Polynomials and Their Graphs: Additional Practice (Set 2)

Take the leading piece of each factor: $(1 - x)^3$ contributes $(-x)^3 = -x^3$, and $(x + 2)^2$ contributes $x^2$.

$(-x^3)(x^2) = -x^5$

The degree $5$ is odd, so the ends disagree, and the leading coefficient $-1$ points the right end down: the graph rises on the left and falls on the right.

Even powers erase the sign of the input: $(-1)^4 = 1$ and $(-1)^2 = 1$.

$f(-1) = 1 + k + 9 = k + 10 = 4$

Solving gives $k = -6$.

The terms $ax^3$ and $bx$ use odd powers, so replacing $2$ by $-2$ flips their signs while the constant $5$ stays put. Adding the two evaluations makes the odd parts cancel:

$f(2) + f(-2) = (8a + 2b + 5) + (-8a - 2b + 5) = 10$

Since $f(2) = 25$, it follows that $f(-2) = 10 - 25 = -15$, with no need to find $a$ or $b$.

The constant term of any polynomial is its value at $0$.

$(2 \cdot 0 - 3)^4 = (-3)^4 = 81$

The even power makes the result positive; $-81$ would be $-3^4$ with the sign left outside the power.

An even-degree graph's far ends agree, so tracing it left to right you finish in the opposite travel state (rising versus falling) from the one you started in. Each turning point is one flip, so the number of flips must be odd.

$\text{even degree} \;\Rightarrow\; \text{odd number of turning points} \le 6 - 1 = 5$

So $1$, $3$, and $5$ are all possible, but $4$ is not.

Even degree with a positive lead makes both far ends rise, but at $x = 0$ the graph sits below the axis at height $-3$.

$f(0) = -3 < 0, \qquad \text{both far ends rise above } 0$

The unbroken curve must therefore cross the axis at least once on each side of $0$: at least $2$ real zeros. Exactly two is achievable, for example $x^4 - 3$, so the least is $2$.

Three turning points force $n - 1 \ge 3$, so $n \ge 4$. Both ends falling means the ends agree, so the degree is even.

$n \ge 4 \text{ and } n \text{ even} \;\Rightarrow\; n = 4 \text{ possible}$

The parity check agrees: an even-degree graph needs an odd number of turning points, and $3$ is odd. An M-shaped quartic such as $-(x^4 - 5x^2 + 4)$ realizes it, with two hills around one valley.

Odd degree makes the two far ends point in opposite directions, so tracing left to right you finish in the same travel state you started in. Each turning point is one flip between rising and falling, so the flip count must be even.

$\text{odd degree} \;\Rightarrow\; \text{even number of turning points} \le 5 - 1 = 4$

So $0$, $2$, and $4$ are possible ($x^5$ itself has none), but $3$ is not.

Test the candidates by evaluation, powers first.

$p(3) = 2(27) - 5(9) + 3 - 12 = 54 - 45 + 3 - 12 = 0$

So $3$ is a zero. The others fail: $p(2) = 16 - 20 + 2 - 12 = -14$, $p(4) = 128 - 80 + 4 - 12 = 40$, and $p(-3) = -54 - 45 - 3 - 12 = -114$.

Composition substitutes $g$ into $f$, so the leading term of $f(g(x))$ comes from squaring the degree-$3$ input: $\left[g(x)\right]^2$ has degree $3 + 3 = 6$ because degrees add in a product.

$\deg f(g(x)) = 2 \cdot 3 = 6$

The lower term of $f$ contributes only degree $3$, which cannot reach the leading term. Composition multiplies degrees, while multiplication adds them.

A polynomial's graph is an unbroken curve, and this one sits below the axis at $x = 1$ and above it at $x = 2$.

$f(1) = -3 < 0 < 5 = f(2)$

An unbroken curve cannot get from below the axis to above it without touching it, so some input between $1$ and $2$ gives output $0$. The degree plays no role; a sign change traps a zero.

An even degree makes both far ends point the same way, so tracing left to right the graph starts in one travel state and must finish in the other (falling then rising, or rising then falling). That takes an odd number of flips, and an odd number is at least one.

$\text{even degree} \;\Rightarrow\; \text{at least } 1 \text{ turning point}$

The parabola shows the bound is tight: exactly one turn. By contrast an odd-degree graph can run through with none, like $y = x^3$.