Skill: Sketch polynomial functions using zeros and multiplicities

Read the polynomial in factored form. Each factor gives an $x$-intercept, and its exponent gives the multiplicity.

• Odd multiplicity: The graph crosses the $x$-axis.
• Even multiplicity: The graph touches the $x$-axis and turns around.
• A larger multiplicity makes the graph flatter near that zero.

Then use the degree and leading coefficient to determine the end behavior.

Worked example

Sketch the polynomial

$$f(x)=-2(x+2)^2(x-1)^3.$$

1. Find the zeros and multiplicities

Set each factor equal to zero:

$$x+2=0 \quad \Rightarrow \quad x=-2$$

The zero $x=-2$ has multiplicity $2$, which is even. The graph touches the $x$-axis and turns around there.

$$x-1=0 \quad \Rightarrow \quad x=1$$

The zero $x=1$ has multiplicity $3$, which is odd. The graph crosses the $x$-axis there, but it is flatter than it would be at a zero of multiplicity $1$.

So the $x$-intercepts are

$$(-2,0) \quad \text{and} \quad (1,0).$$

2. Determine the end behavior

The degree is

$$2+3=5,$$

so the polynomial has odd degree. Its leading coefficient is negative because of the factor $-2$.

An odd-degree polynomial with a negative leading coefficient has this end behavior:

• As $x\to -\infty$, $f(x)\to \infty$.
• As $x\to \infty$, $f(x)\to -\infty$.

Thus, the graph rises on the far left and falls on the far right.

3. Determine whether the graph is above or below the axis

The graph touches at $x=-2$, so it does not change sides there. It crosses at $x=1$, so it changes sides there.

For an additional point, find the $y$-intercept:

$$f(0)=-2(0+2)^2(0-1)^3$$ $$f(0)=-2(4)(-1)=8.$$

Therefore, the graph passes through $(0,8)$.

4. Put the features together

A qualitative sketch should:

1. Start high on the left.
2. Decrease to $(-2,0)$.
3. Touch the $x$-axis at $x=-2$ and turn back upward.
4. Pass through $(0,8)$.
5. Cross the $x$-axis at $x=1$, with a noticeably flatter crossing because the multiplicity is $3$.
6. Continue downward to the right.

The zeros and multiplicities determine the intercept behavior and the overall end behavior. They do not determine the exact locations of all turning points, so the sketch should be qualitative rather than perfectly precise.