Graphing Polynomial Functions: Practice

The crossing needs an odd multiplicity and the touch needs an even one, so the real multiplicities add up to odd plus even, which is odd.

$\underbrace{\text{odd}}_{\text{crossing}} + \underbrace{\text{even}}_{\text{touch}} = \text{odd}$

Every remaining root is non-real, and non-real roots of a real polynomial come in conjugate pairs, so there is an even number of them. An odd number plus an even number is odd, which can never equal the even degree $4$. So the configuration is impossible.

The first suggested example has degree $1 + 2 + 2 = 5$, not $4$. The second touches at both of its zeros and never crosses. The claim about at most three x-intercepts is simply false, since $(x-1)(x-2)(x-3)(x-4)$ has four.

The non-real roots come in conjugate pairs, so they contribute an even total. The multiplicities of the real zeros must therefore add up to a number with the same parity as the degree.

$\text{degree odd} \ \Longrightarrow \ \text{the real multiplicities sum to an odd number}$

An odd sum cannot be zero, and it cannot come from even multiplicities alone, so at least one real zero has odd multiplicity, and the graph crosses there. End behavior gives the same conclusion by a different road: an odd-degree graph sends one arm up and the other down, so it must travel from below the axis to above it.

The other statements can fail. $x^3 - x$ has three real roots and three x-intercepts, while $x^3 + x = x(x^2+1)$ has only one, so neither the count nor the presence of non-real roots is forced.

The zeros are $-2$, $1$, and $4$, so test one point in each of the four intervals they cut out.

$P(-3) = -(1)(-4)(-343) = -1372, \qquad P(0) = -(4)(-1)(-64) = -256$

$P(2) = -(16)(1)(-8) = 128, \qquad P(5) = -(49)(4)(1) = -196$

Only the test at $x = 2$ came out positive, so $P$ is above the axis exactly on $1 < x < 4$. The signs are consistent with the structure: the even multiplicity at $-2$ produces no sign change, while the odd multiplicities at $1$ and $4$ each flip the sign.

Crossing forces the multiplicity to be odd, so it is $1$, $3$, $5$, and so on. Multiplicity $1$ is ruled out by the flattening, because near a simple zero the curve looks like the straight line $y = G(r)(x - r)$, which meets the axis at an angle rather than lying down against it.

$\text{odd and greater than } 1 \ \Longrightarrow \ m \geq 3$

So the smallest possibility is $m = 3$. Be careful with what the picture proves: flattening is a strong hint, never a proof, since $m$ could equally well be $5$ or $7$.

The polynomial is a quadratic in $x^2$, so factor it as one.

$x^4 - 5x^2 + 4 = (x^2 - 1)(x^2 - 4) = (x-1)(x+1)(x-2)(x+2)$

The four real zeros $1$, $-1$, $2$, and $-2$ are all simple, so the graph crosses the axis at all four of them. A quartic cannot do better than that, since it has only $4$ roots in total.

Divide synthetically by $(x-2)$ using the coefficients $1, -5, 8, -4$: bring down $1$, then $-5 + 2 = -3$, then $8 - 6 = 2$, then $-4 + 4 = 0$. The remainder is $0$, and the depressed polynomial is $x^2 - 3x + 2$.

That quotient still vanishes at $2$, since $4 - 6 + 2 = 0$, so the factor $(x-2)$ divides $P$ at least twice.

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

The multiplicity of $2$ is even, so the graph touches the axis at $(2, 0)$ and turns back. Since $P$ is positive on both sides of $2$, that point is a local minimum sitting exactly on the axis, not above it.

The sign of $P$ can change only at a real zero, and at a zero of even multiplicity it does not change. So $P$ never changes sign anywhere on the number line.

That single sign is settled by the end behavior. The degree is even and the leading coefficient is positive, so both arms rise and $P$ is positive far out to the right.

$P(x) > 0 \ \text{ for large } x \ \Longrightarrow \ P(x) \geq 0 \ \text{ for every real } x$

The graph therefore stays on or above the axis, touching it at the two zeros. It crosses nowhere, so the second choice is false, and it does touch, so the third is false. The last choice is not forced: $(x-1)^2(x+1)^2(x^2+1)$ fits the description and has a non-real pair.

This is the one degree where algebra really can find the turning point. A parabola is symmetric about the vertical line halfway between its two roots.

$x = \frac{-1 + 3}{2} = 1$

Evaluate there to get the height.

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

The vertex is $(1, -4)$. The point $(0, -3)$ is only the y-intercept, and $(1, 0)$ is not on the graph at all. For a cubic or higher, this symmetry argument collapses, and the turning points really do become a question for calculus.

The polynomial is a product of two squares, and a square of a real number is never negative.

$P(x) = \underbrace{(x-1)^2}_{\geq \, 0} \cdot \underbrace{(x-3)^2}_{\geq \, 0} \geq 0$

So $P(x) \geq 0$ everywhere, with equality only at $x = 1$ and $x = 3$. The multiplicity rule says the same thing: both zeros are even, so the sign never flips at either one, and the sign it holds is positive.

Turn each zero into a factor with the multiplicity it needs, and allow a scale factor $a$ out front.

$P(x) = a\,x^2(x+4)$

Now use the point $(1, 10)$ to pin $a$ down.

$a(1)^2(1+4) = 5a = 10 \quad \Longrightarrow \quad a = 2$

So $P(x) = 2x^2(x+4)$. The choice $x^2(x+4)$ passes through $(1,5)$ instead, and $2x(x+4)^2$ puts the double zero at the wrong place, touching the axis at $-4$ rather than at $0$.

Check each claim against what the method actually delivers. The sign on $2 < x < 5$ comes from a single test point.

$P(3) = (4)(1)(-2) = -8 < 0$

The y-intercept comes from one evaluation, $P(0) = (1)(-2)(-5) = 10$. The three crossings come from three simple zeros, each of odd multiplicity.

The height of the hump is a different matter. Nothing in this chapter locates a turning point or measures how high the curve climbs between two zeros. You could evaluate $P$ at extra points to bound it, but the sketch alone justifies no such number, and finding the exact maximum is a calculus problem.

An odd-degree polynomial with real coefficients must cross the axis, because its two arms point in opposite directions, so the curve travels from below the axis to above it, and an unbroken curve cannot do that without meeting the axis.

$\deg(x^5 - 3x + 1) = 5, \ \text{ which is odd} \ \Longrightarrow \ \text{at least one crossing}$

The other three all fail. Both $x^4 + 1$ and $x^2 + 4$ are always positive, so their graphs never touch the axis. And $(x-2)^2(x+3)^2$ has only even multiplicities, so it touches the axis at $2$ and at $-3$ but never crosses it.