Chapter 5 — Polynomial and Rational Functions

Example: Find the vertex, axis of symmetry, and max/min of $f(x) = -2x^2 + 12x - 7$.

$a = -2, \; b = 12, \; c = -7$

• Vertex: $(3, 11)$
• Axis of symmetry: $x = 3$
• Since $a = -2 < 0$, the vertex is a maximum. The maximum value is $f(3) = 11$.

Example: Convert $f(x) = 3x^2 - 12x + 5$ to vertex form.

$f(x) = 3(x^2 - 4x) + 5$

Half of $-4$ is $-2$; square it: $4$.

$f(x) = 3(x^2 - 4x + 4 - 4) + 5 = 3\bigl((x-2)^2 - 4\bigr) + 5$

$f(x) = 3(x-2)^2 - 12 + 5 = 3(x - 2)^2 - 7$

Vertex form: $f(x) = 3(x - 2)^2 - 7$ → Vertex $(2, -7)$, opens up (minimum).

Example — Projectile Motion: A ball is thrown upward with height $h(t) = -16t^2 + 64t + 5$ feet after $t$ seconds.

Maximum height: $t = -\dfrac{64}{2(-16)} = 2$ seconds.

$h(2) = -16(4) + 64(2) + 5 = -64 + 128 + 5 = 69$ feet.

When does it hit the ground? Solve $h(t) = 0$:

$-16t^2 + 64t + 5 = 0 \Rightarrow t = \dfrac{-64 \pm \sqrt{64^2 - 4(-16)(5)}}{2(-16)} = \dfrac{-64 \pm \sqrt{4096 + 320}}{-32}$

$= \dfrac{-64 \pm \sqrt{4416}}{-32} = \dfrac{-64 \pm 66.45}{-32}$

Taking the positive root: $t = \dfrac{-64 - 66.45}{-32} \approx 4.08$ seconds.

Example — Maximum Area: A farmer has 200 feet of fencing to enclose a rectangular pen against a barn (only 3 sides need fencing). Maximize the area.

Let $x = $ width (two sides). Length $= 200 - 2x$.

$A(x) = x(200 - 2x) = -2x^2 + 200x$

Maximum at $x = -\dfrac{200}{2(-2)} = 50$ feet.

$A(50) = 50(200 - 100) = 50 \cdot 100 = 5000$ sq. ft.

Example: $f(x) = (x + 2)^3(x - 1)^2(x - 4)$

Degree: $3 + 2 + 1 = 6$ (even). Leading coefficient: positive. End behavior: both ends up.

Maximum turning points: $6 - 1 = 5$.

Example: Show that $f(x) = x^3 - 4x + 1$ has a zero between 0 and 1.

$f(0) = 1 > 0$ and $f(1) = 1 - 4 + 1 = -2 < 0$.

Since $f$ is continuous and changes sign, by the IVT there exists a zero in $(0, 1)$. ✓

Example: Divide $2x^3 + 3x^2 - 5x + 1$ by $x + 2$.

Step 1: $2x^3 \div x = 2x^2$. Multiply: $2x^2(x + 2) = 2x^3 + 4x^2$. Subtract: $-x^2 - 5x + 1$.

Step 2: $-x^2 \div x = -x$. Multiply: $-x(x + 2) = -x^2 - 2x$. Subtract: $-3x + 1$.

Step 3: $-3x \div x = -3$. Multiply: $-3(x + 2) = -3x - 6$. Subtract: $7$.

Quotient: $2x^2 - x - 3$, Remainder: $7$.

Example: Divide $x^3 - 6x^2 + 11x - 6$ by $(x - 2)$.

$c = 2$, Coefficients: $1, -6, 11, -6$

  2 |  1   -6    11   -6
    |       2    -8     6
    |  1   -4     3     0

Quotient: $x^2 - 4x + 3$, Remainder: $0$.

Since remainder $= 0$, $(x - 2)$ is a factor, and $x = 2$ is a zero.

Example: Is $(x + 3)$ a factor of $f(x) = x^3 + 2x^2 - 5x + 6$?

Evaluate $f(-3) = (-3)^3 + 2(-3)^2 - 5(-3) + 6 = -27 + 18 + 15 + 6 = 12 \ne 0$.

No — $(x + 3)$ is not a factor; the remainder upon division is 12.

Example: $f(x) = x^2 + 4$ has degree 2 and no real zeros (discriminant $= 0 - 16 = -16 < 0$).

Its two complex zeros are $x = 2i$ and $x = -2i$ (conjugate pair). ✓

Example: List all possible rational zeros of $f(x) = 2x^3 + 3x^2 - 8x + 3$.

Factors of $a_0 = 3$: $\pm 1, \pm 3$

Factors of $a_n = 2$: $\pm 1, \pm 2$

Possible rational zeros: $\pm 1, \pm 3, \pm \tfrac{1}{2}, \pm \tfrac{3}{2}$

Test $f(1) = 2 + 3 - 8 + 3 = 0$ ✓ → $x = 1$ is a zero.

Synthetic division by $(x - 1)$ gives $2x^2 + 5x - 3 = (2x - 1)(x + 3)$.

All zeros: $x = 1, \; x = \tfrac{1}{2}, \; x = -3$.

Example: $f(x) = x^4 - 3x^3 + x - 5$

Signs of coefficients: $+, -, +, -$ → 3 sign changes → 3 or 1 positive real zeros.

$f(-x) = x^4 + 3x^3 - x - 5$

Signs: $+, +, -, -$ → 1 sign change → exactly 1 negative real zero.

Example: Find all zeros of $f(x) = x^4 - 4x^3 + 6x^2 - 4x + 1$.

Possible rational zeros: $\pm 1$.

$f(1) = 1 - 4 + 6 - 4 + 1 = 0$ ✓

Synthetic division gives $x^3 - 3x^2 + 3x - 1$.

$f_2(1) = 1 - 3 + 3 - 1 = 0$ ✓ → divide again.

This continues: $f(x) = (x - 1)^4$.

The only zero is $x = 1$ with multiplicity 4.

Example: $f(x) = \dfrac{x + 1}{x^2 - 9} = \dfrac{x + 1}{(x-3)(x+3)}$

Denominator $= 0$ when $x = 3$ or $x = -3$.

Domain: $(-\infty, -3) \cup (-3, 3) \cup (3, \infty)$, or equivalently $\mathbb{R} \setminus {-3, 3}$.

Example: $f(x) = \dfrac{3}{x - 2}$

$q(x) = x - 2 = 0 \Rightarrow x = 2$, and $p(2) = 3 \ne 0$.

Vertical asymptote: $x = 2$.

As $x \to 2^+$: $f(x) \to +\infty$. As $x \to 2^-$: $f(x) \to -\infty$.

Example 1: $f(x) = \dfrac{5x}{2x + 1}$ → degrees equal → HA: $y = \dfrac{5}{2}$.

Example 2: $f(x) = \dfrac{3}{x^2 + 1}$ → numerator degree (0) < denominator degree (2) → HA: $y = 0$.

Example 3: $f(x) = \dfrac{x^2 + 1}{x - 1}$ → numerator degree > denominator degree → no HA.

Example: $f(x) = \dfrac{x^2 + 2x + 1}{x - 1}$

Long division: $x^2 + 2x + 1 \div (x - 1)$:

$x^2 \div x = x$. $x(x-1) = x^2 - x$. Subtract: $3x + 1$.

$3x \div x = 3$. $3(x-1) = 3x - 3$. Subtract: $4$.

As $x \to \pm\infty$, $\dfrac{4}{x-1} \to 0$, so the graph approaches the line $y = x + 3$.

Oblique asymptote: $y = x + 3$.

Example: $f(x) = \dfrac{x^2 - 4}{x - 2} = \dfrac{(x-2)(x+2)}{x-2}$

Cancel $(x - 2)$: simplified form $= x + 2$ (for $x \ne 2$).

Hole at $x = 2$: $y = 2 + 2 = 4$. The hole is at the point $(2, 4)$.

The graph looks like the line $y = x + 2$ but with a missing point at $(2, 4)$.

Example: Find the inverse of $f(x) = x^2 + 1$ for $x \ge 0$.

$y = x^2 + 1 \Rightarrow x = y^2 + 1 \Rightarrow y^2 = x - 1 \Rightarrow y = \sqrt{x - 1}$

(Taking positive root since original domain was $x \ge 0$.)

$f^{-1}(x) = \sqrt{x - 1}$, domain: $x \ge 1$.

Example: The distance $d$ a spring stretches varies directly with the force $F$ applied (Hooke’s Law): $d = kF$.

If a 10-lb force stretches the spring 3 inches: $3 = k(10) \Rightarrow k = 0.3$.

Model: $d = 0.3F$. A 25-lb force: $d = 0.3(25) = 7.5$ inches.

Example: The time $t$ to travel a fixed distance varies inversely with speed $s$: $t = \dfrac{k}{s}$.

If $s = 60$ mph gives $t = 3$ hours: $k = 60 \times 3 = 180$, so $t = \dfrac{180}{s}$.

At $s = 90$ mph: $t = \dfrac{180}{90} = 2$ hours.

Example: The volume $V$ of a cylinder varies jointly with the height $h$ and the square of the radius $r$: $V = kr^2h$.

Given $V = 314$ when $r = 5$ and $h = 4$:

$314 = k(25)(4) = 100k \Rightarrow k = 3.14 \approx \pi$ ✓

$V = \pi r^2 h$ — the exact formula!