Comprehensive Test Review Notes: College Algebra

Evaluating Functions Algebraically

• Problem Statement: Given $f(x) = 2x^2 - 3x + 4$, find and simplify $f(x - 2)$.

• Substitution Step: Replace every instance of $x$ in the function definition with the expression $(x - 2)$.   $f(x - 2) = 2(x - 2)^2 - 3(x - 2) + 4$

• Algebraic Expansion:

  • Expand the binomial squared term $(x - 2)^2$:     $(x - 2)^2 = (x - 2)(x - 2) = x^2 - 2x - 2x + 4 = x^2 - 4x + 4$

  • Substitute the binomial expansion back into the function equation:     $f(x - 2) = 2(x^2 - 4x + 4) - 3(x - 2) + 4$

  • Distribute scalar factors across their respective terms:     $2(x^2 - 4x + 4) = 2x^2 - 8x + 8$     $-3(x - 2) = -3x + 6$

  • Recombine all terms:     $f(x - 2) = 2x^2 - 8x + 8 - 3x + 6 + 4$

• Combining Like Terms:

  • Group linear terms: $-8x - 3x = -11x$

  • Group constant terms: $8 + 6 + 4 = 18$

  • Final simplified expression:     $f(x - 2) = 2x^2 - 11x + 18$

Solving One-Variable Rational Equations

• Problem Statement: Solve the rational equation for $x$ and state any domain restrictions:   $\frac{4}{x + 1} - \frac{2}{3} = \frac{2}{3(x + 1)}$

• Domain Restrictions:

  • Set all denominator expressions containing variables equal to zero to find restricted values:     $x + 1 = 0 \Rightarrow x = -1$     $3(x + 1) = 0 \Rightarrow x = -1$

  • Restriction: $x \neq -1$

• Clearing Denominators:

  • Determine the Least Common Denominator (LCD) of the denominators $x + 1$, $3$, and $3(x + 1)$:     $\text{LCD} = 3(x + 1)$

  • Multiply every term on both sides of the equation by the LCD $3(x + 1)$:     $3(x + 1) \cdot \frac{4}{x + 1} - 3(x + 1) \cdot \frac{2}{3} = 3(x + 1) \cdot \frac{2}{3(x + 1)}$

  • Cancel common terms in each fraction:     $3 \cdot 4 - 2(x + 1) = 2$

• Solving the Linear Result:

  • Distribute $-2$ through $(x + 1)$:     $12 - 2x - 2 = 2$

  • Combine like constant terms on the left side:     $10 - 2x = 2$

  • Subtract $10$ from both sides:     $-2x = -8$

  • Divide both sides by $-2$:     $x = 4$

• Restriction Check: The calculated solution $x = 4$ does not match the restricted value $x = -1$, making it a valid solution.

Solving One-Variable Linear Equations

• Problem Statement: Solve for $x$ in the linear equation:   $4(x - 3) + 7 = 2x + 9$

• Step-by-Step Algebraic Procedure:

  • Distribute the scalar factor $4$ on the left side:     $4x - 12 + 7 = 2x + 9$

  • Combine constant terms $-12 + 7$ on the left side:     $4x - 5 = 2x + 9$

  • Subtract $2x$ from both sides:     $2x - 5 = 9$

  • Add $5$ to both sides:     $2x = 14$

  • Divide both sides by $2$:     $x = 7$

Solving Literal Equations

• Problem Statement: Solve the trapezoid area formula for the height variable $h$:   $A = \frac{1}{2} h (b_1 + b_2)$

• Step-by-Step Variable Isolation:

  • Multiply both sides of the literal equation by $2$ to eliminate the fraction:     $2A = h(b_1 + b_2)$

  • Divide both sides by the quantity $(b_1 + b_2)$ to isolate $h$:     $h = \frac{2A}{b_1 + b_2}$

Solving Absolute Value Equations

• Problem Statement: Solve the one-variable absolute value equation for $x$:   $|3x - 4| + 2 = 10$

• Isolating the Absolute Value Term:

  • Subtract $2$ from both sides of the equation:     $|3x - 4| = 8$

• Splitting into Two Linear Equations:

  • Case 1 ($3x - 4$ is positive):     $3x - 4 = 8$     $3x = 12$     $x = 4$

  • Case 2 ($3x - 4$ is negative):     $3x - 4 = -8$     $3x = -4$     $x = -\frac{4}{3}$

• Final Solution Set:   $x = 4 \quad \text{or} \quad x = -\frac{4}{3}$

Solving One-Variable Linear Inequalities

• Problem Statement: Solve the compound inequality and write the solution in interval notation:   $-8 \le -3x + 4 < 13$

• Solving the Compound Inequality:

  • Subtract $4$ from all three parts of the inequality:     $-8 - 4 \le -3x + 4 - 4 < 13 - 4$     $-12 \le -3x < 9$

  • Divide all three parts by $-3$. Reorient the inequality symbols because of division by a negative number:     $\frac{-12}{-3} \ge \frac{-3x}{-3} > \frac{9}{-3}$     $4 \ge x > -3$

  • Re-order the double inequality in standard left-to-right format:     $-3 < x \le 4$

• Interval Notation: The strict inequality at $-3$ requires an open parenthesis, and the inclusive inequality at $4$ requires a closed bracket:   $(-3, 4]$

Solving Absolute Value Inequalities

• Problem Statement: Solve the absolute value inequality and express the solution set in interval notation:   $|2x + 5| \ge 9$

• Splitting into Compound Inequalities:

  • An absolute value inequality of the form $|u| \ge c$ (where $c > 0$) translates to $u \ge c$ or $u \le -c$

  • Case 1:     $2x + 5 \ge 9$     $2x \ge 4$     $x \ge 2$

  • Case 2:     $2x + 5 \le -9$     $2x \le -14$     $x \le -7$

• Interval Notation: Combine the disjoint intervals using the union operator $\cup$:   $(-\infty, -7] \cup [2, \infty)$

One-Variable Word Problems

• Problem Statement: A rectangular garden has a length that is $5\,\text{ft}$ more than its width. The perimeter of the garden is $50\,\text{ft}$. Find the length and width of the garden.

• Variable Setup:

  • Let $x = \text{width of the garden (in feet)}$

  • Let $x + 5 = \text{length of the garden (in feet)}$

• Equation Construction:

  • The perimeter of a rectangle equals the sum of all four sides:     $P = x + (x + 5) + x + (x + 5) = 50$

  • Combine like terms on the left side:     $4x + 10 = 50$

• Algebraic Execution:

  • Subtract $10$ from both sides:     $4x = 40$

  • Divide both sides by $4$:     $x = 10$

• Final Solution:

  • $\text{Width} = x = 10\,\text{ft}$

  • $\text{Length} = x + 5 = 10 + 5 = 15\,\text{ft}$

Identifying Function Domains

• Problem Statement: Find the domain of the radical function and express the answer in interval notation:   $f(x) = \sqrt{x - 2}$

• Domain Requirement: The expression under an even root (the radicand) must be non-negative (greater than or equal to zero):   $x - 2 \ge 0$   $x \ge 2$

• Interval Notation:   $[2, \infty)$

Evaluating Functions Numerically

• Problem Statement: Given $g(x) = 2x^2 + 3x - 4$, evaluate $g(-3)$.

• Substitution and Order of Operations:

  • Substitute $-3$ for $x$ in the function:     $g(-3) = 2(-3)^2 + 3(-3) - 4$

  • Evaluate the exponent: $(-3)^2 = 9$     $g(-3) = 2(9) + 3(-3) - 4$

  • Perform multiplications:     $g(-3) = 18 - 9 - 4$

  • Compute final subtraction:     $g(-3) = 5$

Writing Linear Functions from Point and Slope

• Problem Statement: A linear function $f$ has slope $3$ and satisfies $f(-2) = 5$. Write the equation of $f$ in the slope-intercept form f(x) = mx + b$.\n\n* **Given Values**:\n * Slope m = 3\n * Point (x, y) = (-2, 5)\n\n* **Finding the y-intercept (b)**:\n * Substitute the point and slope into slope-intercept form y = mx + b:\n    5 = 3(-2) + b\n    5 = -6 + b\n * Add 6 to both sides:\n    b = 11\n\n* **Final Function Equation**:\n  f(x) = 3x + 11\n\n# Writing Equations of Parallel Lines\n\n* **Problem Statement**: Write an equation of the line that is parallel to y = -2x + 7$and passes through the point$(3, -1). Express the answer in slope-intercept form.\n\n* **Properties of Parallel Lines**:\n * Parallel lines have identical slopes.\n * Slope of given line: m = -2\n * Slope of parallel line: m = -2\n\n* **Finding the y-intercept (b)**:\n * Substitute m = -2$and$(x, y) = (3, -1)$into$y = mx + b:\n    -1 = -2(3) + b\n    -1 = -6 + b\n * Add 6 to both sides:\n    b = 5\n\n* **Final Line Equation**:\n  y = -2x + 5\n\n# Writing Equations of Perpendicular Lines\n\n* **Problem Statement**: Write an equation of the line that is perpendicular to a line with slope m_1 = \frac{1}{2}$(from$y = \frac{1}{2}x - 6$) and passes through the point$(-2, 5). Express the answer in slope-intercept form.\n\n* **Properties of Perpendicular Lines**:\n * Perpendicular lines have negative reciprocal slopes (m_1 \cdot m_2 = -1).\n * Perpendicular slope: m = -2\n\n* **Finding the y-intercept (b)**:\n * Substitute m = -2$and$(x, y) = (-2, 5)$into$y = mx + b:\n    5 = -2(-2) + b\n    5 = 4 + b\n * Subtract 4 from both sides:\n    b = 1\n\n* **Final Line Equation**:\n  y = -2x + 1\n\n# Solving Systems of Linear Equations\n\n* **Problem Statement**: Solve the system of equations and express the answer as an ordered pair:\n  3x + 2y = 12\n  x - y = -1\n\n* **Method of Elimination**:\n * Multiply the second equation by 2$to equalize coefficients of$y with opposite signs:\n    2(x - y) = 2(-1) \Rightarrow 2x - 2y = -2\n * Add the modified second equation to the first equation:\n    (3x + 2y) + (2x - 2y) = 12 + (-2)\n    5x = 10\n * Divide by 5:\n    x = 2\n\n* **Solving for y**:\n * Substitute x = 2$back into$x - y = -1:\n    2 - y = -1\n    -y = -3\n    y = 3\n\n* **Ordered Pair Solution**:\n  (2, 3)\n\n# Systems of Equations Word Problems\n\n* **Problem Statement**: A school sold 120\,\text{tickets}$to a play. Adult tickets cost$\$8$each and student tickets cost$\$5$each. The school collected a total of$\$780. How many adult tickets and how many student tickets were sold?\n\n* **Defining Variables**:\n * Let a = \text{number of adult tickets sold}\n * Let s = \text{number of student tickets sold}\n\n* **Formulating System of Equations**:\n * Equation 1 (Quantity of tickets):\n    a + s = 120\n * Equation 2 (Revenue total):\n    8a + 5s = 780\n\n* **Solving by Elimination**:\n * Multiply Equation 1 by -5:\n    -5(a + s) = -5(120) \Rightarrow -5a - 5s = -600\n * Add the modified equation to Equation 2:\n    (-5a - 5s) + (8a + 5s) = -600 + 780\n    3a = 180\n * Divide both sides by 3:\n    a = 60\n\n* **Solving for Remaining Variable**:\n * Substitute a = 60 back into Equation 1:\n    60 + s = 120\n    s = 60\n\n* **Final Answer with Units**:\n * \text{Adult tickets} = 60\,\text{tickets}\n * \text{Student tickets} = 60\,\text{tickets}$$