Graphing Polynomial Functions: Additional Practice (Set 2)

Two double zeros use up $2 + 2 = 4$ degrees, which is the whole quartic, so the polynomial is completely determined by the leading coefficient.

$P(x) = 2(x+1)^2(x-3)^2$

Evaluate at $x = 0$.

$P(0) = 2(1)^2(-3)^2 = 2(1)(9) = 18$

The y-intercept is $(0, 18)$. It has to be positive: both squared factors are positive at $x = 0$, and the leading coefficient is positive.

If the graph only ever touched the axis, every real zero would have even multiplicity, so the real multiplicities would add up to an even number. The non-real roots come in conjugate pairs, so they also add up to an even number.

$\text{even} + \text{even} = \text{even} \neq 5$

An odd degree can never be built that way, so the situation is impossible: an odd-degree polynomial always has a real zero of odd multiplicity, and it crosses the axis there. The first suggested example crosses at $x = -1$, where the multiplicity is $3$. The second has degree $2 + 2 + 2 = 6$, not $5$. And the claim about five x-intercepts is false, since $x^5 + x = x(x^4+1)$ has only one.

A crossing happens exactly at a real zero of odd multiplicity, so sort the three zeros by parity.

$x = -1 \ (m = 2, \text{ even}), \qquad x = 2 \ (m = 3, \text{ odd}), \qquad x = 5 \ (m = 1, \text{ odd})$

The zeros $2$ and $5$ have odd multiplicity, so the curve crosses at those two points. At $x = -1$ it touches and turns back. The degree is $2 + 3 + 1 = 6$, but the degree counts roots, not crossings.

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

$1 + 2 + 1 = 4$

So the degree is at least $4$. That value is also consistent with the arms: both arms rise only for an even degree, and $4$ is even. An example that fits is $P(x) = (x+3)x^2(x-2)$, so the minimum is achieved.

The x-intercepts are the real zeros, so those are visible. The two arms show the sign of the leading coefficient, and whether they agree or disagree shows the parity of the degree.

$\text{minimum degree} = \sum_{\text{crossings}} 1 + \sum_{\text{touches}} 2$

But that is only a minimum. A crossing might have multiplicity $1$ or $3$ or $5$, a touch might have multiplicity $2$ or $4$, and the polynomial might carry any number of invisible conjugate pairs of non-real roots. Every one of those possibilities raises the degree by an even amount, so the picture pins down the parity of the degree and a lower bound for it, never its exact value.

The zeros are $-2$, $1$ (a double zero), and $4$. Test a point inside each candidate interval.

$P(0) = 3(1)(2)(-4) = -24, \qquad P(2) = 3(1)(4)(-2) = -24, \qquad P(5) = 3(16)(7)(1) = 336$

Only the test beyond $4$ came out positive. Note what happened across the double zero at $x = 1$: the sign stayed negative, exactly as an even multiplicity requires. The interval $-2 < x < 4$ fails for the extra reason that it contains that zero.

Two bounds squeeze the answer from both sides. Between two consecutive real zeros the curve leaves the axis and comes back to it, so it must turn at least once in each gap, and four zeros in a row leave three gaps.

$\text{at least } 3 \ \text{turning points}$

A polynomial of degree $n$ has at most $n - 1$ turning points, and here $n - 1 = 3$.

$\text{at most } 3 \ \text{turning points}$

The two bounds meet, so the graph has exactly $3$. Notice what this argument does not give you: it never says where those three turning points are, and finding that would need calculus.

An x-intercept is a real zero, and a graph that never meets the axis has no x-intercept, so it has no real zero at all.

$P(x) > 0 \ \text{ for every real } x \ \Longrightarrow \ P(x) = 0 \ \text{ has no real solution}$

The other statements fail. The degree must in fact be even, not odd, since an odd-degree graph always crosses. The leading coefficient must be positive, or the arms would head downward and the curve would drop below the axis. And the number of turning points is not forced: $x^2 + 1$ turns once, while $x^4 - x^2 + 1$ turns three times. Its values fall, rise, fall, and rise again as $x$ increases through $-1.2, -0.7, 0, 0.7, 1.2$, which forces at least three turns, and a degree $4$ polynomial has at most $4 - 1 = 3$ turning points, so it makes exactly three.

For the graph never to rise above the axis, the polynomial must satisfy $P(x) \leq 0$ for every real $x$, so it can have no real zero of odd multiplicity, which would flip the sign. Non-real roots are unconstrained, because only a real zero can change the sign of $P$ along the x-axis, and a factor such as $x^2 + 2$ contributes a conjugate pair of non-real roots while staying positive for every real $x$.

$-\underbrace{(x-1)^2}_{\geq \, 0} \cdot \underbrace{(x^2+2)}_{> \, 0} \leq 0$

The square is never negative and the quadratic factor is always positive, so the minus sign makes the whole product negative or zero, with equality only at $x = 1$. The version without the minus sign is never negative instead. The other two both have a zero of odd multiplicity at $x = 1$, so their graphs sit above the axis on one side of it.

The touch needs an even multiplicity and each crossing needs an odd one. The smallest choices are $2$, $1$, and $1$, and they add to $4$, which is exactly the degree available, so the factors are forced.

$P(x) = a(x+2)^2(x-1)(x-3)$

Now use the point $(0, 12)$.

$a(2)^2(-1)(-3) = 12a = 12 \quad \Longrightarrow \quad a = 1$

The leading coefficient is $1$. It is positive, which fits the even degree and the fact that the curve is above the axis at $x = 0$, between the touch and the first crossing.

The Fundamental Theorem of Algebra gives $5$ roots counted with multiplicity, and the single real zero accounts for $1$ of them.

$5 - 1 = 4$

So $4$ roots are non-real, forming two conjugate pairs. The count had to be even, and the parity works out: an odd degree needs the real multiplicities to sum to an odd number, and $1$ is odd. The graph has exactly one x-intercept, and it crosses there.

Five simple zeros in a row leave four gaps between consecutive zeros, and in each gap the curve leaves the axis and returns to it, so it must turn at least once there.

$\text{at least } 4 \ \text{turning points}$

The ceiling comes from the degree: a polynomial of degree $n$ has at most $n - 1$ turning points, and here that is $5 - 1 = 4$.

$\text{at most } 4 \ \text{turning points}$

The bounds agree, so there are exactly $4$. Where they sit is another matter entirely, and algebra alone will not tell you.