Graphing polynomial functions — zeroes, end behavior, turning points

Question

How do we sketch the graph of a polynomial function without plotting dozens of points? Use the polynomial $p(x) = x^3 - 3x^2 - x + 3$ to demonstrate the process.

(CBSE 10 Board pattern)

Solution — Step by Step

Set $p(x) = 0$ : $x^3 - 3x^2 - x + 3 = 0$

Try $x = 1$ : $1 - 3 - 1 + 3 = 0$ . Yes, $x = 1$ is a root.

Factor out $(x - 1)$ by division: $p(x) = (x - 1)(x^2 - 2x - 3) = (x - 1)(x - 3)(x + 1)$

Zeroes: $x = -1, 1, 3$ . These are the points where the graph crosses the x-axis.

The leading term is $x^3$ (odd degree, positive coefficient).

As $x \to +\infty$ , $p(x) \to +\infty$
As $x \to -\infty$ , $p(x) \to -\infty$

So the graph goes from bottom-left to top-right.

Maximum number of turning points = degree $- 1 = 2$ .

Take $p'(x) = 3x^2 - 6x - 1 = 0$ . Using the quadratic formula:

x = \frac{6 \pm \sqrt{36 + 12}}{6} = \frac{6 \pm \sqrt{48}}{6} = 1 \pm \frac{2\sqrt{3}}{3}

So approximately $x \approx -0.15$ and $x \approx 2.15$ . These are the peaks and valleys.

$p(0) = 0 - 0 - 0 + 3 = 3$ . The y-intercept is $(0, 3)$ .

Now connect the dots: start from bottom-left, cross x-axis at $x = -1$ , rise to a local max near $x \approx -0.15$ , fall to a local min near $x \approx 2.15$ , cross at $x = 1$ , dip down, cross at $x = 3$ , then rise to top-right.

flowchart TD
    A["Given polynomial p(x)"] --> B["Step 1: Find zeroes - set p(x) = 0"]
    B --> C["Step 2: Check leading term"]
    C --> D{"Degree odd or even?"}
    D -- Odd --> E["Opposite ends: one up, one down"]
    D -- Even --> F["Same ends: both up or both down"]
    E --> G["Step 3: Find p'(x) = 0 for turning points"]
    F --> G
    G --> H["Step 4: Find y-intercept p(0)"]
    H --> I["Step 5: Plot zeroes, turning points, y-intercept"]
    I --> J["Connect smoothly following end behavior"]

Why This Works

A polynomial of degree $n$ can cross the x-axis at most $n$ times and can turn at most $n - 1$ times. The leading term dominates the behavior for very large and very small values of $x$ , so it determines the “arms” of the graph. Between the zeroes, the sign of $p(x)$ alternates (for simple roots), which tells us whether the curve is above or below the x-axis.

By combining zeroes, end behavior, turning points, and the y-intercept, we have enough information to sketch an accurate shape without plotting a large number of points.

Alternative Method

Instead of calculus (finding $p'(x)$ ), you can use a sign analysis table:

Interval	$(x+1)$	$(x-1)$	$(x-3)$	$p(x)$ sign
$x < -1$	$-$	$-$	$-$	$-$
$-1 < x < 1$	$+$	$-$	$-$	$+$
$1 < x < 3$	$+$	$+$	$-$	$-$
$x > 3$	$+$	$+$	$+$	$+$

This tells you exactly where the graph is above or below the x-axis — no derivatives needed.

For CBSE 10, you will not need calculus to find turning points. The sign analysis method is sufficient and expected. Just know the zeroes, the end behavior, and where the curve is positive vs negative.

Common Mistake

Students often forget that a repeated root (like a double root) means the graph touches the x-axis but does not cross it. For example, $p(x) = (x-2)^2(x+1)$ touches the x-axis at $x = 2$ and crosses at $x = -1$ . If you treat all roots the same, your sketch will be wrong.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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