiplicity \(m\), the polynomial can be written as \(f(x)=(x-a)^m q(x)\), where \(q\) is a polynomial and \(q(a)\ne 0\). Polynomials are continuous, so because \(q(a)\) is nonzero, \(q(x)\) has the same sign as \(q(a)\) for all \(x\) sufficiently close to \(a\). Thus the sign change near \(a\) is determined by \((x-a)^m\).
If \(m\) is even, \((x-a)^m\) is positive on both sides of \(a\), so \(f(x)\) has the same sign on both sides. If \(m\) is odd, \((x-a)^m\) is negative to the left and positive to the right, so \(f(x)\) changes sign. This proves the rule locally at each zero.
Determine signs between the zeros
List the distinct real zeros in increasing order. They divide the real line into intervals. On any one of these intervals, a polynomial with no zeros cannot change sign: if it took both positive and negative values there, continuity and the Intermediate Value Theorem would force it to equal zero somewhere in between.
To determine the sign on each interval, start on one interval where the sign is easy to find—for example, far to the right using the leading coefficient—and move across the zeros. Keep the sign at even-multiplicity zeros; reverse it at odd-multiplicity zeros. This gives the signs throughout the real line without testing a value in every interval.
Use degree and leading coefficient for end behavior
For a polynomial of degree \(n\) with leading coefficient \(a_n\ne 0\), the leading term \(a_nx^n\) determines the end behavior. If \(n\) is even, both ends point in the same direction: both rise when \(a_n>0\), and both fall when \(a_n<0\). If \(n\) is odd, the ends point in opposite directions: the left end rises and the right end falls when \(a_n>0\); the left end falls and the right end rises when \(a_n<0\).
When the polynomial is already factored into linear factors, its degree is the sum of the multiplicities, and its leading coefficient is the coefficient multiplying those factors. The vertical intercept is found separately by evaluating \(f(0)\), if that value is useful for the sketch.
Place each zero on the horizontal axis and record its multiplicity.
Use the degree and leading coefficient to determine the direction of both ends.
Start with a known interval sign and change sign at odd-multiplicity zeros only.
Calculate \(f(0)\) and plot \((0,f(0))\) when it adds information.
Draw a continuous curve that respects the signs, crossings or touches, and end behavior.
Worked sketches
Worked Example: A touch between two crossings
Sketch the key features of \(f(x)=(x+2)^2(x-1)(x-3)\).
1. Find the zeros and multiplicities. The factor \((x+2)^2\) gives the zero \(x=-2\) with multiplicity \(2\). The factors \(x-1\) and \(x-3\) give simple zeros \(x=1\) and \(x=3\). The graph touches at \(-2\) and crosses at \(1\) and \(3\).
2. Determine the ends. The degree is \(2+1+1=4\), which is even. The leading coefficient is positive. Therefore, both ends rise.
3. Track signs from right to left. For \(x>3\), all factors are positive, so \(f(x)>0\). Crossing \(3\), a simple zero, changes the sign, so \(f(x)<0\) on \((1,3)\). Crossing \(1\) changes it again, so \(f(x)>0\) on \((-2,1)\). At \(-2\), the multiplicity is even, so the sign stays positive on \((-\infty,-2)\).
4. Find the vertical intercept. Substitution gives \(f(0)=(2)^2(-1)(-3)=4\cdot 3=12\). The graph passes through \((0,12)\).
Check. At \(x=2\), \(f(2)=4^2(1)(-1)=-16\), confirming the negative sign on \((1,3)\). At \(x=0\), the factored calculation gives \(12\), confirming the vertical intercept. The completed sketch must rise at both ends, touch the axis at \(-2\), cross at \(1\) and \(3\), and pass through \((0,12)\).
Worked Example: An odd-degree graph with a flattened crossing
Sketch the key features of \(g(x)=-2(x+1)^3(x-2)^2\).
1. Find the zeros and multiplicities. The zero \(-1\) has multiplicity \(3\), so the graph crosses there. The zero \(2\) has multiplicity \(2\), so the graph touches there.
2. Determine the ends. The degree is \(3+2=5\), which is odd, and the leading coefficient is \(-2\). Thus the left end rises and the right end falls.
3. Track signs. For \(x>2\), the factors \(x+1\) and \(x-2\) are positive, so the negative coefficient makes \(g(x)<0\). The even-multiplicity zero at \(2\) does not change the sign, so \(g(x)<0\) on \((-1,2)\). Crossing \(-1\), whose multiplicity is odd, changes the sign; therefore \(g(x)>0\) on \((-\infty,-1)\).
4. Find the vertical intercept. \(g(0)=-2(1)^3(-2)^2=-2\cdot1\cdot4=-8\). The graph passes through \((0,-8)\).
Check. At \(x=-2\), \(g(-2)=-2(-1)^3(-4)^2=-2(-1)(16)=32>0\), matching the left interval. At \(x=3\), \(g(3)=-2(4)^3(1)^2=-128<0\), matching the right interval. The sketch rises on the left, crosses at \(-1\), stays below the axis while touching at \(2\), and falls on the right.
Worked Example: Combine an intercept with the interval signs
Sketch the key features of \(h(x)=\frac{1}{2}(x+3)(x-1)^2(x-4)\).
1. Find the zeros and multiplicities. There is a simple zero at \(-3\), a double zero at \(1\), and a simple zero at \(4\). The graph crosses at \(-3\) and \(4\), and touches at \(1\).
2. Determine the ends. The degree is \(1+2+1=4\), and the leading coefficient is \(\frac{1}{2}>0\). Both ends rise.
3. Find the signs. For \(x>4\), every variable factor is positive, so \(h(x)>0\). Crossing \(4\) reverses the sign, giving \(h(x)<0\) on \((1,4)\). At \(1\), the even multiplicity preserves the sign, so \(h(x)<0\) on \((-3,1)\). Crossing \(-3\) reverses the sign, giving \(h(x)>0\) on \((-\infty,-3)\).
4. Find the vertical intercept. \(h(0)=\frac{1}{2}(3)(-1)^2(-4)=\frac{1}{2}\cdot3\cdot1\cdot(-4)=-6\). The graph passes through \((0,-6)\).
Check. At \(x=2\), \(h(2)=\frac{1}{2}(5)(1)^2(-2)=-5<0\), confirming the sign on \((1,4)\). At \(x=-4\), \(h(-4)=\frac{1}{2}(-1)(-5)^2(-8)=100>0\), confirming the sign on \((-\infty,-3)\). These values agree with the end behavior and the intercept.
A common mistake: treating every zero as a crossing
Consider \(p(x)=(x-2)^2(x+1)\). A mistaken sign chart might say that the sign reverses at both zeros just because the graph meets the axis at both. That gives the wrong conclusion: starting with a positive sign for \(x>2\), it would incorrectly assign a negative sign to \((-1,2)\).
The correction is to use multiplicity. The zero \(2\) has even multiplicity, so the sign does not change there. For example, \(p(3)=(1)^2(4)=4>0\) and \(p(0)=(-2)^2(1)=4>0\); the polynomial is positive on both sides of \(2\). The zero \(-1\) has odd multiplicity, so the sign changes there. The multiplicity, not merely the presence of a zero, determines whether a crossing occurs.
What a qualitative sketch does—and does not—show
A feature-based sketch is useful because it brings several algebraic facts together in one picture. It makes it possible to see where a polynomial is positive or negative, where it meets the horizontal axis, and how its ends behave. These are often precisely the features needed to interpret a polynomial model or check that an algebraic result is plausible.
Do not treat the sketch as a claim about the exact curve between known points. The features used here do not determine the precise location of every high or low point, nor do they give exact coordinates for all parts of the curve. Draw a smooth, continuous curve consistent with the established information, and avoid adding unsupported numerical detail.
Check Your Understanding
Use the algebraic features in each question to reason about the graph. No precise curve coordinates are required unless requested.
- For \(f(x)=(x-1)^2(x+2)\), identify each real zero and say whether the graph crosses or touches at it.
- For \(g(x)=-3(x+1)(x-2)^4\), determine the direction of each end and state whether the sign changes at each zero.
- If a polynomial has odd multiplicity at each of its three distinct real zeros, how does its sign change as \(x\) moves from left to right across those zeros?
- Why can a polynomial with no zero in an interval not take both positive and negative values on that interval?
- For \(h(x)=(x+2)^2(x-1)\), calculate the vertical intercept and use the multiplicities to describe the sign on either side of each zero.
