Polynomial Graph Analysis

In Algebra 2, you learned to factor and graph quadratics and simple cubics. College algebra extends that analysis to any degree polynomial. We develop a systematic toolkit: end behavior, zeros with multiplicity, turning points, sign analysis, and the intermediate value theorem. Together, these tools let you sketch accurate graphs by hand.

End Behavior

The end behavior of a polynomial describes what happens to $f(x)$ as $x \to +\infty$ and $x \to -\infty$. End behavior depends only on the leading term $a_n x^n$.

Memory aid: Even degree = both ends go the same direction. Odd degree = ends go opposite directions. Positive leading coefficient = right end goes up.

Worked Example 1: End Behavior

Describe the end behavior of $f(x) = -2x^5 + 3x^3 - x + 7$.

The leading term is $-2x^5$: odd degree with negative coefficient.

• As $x \to +\infty$: $f(x) \to -\infty$ (right end falls)
• As $x \to -\infty$: $f(x) \to +\infty$ (left end rises)

Zeros and Multiplicity

A zero of a polynomial is a value $c$ where $f(c) = 0$. If $(x - c)^m$ is a factor but $(x - c)^{m+1}$ is not, then $c$ has multiplicity $m$.

The multiplicity determines the graph’s behavior at the zero:

• Odd multiplicity (1, 3, 5, …): the graph crosses the $x$-axis at $c$
• Even multiplicity (2, 4, 6, …): the graph touches (bounces off) the $x$-axis at $c$
• Multiplicity 1: crosses like a line (straight through)
• Multiplicity 2: touches and turns (parabolic tangency)
• Multiplicity 3: crosses with an inflection (flattens before crossing)

Worked Example 2: Complete Zero Analysis

Analyze the zeros of $f(x) = 2(x + 1)^3(x - 2)^2(x - 4)$.

The degree is $3 + 2 + 1 = 6$ (even) with leading coefficient $2 > 0$, so both ends go up.

Turning Points

A turning point is where the graph changes from increasing to decreasing or vice versa. A polynomial of degree $n$ has at most $n - 1$ turning points.

Example: A degree 4 polynomial can have at most 3 turning points. It might have fewer (1 or 3, specifically — a polynomial of degree $n$ always has an odd/even number of turning points matching whether $n$ is odd/even minus one).

The minimum number of turning points can be deduced from the zeros and end behavior. If the graph must cross the $x$-axis at certain points and match the end behavior, there must be enough turning points to accommodate those changes.

The Intermediate Value Theorem (IVT)

The intermediate value theorem states: if $f$ is continuous on $[a, b]$ and $f(a)$ and $f(b)$ have opposite signs, then there exists at least one $c$ in $(a, b)$ where $f(c) = 0$.

Since all polynomials are continuous, this always applies. The IVT is a powerful tool for locating zeros when you cannot factor.

Worked Example 3: Using the IVT

Show that $f(x) = x^3 - 4x + 1$ has a zero between $x = 1$ and $x = 2$.

$f(1) = 1 - 4 + 1 = -2 < 0$ $f(2) = 8 - 8 + 1 = 1 > 0$

Since $f(1) < 0$ and $f(2) > 0$, by the IVT there is at least one zero in $(1, 2)$.

We can narrow the interval: $f(1.5) = 3.375 - 6 + 1 = -1.625 < 0$, so the zero is in $(1.5, 2)$.

Further: $f(1.75) = 5.359 - 7 + 1 = -0.641 < 0$, so the zero is in $(1.75, 2)$.

And: $f(1.9) = 6.859 - 7.6 + 1 = 0.259 > 0$, so the zero is in $(1.75, 1.9)$.

Sign Analysis with Test Intervals

To determine where a polynomial is positive or negative:

1. Find all real zeros
2. Plot them on a number line, dividing it into intervals
3. Test one point in each interval to determine the sign

Worked Example 4: Complete Sign Analysis

Determine the sign of $f(x) = (x + 3)(x - 1)^2(x - 5)$.

Zeros: $x = -3$ (mult. 1), $x = 1$ (mult. 2), $x = 5$ (mult. 1).

Test intervals:

Notice: the sign does not change at $x = 1$ because the multiplicity is even (the graph bounces there).

SVG: Polynomial Graph Analysis

The graph shows $f(x) = (x + 2)(x - 1)^2(x - 3)$, a degree-4 polynomial with positive leading coefficient. It crosses at $x = -2$ and $x = 3$ (multiplicity 1), bounces at $x = 1$ (multiplicity 2), and both ends rise.

Complete Graphing Strategy

To sketch the graph of a polynomial by hand:

1. Determine the degree and leading coefficient for end behavior
2. Find all zeros (factor, use rational root theorem, etc.) and determine multiplicity
3. Plot the zeros and mark crossing/bouncing behavior
4. Find the $y$-intercept ($f(0)$)
5. Determine sign in each interval between zeros
6. Estimate turning points — there are at most $n - 1$
7. Connect the dots smoothly, respecting end behavior, signs, and zero behavior

Worked Example 5: Complete Graph Sketch

Sketch $g(x) = -x^3 + 4x = -x(x^2 - 4) = -x(x - 2)(x + 2)$.

End behavior: Degree 3, negative leading coefficient: left end up, right end down.

Zeros: $x = 0$, $x = 2$, $x = -2$ — all multiplicity 1 (graph crosses at each).

$y$-intercept: $g(0) = 0$.

Sign analysis:

Turning points: At most $3 - 1 = 2$. One between $x = -2$ and $x = 0$, another between $x = 0$ and $x = 2$. Using calculus (or symmetry — this function is odd): turning points are at $x = \pm\frac{2}{\sqrt{3}} \approx \pm 1.15$, with $g\!\left(\frac{2}{\sqrt{3}}\right) \approx 3.08$.

Real-World Application: Structural Analysis

In civil engineering, the bending moment of a beam under multiple point loads can be modeled as a polynomial. For a beam with supports at $x = 0$ and $x = L$ with loads at positions $a_1, a_2, \ldots$, the moment diagram is piecewise polynomial. The zeros of the moment polynomial indicate where the beam has zero bending — and the maximum of the moment (a turning point) indicates where the beam is under greatest stress.

For a uniformly loaded beam of length 10 m, the moment is approximately:

$M(x) = -0.5x^2 + 5x$

This is a quadratic with zeros at $x = 0$ and $x = 10$, and maximum at $x = 5$ where $M(5) = 12.5$ kN-m.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• End behavior depends only on the degree and leading coefficient — even degree means same direction on both ends; odd means opposite
• Multiplicity determines crossing vs. bouncing: odd multiplicities cross, even multiplicities bounce
• A degree-$n$ polynomial has at most $n$ real zeros and at most $n - 1$ turning points
• The intermediate value theorem guarantees a zero between any two points where the function changes sign
• Sign analysis divides the number line at zeros and tests each interval
• The complete graphing strategy combines end behavior, zeros, sign analysis, and turning points

Return to College Algebra for more topics in this section.