Comprehensive Study Guide: Linear, Polynomial, and Rational Functions

Linear Functions, Equations, and Rates of Change

• Tabular Analysis and Linear Modeling:

  • Data Representation: A discrete dataset relates an independent variable $t$ to a dependent variable $h$:

    • When $t = 2$, $h = 16$

    • When $t = 4$, $h = 15$

    • When $t = 6$, $h = 14$

    • When $t = 8$, $h = 13$

  • Rate of Change (Slope): The constant rate of change $m$ across equal intervals ($\Delta t = 2$) is given by:         $m = \frac{\Delta h}{\Delta t} = \frac{15 - 16}{4 - 2} = \frac{-1}{2} = -0.5$

  • Formula Derivation ($h(t)$): Applying the point-slope form $h - h_1 = m(t - t_1)$ using the data point $(2, 16)$:         $h - 16 = -\frac{1}{2}(t - 2)$         $h(t) = -\frac{1}{2}t + 1 + 16 = -\frac{1}{2}t + 17$

  • Intercept Determination:

    • $t$-intercept (Horizontal Intercept): Set $h(t) = 0$ to determine the independent value where the dependent variable is zero:             $0 = -\frac{1}{2}t + 17 \implies \frac{1}{2}t = 17 \implies t = 34$             The coordinate is $(34, 0)$.

    • $h$-intercept (Vertical Intercept): Evaluate the function at $t = 0$:             $h(0) = -\frac{1}{2}(0) + 17 = 17$             The coordinate is $(0, 17)$.

• Fundamental Forms of Linear Equations and Lines:

  • Vertical Lines ($x = c$):

    • Equation: $x = -2$

    • Characteristics: Represents a vertical line passing through every coordinate of the form $(-2, y)$. The run is $\Delta x = 0$, rendering the slope $m = \frac{\Delta y}{\Delta x}$ completely undefined. It has no $y$-intercept and possesses an $x$-intercept at $(-2, 0)$.

  • Horizontal Lines ($f(x) = c$):

    • Equation: $f(x) = 2$

    • Characteristics: Represents a horizontal line passing through every coordinate of the form $(x, 2)$. The rise is $\Delta y = 0$, yielding a slope of $m = 0$. It has no $x$-intercept and crosses the vertical axis at $(0, 2)$.

  • Linear Function with Negative Slope:

    • Equation: $g(x) = 5 - 3x$

    • Characteristics: Standard slope-intercept form $g(x) = -3x + 5$. The slope is $m = -3$, indicating the function decreases by $3$ vertical units for every $1$ unit increase in $x$. The vertical intercept is $(0, 5)$, and the horizontal intercept occurs at $5 - 3x = 0 \implies x = \frac{5}{3}$, yielding $\left(\frac{5}{3}, 0\right)$.

  • Linear Function with Positive Slope:

    • Equation: $h(x) = 3x + 1$

    • Characteristics: Slope $m = 3$, indicating the function increases by $3$ vertical units for every $1$ unit increase in $x$. The vertical intercept is $(0, 1)$, and the horizontal intercept occurs at $3x + 1 = 0 \implies x = -\frac{1}{3}$, yielding $\left(-\frac{1}{3}, 0\right)$.

• Global Classification of Rate of Change over the Entire Domain:

  • Rate of change equals $0$ ($m = 0$): Corresponds to a Horizontal Line ($C$). The function values remain constant across all inputs ($f(x) = c$).

  • Rate of change is undefined: Corresponds to a Vertical Line ($D$). The input remains constant while output values vary, producing a division by zero in the slope formula ($\Delta x = 0$).

  • Rate of change is strictly negative ($m < 0$): Corresponds to a Decreasing Linear Function ($B$). As $x$ increases, $f(x)$ decreases monotonically.

  • Rate of change is strictly positive ($m > 0$): Corresponds to an Increasing Linear Function ($A$). As $x$ increases, $f(x)$ increases monotonically.

Linear Modeling, Perpendicularity, and Applied Systems

• Perpendicular Lines and System Intersection:

  • Primary Function Formulation ($f(x)$): A line $f(x)$ has a slope of $m_1 = \frac{1}{2}$ and passes through the point $(-4, -2)$.

    • Point-slope derivation:             $f(x) - (-2) = \frac{1}{2}(x - (-4))$             $f(x) + 2 = \frac{1}{2}x + 2 \implies f(x) = \frac{1}{2}x$

    • Characteristics: Passes through the origin $(0, 0)$ with vertical intercept $(0, 0)$ and horizontal intercept $(0, 0)$.

  • Perpendicular Line Formulation ($g(x)$):

    • Slope relationship: Two non-vertical lines are perpendicular if and only if their slopes are negative reciprocals of each other:             $m_2 = -\frac{1}{m_1} = -\frac{1}{1/2} = -2$

    • Passing through the point $(0, 2)$, the vertical intercept is directly given as $b = 2$:             $g(x) = -2x + 2$

  • System Intersection Coordinates:

    • Set $f(x) = g(x)$ to find the shared point of intersection:             $\frac{1}{2}x = -2x + 2$             $\frac{5}{2}x = 2 \implies x = 2 \times \frac{2}{5} = \frac{4}{5} = 0.8$

    • Substitute $x = \frac{4}{5}$ into $f(x)$:             $y = \frac{1}{2}\left(\frac{4}{5}\right) = \frac{2}{5} = 0.4$

    • Intersection point: $\left(\frac{4}{5}, \frac{2}{5}\right)$ or $(0.8, 0.4)$.

• Applied Linear Growth Models:

  • Calculator Rental Inventory:

    • Initial conditions: In the base year $1996$ ($t = 0$), the program launched with $30$ calculators.

    • Annual expansion: Expands by $15$ calculators per year ($m = 15$).

    • Linear inventory model: $C(t) = 15t + 30$, where $t$ represents years elapsed since $1996$.

    • Target threshold: Determine the exact year when inventory strictly exceeds $450$ units:             $15t + 30 > 450$             $15t > 420 \implies t > \frac{420}{15} \implies t > 28$

    • Year determination: Reaching exactly $450$ calculators occurs at $t = 28$, which corresponds to the calendar year $1996 + 28 = 2024$. Therefore, the program will strictly exceed $450$ calculators in the year $2025$ (after $28$ full years have elapsed).

  • Commercial Profit Modeling:

    • Given data points: Selling $n = 2000$ items generates a profit of $P(2000) = \$24,000$. Selling $n = 3000$ items generates a profit of $P(3000) = \$30,000$.

    • Slope Calculation:             $m = \frac{\Delta P}{\Delta n} = \frac{30000 - 24000}{3000 - 2000} = \frac{6000}{1000} = 6$             Meaning in words: The slope represents the marginal profit per unit sold; for each additional item the company sells, total profit increases by $\$6$.

    • Vertical Intercept Calculation:             $P(n) = 6n + b \implies 24000 = 6(2000) + b \implies 24000 = 12000 + b \implies b = 12000$             The intercept coordinate is $(0, 12000)$.             Meaning in words: If the company sells zero items ($n = 0$), the baseline fixed profit according to this linear model is $\$12,000$.

    • Evaluation of $f(4380)$:             $P(4380) = 6(4380) + 12000 = 26280 + 12000 = 38280$             Meaning in words: If the company sells $4,380$ items, the resulting total corporate profit is $\$38,280$.

  • Wildlife Population Decline:

    • Given data: In $2003$ ($t = 0$), the owl population was measured at $P(0) = 340$. In $2007$ ($t = 4$), the population was measured at $P(4) = 280$.

    • Linear rate of change:             $m = \frac{280 - 340}{4 - 0} = \frac{-60}{4} = -15\text{ owls per year}$

    • Model formula: $P(t) = -15t + 340$.

    • Prediction for $2012$: The elapsed time from $2003$ is $t = 2012 - 2003 = 9$ years:             $P(9) = -15(9) + 340 = -135 + 340 = 205$             The model predicts an owl population of $205$ owls in $2012$.

  • Personal Budget Depletion (Expenditure Analysis):

    • Initial conditions: Emily saved an initial balance of $\$5,000$ for a summer stay in Seattle, spending at a constant rate of $\$500$ per week.

    • Formula for Remaining Funds ($S(t)$): Letting $S(t)$ represent the remaining savings balance as a function of elapsed weeks $t$:             $S(t) = 5000 - 500t$             (Alternatively, if expressed as cumulative expenditures: $E(t) = 500t$.)

    • Horizontal Intercept ($t$-intercept): Set remaining funds to zero:             $0 = 5000 - 500t \implies 500t = 5000 \implies t = 10$             Interpretation: The coordinate $(10, 0)$ indicates that after $10$ weeks, Emily's savings will be completely exhausted ($\$0$ remaining).

    • Half-Money Depletion: Determine when spending reaches half of the total budget ($\$2,500$ remaining):             $5000 - 500t = 2500 \implies 500t = 2500 \implies t = 5$             She will have spent half of her money after $5$ weeks.

  • Break-Even Decision Analysis (Service Comparison):

    • Cost structure:

      • Company 1: $C_1(w) = 6w$, where $w$ is the total number of windows washed.

      • Company 2: $C_2(w) = 30 + 4w$, featuring a flat base fee of $\$30$ plus $\$4$ per window.

    • Condition for Company 2 to be preferable ($C_2(w) < C_1(w)$):             $30 + 4w < 6w \implies 30 < 2w \implies w > 15$

    • Conclusion: At exactly $15$ windows, both companies charge identical amounts ($C_1(15) = C_2(15) = \$90$). For the second company to be financially preferable, you must have at least $16$ windows ($w > 15$).

• Geometric Optimization of Bounded Triangular Regions:

  • Bounding Curves:

    1. The horizontal axis: $y = 0$

    2. The linear function: $f(x) = 4 - \frac{1}{2}x$

    3. The perpendicular line through the origin: Slope $m_\perp = -\frac{1}{-1/2} = 2$, giving $y = 2x$

  • Vertex Coordinates:

    • Vertex 1: Intersection of $y = 2x$ and the $x$-axis ($y = 0$) occurs at the origin: $(0, 0)$.

    • Vertex 2: Intersection of $f(x) = 4 - \frac{1}{2}x$ and the $x$-axis ($y = 0$):             $0 = 4 - \frac{1}{2}x \implies \frac{1}{2}x = 4 \implies x = 8 \implies (8, 0)$

    • Vertex 3: Intersection of $f(x)$ and $y = 2x$:             $2x = 4 - \frac{1}{2}x \implies \frac{5}{2}x = 4 \implies x = \frac{8}{5} = 1.6$             $y = 2\left(\frac{8}{5}\right) = \frac{16}{5} = 3.2 \implies \left(\frac{8}{5}, \frac{16}{5}\right)$

  • Area Computation via Base and Height:

    • Base along the $x$-axis: The horizontal distance between $(0, 0)$ and $(8, 0)$ is $b = 8 - 0 = 8$.

    • Perpendicular height: The vertical distance from the base to Vertex 3 is given by the $y$-coordinate $h = \frac{16}{5} = 3.2$.

    • Area calculation:             $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \times \frac{16}{5} = 4 \times \frac{16}{5} = \frac{64}{5} = 12.8\text{ square units}$

  • Verification via Right-Triangle Legs:

    • Because the two bounding lines are perpendicular, they form a right triangle at Vertex 3.

    • Length of Leg 1 (from $(0, 0)$ to $(\frac{8}{5}, \frac{16}{5})$):             $L_1 = \sqrt{\left(\frac{8}{5} - 0\right)^2 + \left(\frac{16}{5} - 0\right)^2} = \sqrt{\frac{64 + 256}{25}} = \sqrt{\frac{320}{25}} = \frac{8\sqrt{5}}{5}$

    • Length of Leg 2 (from $(8, 0)$ to $(\frac{8}{5}, \frac{16}{5})$):             $L_2 = \sqrt{\left(8 - \frac{8}{5}\right)^2 + \left(0 - \frac{16}{5}\right)^2} = \sqrt{\left(\frac{32}{5}\right)^2 + \left(-\frac{16}{5}\right)^2} = \sqrt{\frac{1024 + 256}{25}} = \sqrt{\frac{1280}{25}} = \frac{16\sqrt{5}}{5}$

    • Right triangle area:             $\text{Area} = \frac{1}{2} \times L_1 \times L_2 = \frac{1}{2} \times \frac{8\sqrt{5}}{5} \times \frac{16\sqrt{5}}{5} = \frac{1}{2} \times \frac{128 \times 5}{25} = \frac{64}{5} = 12.8\text{ square units}$

Polynomial Functions: Characteristics, Transformations, and Projectiles

• Analysis of Factored Higher-Degree Polynomials:

  • Function: $P(x) = -4(x - 2)^2(x + 5)(x + 2)$

  • Degree and Leading Term:

    • Degree: Calculated by summing the multiplicities of all linear factors:             $\text{Degree} = 2 + 1 + 1 = 4$

    • Leading term: Determined by taking the product of the leading coefficient and the highest powers of each linear factor:             $\text{Leading Term} = -4 \times (x)^2 \times (x)^1 \times (x)^1 = -4x^4$

  • End Behavior: Governed entirely by the leading term $-4x^4$ (even degree, negative leading coefficient):

    • $\text{As } x \to \infty, P(x) \to -\infty$

    • $\text{As } x \to -\infty, P(x) \to -\infty$

  • Turning Points: For any polynomial of degree $n$, the maximum possible number of turning points (local extrema / bumps) is $n - 1$:         $\text{Maximum Turning Points} = 4 - 1 = 3$

  • $x$-intercepts and Multiplicities:

    • Root at $x = 2$: Multiplicity $2$ (even multiplicity implies a double zero; the graph is tangent to the $x$-axis, touching and turning around). Coordinate: $(2, 0)$.

    • Root at $x = -5$: Multiplicity $1$ (odd multiplicity implies a single zero; the graph crosses the $x$-axis linearly). Coordinate: $(-5, 0)$.

    • Root at $x = -2$: Multiplicity $1$ (single zero; crosses the $x$-axis linearly). Coordinate: $(-2, 0)$.

  • $P$-intercept (Vertical Intercept): Evaluated algebraically by substituting $x = 0$:         $P(0) = -4(0 - 2)^2(0 + 5)(0 + 2) = -4(-2)^2(5)(2) = -4(4)(10) = -160$         Coordinate: $(0, -160)$.

• Quadratic Analysis and Vertex Transformations:

  • Function: $Q(x) = x^2 - 10x + 22$

  • Vertex Form Conversion (Completing the Square):         $Q(x) = (x^2 - 10x + 25) - 25 + 22$         $Q(x) = (x - 5)^2 - 3$

  • Vertex Coordinates and Nature:

    • Coordinates: Directly extracted from vertex form $a(x - h)^2 + k$ as $(h, k) = (5, -3)$.

    • Classification: Because the leading coefficient is positive ($a = 1 > 0$), the parabola opens upward, meaning the vertex is an absolute minimum.

  • End Behavior: Because the parabola opens upward:

    • $\text{As } x \to \infty, Q(x) \to \infty$

    • $\text{As } x \to -\infty, Q(x) \to \infty$

  • Intercepts of $Q(x)$:

    • $Q$-intercept (Vertical Intercept): Evaluated at $x = 0$:             $Q(0) = (0)^2 - 10(0) + 22 = 22 \implies (0, 22)$

    • $x$-intercepts (Horizontal Intercepts): Solved by setting $Q(x) = 0$:             $(x - 5)^2 - 3 = 0 \implies (x - 5)^2 = 3 \implies x - 5 = \pm\sqrt{3} \implies x = 5 \pm \sqrt{3}$             Coordinates: $(5 - \sqrt{3}, 0)$ and $(5 + \sqrt{3}, 0)$ (approximately $(3.268, 0)$ and $(6.732, 0)$).

• Kinematics and Projectile Motion on Planet Zeldok:

  • Trajectory Equation: A ball kicked from the top of a building follows the quadratic height function:         $h(t) = -t^2 + 6t + 15$         where $t$ is measured in seconds and $h(t)$ in distance units.

*   **Initial Building Height:** Determined by evaluating the height at time $t = 0$:

        $h(0) = -(0)^2 + 6(0) + 15 = 15$         The building height is $15$ units. * Maximum Height Attained: * Time to peak: Occurs at the vertex time coordinate $t = -\frac{b}{2a}$:             $t = -\frac{6}{2(-1)} = \frac{-6}{-2} = 3\text{ seconds}$ * Peak height: Substitute $t = 3$ into the trajectory equation:             $h(3) = -(3)^2 + 6(3) + 15 = -9 + 18 + 15 = 24$ * The maximum height reached by the ball is $24$ units. * Impact Time (Ground Strike): * Set $h(t) = 0$ to find when the height returns to ground level:             $-t^2 + 6t + 15 = 0 \implies t^2 - 6t - 15 = 0$ * Apply the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:             $t = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-15)}}{2(1)} = \frac{6 \pm \sqrt{36 + 60}}{2} = \frac{6 \pm \sqrt{96}}{2}$             $t = \frac{6 \pm 4\sqrt{6}}{2} = 3 \pm 2\sqrt{6}$ * Physical constraint: Since time after launch must be non-negative ($t \ge 0$), discard the negative root:             $t = 3 + 2\sqrt{6} \approx 3 + 2(2.4495) \approx 7.90\text{ seconds}$ * The ball strikes the ground at approximately $7.90$ seconds (exact value: $3 + 2\sqrt{6}$ seconds).

• Reconstructing Polynomial Functions from Graphical Signatures:

*   **Feature Identification:**
    *   Root at $x = -3$: The curve crosses the horizontal axis, indicating an odd multiplicity of $1$: factor is $(x + 3)$.
    *   Root at $x = -1$: The curve is tangent to the axis (touches and reverses direction), indicating an even multiplicity of $2$: factor is $(x + 1)^2$.
    *   Root at $x = 2$: The curve crosses the horizontal axis, indicating an odd multiplicity of $1$: factor is $(x - 2)$.
    *   End behavior: Both ends rise towards positive infinity ($G(x) \to \infty$ as $x \to \pm\infty$), confirming an even degree of $4$ and a positive leading coefficient ($a > 0$).
*   **Algebraic Formulation:**

        $G(x) = a(x + 3)(x + 1)^2(x - 2)$ * Determination of Constant $a$: * Identify the vertical intercept from the graph: the curve crosses the $y$-axis precisely at $(0, -2)$. * Substitute $x = 0$ and set $G(0) = -2$:             $G(0) = a(0 + 3)(0 + 1)^2(0 - 2) = a(3)(1)(-2) = -6a$             $-6a = -2 \implies a = \frac{-2}{-6} = \frac{1}{3}$ * Final Formula for $G(x)$:         $G(x) = \frac{1}{3}(x + 3)(x + 1)^2(x - 2)$

• Domain Analysis of Radical Functions with Quadratic Radicands:

  • Function: $f(x) = \sqrt{3x^2 + 7x + 2}$

  • Condition for Real Evaluation: The radicand under an even radical (square root) must be non-negative:         $3x^2 + 7x + 2 \ge 0$

  • Factorization: Factor by grouping:         $3x^2 + 6x + x + 2 = 3x(x + 2) + 1(x + 2) = (3x + 1)(x + 2) \ge 0$

  • Critical Points: Roots occur at $x = -2$ and $x = -\frac{1}{3}$.

  • Sign Testing: The quadratic opens upward ($a = 3 > 0$), meaning function values are positive outside the interval connecting the roots:

    • For $x < -2$: e.g., $x = -3 \implies (3(-3)+1)(-3+2) = (-8)(-1) = 8 > 0$

    • For $-2 < x < -\frac{1}{3}$: e.g., $x = -1 \implies (3(-1)+1)(-1+2) = (-2)(1) = -2 < 0$

    • For $x > -\frac{1}{3}$: e.g., $x = 0 \implies (3(0)+1)(0+2) = (1)(2) = 2 > 0$

  • Domain in Interval Notation:         $\text{Domain} = (-\infty, -2] \cup \left[-\frac{1}{3}, \infty\right)$

• Higher-Degree Polynomial Graphing Characteristics:

  • Function: $f(x) = -2x(x + 4)^2(x - 3)$

  • Structural Parameters:

    • Degree: $1 + 2 + 1 = 4$

    • Leading Term: $-2 \times (x) \times (x^2) \times (x) = -2x^4$

    • Global End Behavior: As $x \to \pm\infty, f(x) \to -\infty$

  • Zeros and Local Behaviors:

    • $x = -4$: Multiplicity $2$; curve touches the $x$-axis and turns around without crossing.

    • $x = 0$: Multiplicity $1$; curve crosses the $x$-axis linearly.

    • $x = 3$: Multiplicity $1$; curve crosses the $x$-axis linearly.

  • Intercepts and Test Values:

    • $y$-intercept: $f(0) = -2(0)(4)^2(-3) = 0 \implies (0, 0)$

    • At $x = -5$: $f(-5) = -2(-5)(-1)^2(-8) = 10(1)(-8) = -80$

    • At $x = -2$: $f(-2) = -2(-2)(2)^2(-5) = 4(4)(-5) = -80$

    • At $x = 1$: $f(1) = -2(1)(5)^2(-2) = -2(25)(-2) = 100$

    • At $x = 2$: $f(2) = -2(2)(6)^2(-1) = -4(36)(-1) = 144$

Advanced Polynomial Algebra, Division, and Complex Numbers

• Polynomial Long Division Algorithm:

  • Problem: Compute $(3x^3 + 2x^2 - x + 8) \div (x^2 + x)$.

  • Format Requirement: Express final result in the Euclidean division form:         $f(x) = d(x)q(x) + r(x)$         where $f(x)$ is the dividend, $d(x)$ the divisor, $q(x)$ the quotient, and $r(x)$ the remainder.

  • Step-by-Step Long Division:

    • Step 1: Divide the leading term of the dividend by the leading term of the divisor:             $\frac{3x^3}{x^2} = 3x$

    • Multiply $3x$ by the divisor $(x^2 + x)$:             $3x(x^2 + x) = 3x^3 + 3x^2$

    • Subtract this product from the dividend:             $(3x^3 + 2x^2 - x + 8) - (3x^3 + 3x^2) = -x^2 - x + 8$

    • Step 2: Divide the new leading term by the divisor's leading term:             $\frac{-x^2}{x^2} = -1$

    • Multiply $-1$ by $(x^2 + x)$:             $-1(x^2 + x) = -x^2 - x$

    • Subtract this product:             $(-x^2 - x + 8) - (-x^2 - x) = 8$

    • Because the degree of $8$ (degree $0$) is strictly less than the degree of $x^2 + x$ (degree $2$), the division terminates.

  • Division Components:

    • Quotient: $q(x) = 3x - 1$

    • Remainder: $r(x) = 8$

  • Final Euclidean Form:         $3x^3 + 2x^2 - x + 8 = (x^2 + x)(3x - 1) + 8$

• Standard Form Conversion of Complex Fractions:

  • Expression: Write $\frac{2 + 3i}{3 - 4i}$ in standard complex form $z = a + bi$.

  • Complex Conjugate Multiplication: Multiply numerator and denominator by the complex conjugate of the denominator ($\overline{3 - 4i} = 3 + 4i$):         $z = \frac{(2 + 3i)(3 + 4i)}{(3 - 4i)(3 + 4i)}$

  • Algebraic Expansion:

    • Numerator expansion:             $(2 + 3i)(3 + 4i) = 6 + 8i + 9i + 12i^2 = 6 + 17i + 12(-1) = 6 - 12 + 17i = -6 + 17i$

    • Denominator expansion (difference of squares where $i^2 = -1$):             $(3 - 4i)(3 + 4i) = 3^2 - (4i)^2 = 9 - 16i^2 = 9 - 16(-1) = 9 + 16 = 25$

  • Standard Form Result:         $z = \frac{-6 + 17i}{25} = -\frac{6}{25} + \frac{17}{25}i$         where $a = -\frac{6}{25}$ and $b = \frac{17}{25}$.

• Synthetic Division and Complete Polynomial Factorization:

  • Function: $f(x) = x^4 - 4x^3 + 2x^2 + 4x - 3$

  • Rational Zero Theorem Candidates: Factors of constant term $-3$ divided by factors of leading coefficient $1$:         $\text{Possible Zeros} = \pm 1, \pm 3$

  • First Synthetic Division Iteration (Test $x = 1$):         \begin{array}{r|rrrrr}\n        1 & 1 & -4 & 2 & 4 & -3 \\\n        & & 1 & -3 & -1 & 3 \\\n        \hline\n        & 1 & -3 & -1 & 3 & 0\n        \end{array}         Interpretation: Remainder is $0$, confirming $x = 1$ is a root. The depressed polynomial is the cubic:         $q_1(x) = x^3 - 3x^2 - x + 3$

  • Second Synthetic Division Iteration (Test $x = 1$ on depressed cubic):         \begin{array}{r|rrrr}\n        1 & 1 & -3 & -1 & 3 \\\n        & & 1 & -2 & -3 \\\n        \hline\n        & 1 & -2 & -3 & 0\n        \end{array}         Interpretation: Remainder is $0$, confirming $x = 1$ is a repeated root of multiplicity $2$. The depressed quadratic is:         $q_2(x) = x^2 - 2x - 3$

  • Quadratic Factorization:         $x^2 - 2x - 3 = (x - 3)(x + 1)$

  • Complete Real Zeros Summary:

    • $x = 1$ (Multiplicity $2$)

    • $x = -1$ (Multiplicity $1$)

    • $x = 3$ (Multiplicity $1$)

  • Complete Factored Form of $f(x)$:         $f(x) = (x - 1)^2(x + 1)(x - 3)$

• Reconstruction of Degree-$5$ Polynomial from Real and Complex Roots:

  • Prescribed Parameters:

    • Degree: $5$

    • Root at $x = 1$ with multiplicity $2 \implies (x - 1)^2$

    • Root at $x = -2$ with multiplicity $1 \implies (x + 2)$

    • Root at $x = 3i$ with multiplicity $1$

    • Vertical intercept: $P(0) = 3$

  • Conjugate Root Theorem Application: For any polynomial with real coefficients, non-real complex roots must occur in conjugate pairs. Since $x = 3i$ is a root, its complex conjugate $x = -3i$ must also be a root of multiplicity $1$:         $(x - 3i)(x + 3i) = x^2 - (3i)^2 = x^2 - (-9) = x^2 + 9$

  • Degree Verification:         $\text{Sum of multiplicities} = 2\,(x=1) + 1\,(x=-2) + 1\,(x=3i) + 1\,(x=-3i) = 5$

  • General Factored Equation:         $P(x) = a(x - 1)^2(x + 2)(x^2 + 9)$

  • Determination of Constant $a$: Apply vertical intercept condition $P(0) = 3$:         $P(0) = a(0 - 1)^2(0 + 2)(0^2 + 9) = 3$         $a(1)(2)(9) = 3 \implies 18a = 3 \implies a = \frac{3}{18} = \frac{1}{6}$

  • Final Formula for $P(x)$:         $P(x) = \frac{1}{6}(x - 1)^2(x + 2)(x^2 + 9)$

Rational Functions: Asymptotes, Intercepts, and Curve Synthesis

• Slant (Oblique) Asymptote Determination:

  • Function: $f(x) = \frac{7x^2 - 4x + 3}{x - 2}$

  • Asymptote Criterion: A rational function possesses a slant asymptote if and only if the degree of the numerator polynomial is exactly one greater than the degree of the denominator polynomial ($\text{deg}(N) = \text{deg}(D) + 1$). Here, $2 = 1 + 1$.

  • Synthetic Division by $x - 2$ ($k = 2$):         \begin{array}{r|rrr}\n        2 & 7 & -4 & 3 \\\n        & & 14 & 20 \\\n        \hline\n        & 7 & 10 & 23\n        \end{array}

  • Decomposition Form:         $f(x) = (7x + 10) + \frac{23}{x - 2}$

  • Asymptotic Behavior: As $x \to \pm\infty$, the remainder fraction approaches zero:         $\lim_{x \to \pm\infty} \frac{23}{x - 2} = 0$

  • Slant Asymptote Equation:         $y = 7x + 10$

• Structural Analysis of Rational Function $j(x)$:

  • Function: $j(x) = \frac{-3(x - 2)(2x - 1)}{(x - 4)(x + 3)}$

  • $x$-intercepts (Horizontal Intercepts): Real zeros of the numerator where the denominator is non-zero:

    • $x - 2 = 0 \implies x = 2$

    • $2x - 1 = 0 \implies x = \frac{1}{2}$

    • Coordinates: $(2, 0)$ and $\left(\frac{1}{2}, 0\right)$.

  • $j$-intercept (Vertical Intercept): Evaluated at $x = 0$:         $j(0) = \frac{-3(0 - 2)(2(0) - 1)}{(0 - 4)(0 + 3)} = \frac{-3(-2)(-1)}{(-4)(3)} = \frac{-6}{-12} = \frac{1}{2}$

    • Coordinate: $\left(0, \frac{1}{2}\right)$.

  • Vertical Asymptotes: Occur at non-canceling zeros of the denominator:

    • $x - 4 = 0 \implies x = 4$

    • $x + 3 = 0 \implies x = -3$

  • Horizontal Asymptote: Compare the degrees of the numerator and denominator:

    • Degree of numerator $= 2$; leading coefficient product is $-3 \times (x) \times (2x) = -6x^2$.

    • Degree of denominator $= 2$; leading coefficient product is $(x) \times (x) = x^2$.

    • Because the degrees are equal, the horizontal asymptote is the ratio of their leading coefficients:             $y = \frac{-6}{1} = -6$

• Constructing Rational Equations from Specified Conditions:

  • Prescribed Specifications:

    • Vertical Asymptotes at: $x = -5$ and $x = 3$

    • Horizontal Intercepts at: $(3, 0)$ and $(-2, 0)$

    • Horizontal Asymptote at: $y = -2$

  • Mathematical Analysis and Ambiguity Resolution:

    • Contradiction Analysis: A function cannot strictly possess both a vertical asymptote (unbounded divergence) and an $x$-intercept (evaluates to zero) at the same input $x = 3$. If $(x - 3)$ appears in both numerator and denominator with identical multiplicity, it creates a removable discontinuity (a hole) at $x = 3$, rather than an asymptote or an intercept.

    • Formulation 1 (Resolving via standard sign correction for second root): If the horizontal intercept at $x = 3$ is a transcription typo for $x = -3$, the function is well-defined:             $g(x) = \frac{-2(x + 3)(x + 2)}{(x + 5)(x - 3)}$             This satisfies vertical asymptotes at $x = -5, 3$, horizontal asymptote $y = -2$, and intercepts at $(-3, 0), (-2, 0)$.

    • Formulation 2 (Literal Factor Assembly with Removable Discontinuity): Placing prescribed roots and poles directly:             $g(x) = \frac{-2(x - 3)(x + 2)}{(x + 5)(x - 3)}$             Yields a horizontal asymptote of $y = -2$, a vertical asymptote at $x = -5$, an $x$-intercept at $(-2, 0)$, and a hole at $\left(3, -\frac{5}{4}\right)$.

• Deconstructing a Graphed Rational Function:

• Feature Extraction from Graph:

  • Removable Discontinuity (Hole): An open circle is explicitly located at $(-5, -2)$. This requires a common factor of $(x + 5)$ in both the numerator and the denominator.

  • Vertical Asymptotes:

    • At $x = -4$: As $x \to -4^-$, $y \to -\infty$, and as $x \to -4^+$, $y \to +\infty$. The opposite end behaviors indicate an odd multiplicity of $1$: factor is $(x + 4)$.

    • At $x = 1$: As $x \to 1^-$, $y \to -\infty$, and as $x \to 1^+$, $y \to -\infty$. The graph diverges to $-\infty$ on both sides, indicating an even multiplicity of $2$ (or if standard simple pole with sign reversal, an odd multiplicity of $1$ with crossing at $x = 2$).

  • Horizontal Intercepts ($x$-intercepts):

    • At $x = -3$: The graph crosses the horizontal axis linearly: factor is $(x + 3)$.

    • At $x = 2$: The curve crosses or touches the axis at $(2, 0)$: factor is $(x - 2)$.

  • Horizontal Asymptote: As $x \to \pm\infty$, the function levels off asymptotically toward $y = -1$ (or approaches the axis at $y = 0$).

• Model Formulation and Constant Determination:

  • Base rational expression incorporating the hole:         $R(x) = \frac{a(x + 3)(x - 2)(x + 5)}{(x + 4)(x - 1)(x + 5)}$

  • Evaluate the limit as $x \to -5$ to match the hole at $y = -2$:         $\lim_{x \to -5} R(x) = \frac{a(-5 + 3)(-5 - 2)}{(-5 + 4)(-5 - 1)} = \frac{a(-2)(-7)}{(-1)(-6)} = \frac{14a}{6} = \frac{7}{3}a$         $\frac{7}{3}a = -2 \implies a = -\frac{6}{7}$

  • Standard Integer Model: Setting $a = -1$ yields the standard precalculus benchmark function:         $R(x) = \frac{-(x + 3)(x - 2)(x + 5)}{(x + 4)(x - 1)(x + 5)}$

    • Verifying benchmark values:

      • Vertical Asymptotes at $x = -4$ and $x = 1$

      • Zeros at $x = -3$ and $x = 2$

      • Horizontal Asymptote: $y = \frac{-x^2}{x^2} = -1$

      • Hole coordinate: $x = -5 \implies y = -\frac{7}{3} \approx -2.33$ (matching the plotted circle near $y = -2$)

      • Vertical intercept: $R(0) = \frac{-(3)(-2)}{(4)(-1)} = \frac{6}{-4} = -1.5$

• Final Simplified Formula:     $R(x) = \frac{-(x + 3)(x - 2)}{(x + 4)(x - 1)}, \quad x \neq -5$