Graphing Polynomial Functions: Practice

Read the three features the picture shows. The curve crosses at $x = -2$, so that zero has odd multiplicity. It touches and turns back at $x = 1$, so that zero has even multiplicity. The right arm rises, and for a cubic that means a positive leading coefficient.

$P(x) = (x+2)^1(x-1)^2$

The first choice has the roles reversed, touching at $-2$ and crossing at $1$. The one with $(x-2)$ and $(x+1)^2$ has its zeros at $2$ and $-1$, in the wrong places. The negative version has the right zeros but sends the right arm downward. Only $(x+2)(x-1)^2$ matches all three features.

The zeros are $0$, $2$, and $-3$, so the interval $0 < x < 2$ is free of zeros and $P$ keeps one sign on it. Test $x = 1$.

$P(1) = (1)(1-2)(1+3) = (1)(-1)(4) = -4 < 0$

Exactly one of the three factors is negative there, so the product is negative and the curve lies below the axis.

Divide by $(x-2)$ synthetically, using the coefficients $1, -3, 0, 4$. Bring down $1$, multiply by $2$ and add to get $-1$, multiply by $2$ and add to get $-2$, multiply by $2$ and add to get $0$. The remainder is $0$, so the depressed polynomial is $x^2 - x - 2$.

Now divide by $(x-2)$ again, since a second zero remainder would mean $x = 2$ repeats. Testing the depressed polynomial at $2$ gives $4 - 2 - 2 = 0$, so it does.

$P(x) = (x-2)^2(x+1)$

The multiplicity of $x = 2$ is $2$, which is even, so the graph touches the axis there and turns back.

Each crossing needs an odd multiplicity, and the smallest odd value is $1$. The touch needs an even multiplicity, and the smallest even value is $2$.

$1 + 2 + 1 = 4$

So the degree is at least $4$, and $P(x) = (x+1)(x-1)^2(x-4)$ reaches it. The true degree could be $6$ or $8$, since extra multiplicity comes in steps of two and non-real roots come in pairs, but it cannot be smaller than $4$ and it cannot be odd.

Substitute $x = 0$ into every factor, and do not forget the constant $3$ out front.

$P(0) = 3(0-1)(0+2)^2 = 3(-1)(4) = -12$

The graph passes through $(0, -12)$.

The zeros fix the factors, and the extra point fixes the scale factor $a$. Substitute $x = 0$ and set the result equal to $6$.

$a(0-1)(0+2) = -2a = 6 \quad \Longrightarrow \quad a = -3$

So $P(x) = -3(x-1)(x+2)$. Because $a$ is negative, this parabola opens downward, which the two zeros alone could never have told you.

By the Fundamental Theorem of Algebra there are exactly $5$ roots counted with multiplicity. The real zeros use up $1 + 2 = 3$ of them.

$5 - 3 = 2$

So $2$ roots are non-real, which is exactly one conjugate pair. The answer had to be even, since non-real roots of a real polynomial always come in conjugate pairs. Those two roots contribute no x-intercept, so the graph meets the axis at just two points.

The factor $x^2 + 1$ has no real zero, since $x^2 = -1$ has no real solution, so it contributes no x-intercept. The factor $(x-3)^2$ gives the real zero $3$ with multiplicity $2$.

$P(x) = \underbrace{(x^2+1)}_{> \, 0 \ \text{always}} (x-3)^2 \geq 0$

The multiplicity is even, so the curve touches the axis at $x = 3$ and turns back rather than crossing. Since $P(x) \geq 0$ for every real $x$, with equality only at $3$, the graph meets the axis exactly once.

Near a zero $r$ of multiplicity $m$, the size of $P$ is governed by the factor $|x-r|^m$, and a bigger $m$ makes that factor far smaller at the same distance from $r$.

$(0.2)^1 = 0.2 \qquad \text{versus} \qquad (0.2)^5 = 0.000\,32$

The zero $x = -2$ has multiplicity $5$ against multiplicity $1$ at $x = 1$, so the curve hugs the axis far more closely at $-2$ before pulling away. It still crosses there, because $5$ is odd.

Monic means the leading coefficient is $1$, and the three simple zeros give the three factors, so the polynomial is fully determined.

$P(x) = (x-1)(x-2)(x-3), \qquad P(0) = (-1)(-2)(-3) = -6$

Three negative factors multiply to a negative number, so the y-intercept is $(0, -6)$.

Each distinct real zero contributes at least $1$ to the total multiplicity, and the total multiplicity is exactly the degree.

$\text{number of distinct real zeros} \leq \deg P = 4$

So there can be at most $4$ x-intercepts, and $P(x) = (x-1)(x-2)(x-3)(x-4)$ achieves it. There can be fewer, if some zeros repeat or some roots are non-real.

The factor $(x-2)$ gives the real zero $2$. For the quadratic factor, compute the discriminant.

$b^2 - 4ac = (-4)^2 - 4(1)(5) = 16 - 20 = -4 < 0$

A negative discriminant means the quadratic has a non-real conjugate pair of roots and no real zero, so it contributes no x-intercept. The graph therefore meets the x-axis exactly once, at $x = 2$.